algebra
function
domain
Stanford
college
geometry
geometric transformation

Science of Morality, Anyone?

Where, in American culture, is the domain of knowledge that we would identify as morality studied and taught?

I suspect that if we do not quickly develop a science of morality that will make it possible for us to live together on this planet in a more harmonious manner our technology will help us to destroy the species and perhaps the planet soon.

It seems to me that we have given the subject matter of morality primarily over to religion.  It also seems to me that if we ask the question ‘why do humans treat one another so terribly?’ we will find the answer in this moral aspect of human culture.

The ‘man of maxims’ “is the popular representative of the minds that are guided in their moral judgment solely by general rules, thinking that these will lead them to justice by a ready-made patent method, without the trouble of exerting patience, discrimination, impartiality—without any care to assure themselves whether they have the insight that comes from a hardly-earned estimate of temptation, or from a life vivid and intense enough to have created a wide fellow-feeling with all that is human.” George Eliot [i]The Mill on the Floss[/i]

I agree to the point of saying that we have moral instincts, i.e. we have moral emotions.  Without these moral emotions we could not function as social creatures.  These moral emotions are an act of evolution.  I would ague that the instinct for grooming that we see in monkeys is one example of this moral emotion.

We can no longer leave this important matter in the hands of the Sunday-school. Morality must become a top priority for scientific study.

I wouldn't think it isn't a scientific study. Particularly in the field of psychology, many have studied why people are kind and why people do bad things. Morality is a social norm. It isn't dictated by religion, although many like to think it is. Humans have morals because we see other people doing certain things. We form relationships between whether those things are good or bad and we decide whether or not to do those things in the future. This process started long before religion, long before humans even existed! Religion does not dictate morality; I know plenty of non-religious people (including myself) who are completely moral. For some good discussion, read up on the [url=http://en.wikipedia.org/wiki/Stanford_prison_experiment]Stanford prison experiment[/url] and the [url=http://en.wikipedia.org/wiki/Milgram_experiment]Milgram shock experiment[/url]. They show how role-playing, and authority figures can completely change one's morals and behavior. Additionally, they provide insight into how a disaster such as World War II happened.

There is a "science" of morality, and it's studied in certain areas of philosophy. Likewise, the concept of building a secular, universal set of morality has been done before.

http://en.wikipedia.org/wiki/Existentialism

For example, in most (if not all) existential philosophies, the concept of granting individual freedom superseeds all attempts of purely sociaetal benefits. This likewise, reflects most contemporary attitudes regarding morality.

[quote="coberst"] Morality must become a top priority for scientific study.[/quote]

On the contrary... science aims to discover natural truths in an empirical manner. Most aspects of morality cannot, by modern secular standards, be described as natural truths (or otherwise no progress in determining the natural truths of morality whatsoever can be made, therefore no one studies it):

-How should people with good morals behave? Why and for what should people have good morals?
That's religion and philosophy. And politics.

-Why do people behave the way they do, with respect to societal moral standards?
Psychology. Social sciences. Stuff that weren't considered strictly sciences a century ago. Maybe not even today.

-How have morals in society developed and changed? (This questions assumes the secular view of morality, that is, that morality is a cultural construct.)
Anthropology? More topics that weren't considered strictly sciences a century ago.

[quote="coberst"]Morality must become a top priority for scientific study.[/quote]I'm all for scientific improvement, but my question is why? Why should morality go above
cures for AIDS, or cancer, or any other disease on the priority scale? As far as I can
see, there is no helpful benefit to learning about morality and the only thing I can
think of is using this to tweak people's minds; something I am not for. On the other
hand, if we put money into [i]other[/i] fields of study, we can see [b]real[/b] results.

What do you [u]do[/u] with a science of morality?

Who would develop it; why should we trust these people over others?

Who would enforce it; would it be unduly stringent; is it ethical to impose such a high degree of morality on people?

Why is grooming, as displayed by monkeys, an indication of moral emotions?

Emotions are instincts; they are something that is part of our genes.  They are part of our genetic makeup because they were necessary for the survival of the social species.  Some species are loners but some are naturally social.  The social species needed emotions that facilitated social unity.  Mutual grooming is one means for bonding between individuals and the group.

Would morals count as knowledge?  Do emotions count as knowledge?  Directly I must say that the emotion of fear is not knowledge.  The emotion leads to a feeling and the consciousness of the feeling becomes knowledge.  Morality is about relationships, i.e. certain instincts make a social group possible.

Without social cohesion social groups cannot survive.  Reasoning about facts is a human means for survival and thriving.  The more we know and understand about relationships the better will be our lives.  In fact, because we have developed such powerful technology and thus have placed in the hands of people such power that if we do not do a better job about relationships our species cannot long survive.


Find all ordered pairs, that are non-negative integers, (x,y,z) such that 2x+3y+6z=30.

Can someone explain how to do these?

Can you post a solution please, using the method you just stated?  :)

[hide]I'm assuming he means by letting $ z\equal{}0$, which gives $ 2x\plus{}3y\equal{}30$. Then taking this equation in mod 2, we get that $ y\equiv 0\bmod{2}$. This implies that $ y\equal{}2k$ for an integer $ k$. Plugging this into our equation we have that $ 2x\plus{}6k\equal{}30$, so $ x\equal{}15\minus{}3k$. Since x has to be non-negative, the maximum value for k in this case is 5. So from letting z=0 we get the following ordered pairs: $ (15,0,0), (12,2,0), (9,4,0), (6,6,0), (3,8,0), (0,10,0)$

Letting $ y\equal{}0$ gives us $ 2x\plus{}6z\equal{}30$, or $ x\plus{}3z\equal{}15$. Taking this equation in mod 3 we have that $ x\equiv 0\bmod{3}$, therefore this implies that $ x\equal{}3l$ for some integer $ l$. Plugging this back we get that $ 3l\plus{}3z\equal{}15$, so $ z\equal{}5\minus{}l$, and the maximum value $ l$ can take is 5, but since this gives us the same ordered pair from the beginning we can take the max of $ l$ to be 4 to get the following ordered pairs: $ (12,0,1), (9,0,2), (6,0,3), (3,0,4), (0,0,5)$

The final case is by letting $ x\equal{}0$ to get $ 3y\plus{}6z\equal{}30\equal{}\equal{}>y\plus{}2z\equal{}10$ Taking this in mod 2 we have that $ y\equiv 0\bmod{2}$, so $ y\equal{}2m$ for some integer $ m$. This means that $ z\equal{}5\minus{}m$, giving us the following solutions: $ (0,2,4), (0,4,3), (0,6,2), (0,8,1)$[/hide]
I hope I didn't miss any :o

Hi. I get the $ 16$ solutions above and a further $ 5$:

$ 3$ divides $ x$, so $ x\equal{}3a$

$ 2$ divides $ y$, so $ y\equal{}2b$

and then $ z\equal{}5\minus{}a\minus{}b$.

Ckecking the possible cases the other five solutions are :

$ (x, y, z) \equal{} (3, 4, 2), (3, 6, 1), (6, 4, 1), (9, 2, 1)$ and $ (6, 2, 2)$

Changing your solution a tiny bit can make it very fast. We know $ a \plus{} b \plus{} z \equal{} 5$, where $ 3a \equal{} x$ and $ 2b \equal{} y$, so using Balls and Urns, we get the answer as $ \binom{7}{2} \equal{} 21$.


algorithm
number theory
Euclidean algorithm

Find all positive integers $n$ such that $\frac{n^2+2}{2n+1}$ is an integer.

[hide="Hint for the first method"]Euclidean Algorithm ;) [/hide]
[hide="Hint for the second method"]Synthetic division ;) [/hide]

$(n^2+2,2n+1)=(4n^2+8,2n+1)=(9,2n+1)$

Hence we only need to check whether $2n+1=1,3,5,7,9$  :)

and only $n=1,4$ works  :D

*edit mistake
*edit again  for $(9,2n+1)$ :P

[quote="shyong"][quote="Adunakhor"][hide="division"]
$n^2+2 = (n+1)\displaystyle \left(n - 1 + \frac{3}{n+1}\right)$

$\frac{3}{n+1}$ must be an integer, $n=2$.

[/hide][/quote]

but the denominator is $2n+1$ not $n+1$ ;)[/quote]

Oh, misread the question  :blush: Though $2$ doesn't seem to work either, $\frac{2^2+2}{2(2)+1}=\frac{6}{5}$

[hide]$\frac{n^2+2}{2n+1}=\frac{1}{2}n+\frac{9}{4(2n+1)}-\frac{1}{4}$

But I don't see how this helps.
[/hide]

For the fraction to be an integer, the denominator must evenly divide the numerator.  chess64, since you already divided the expression out, all you have to do is find what values of $n$ there are such that $4(2n+1)|9$.  $n\geq0$ and the only postive divisors of $9$ are $1$, $3$, and $9$, and $n>0$ only for $4(2n+1)=9$.  But wait, then $n\notin \mathbb{N}$...

Well I wouldn't really consider $\frac{5}{8}$ to be an integer, and also notice the $\frac{1}{2}n$ and $-\frac{1}{4}$ terms. . .

Using Euclidean Algorithm, you get that the only solution is $n=1$.

Masoud Zargar

EDIT: Damn it! I just saw Shyong's edit.  :mad:

[quote="chess64"][quote="shyong"]$(4n^2+8,2n+1)=(7,2n+1)$[/quote]

How?[/quote]

Ah, Shyong was incorrect, it should be

$4n^2 + 8 = (2n+1)(2n-1) + 9$

And hence

$(9, 2n+1)$

Instead, agreeing with the synthetic division.  :)

Opps , edit again .... the answer should be $n=1$ or $4$ ... :blush:

Correct answers are when $n=1$ and $n=4$


Find all value of m such that :
$ 9x^2 \plus{} 20y^2 \plus{} 4z^2 \minus{} 12xy \plus{} 6xz \plus{} myz > 0$
with x,y,z not simultaneous equal 0.

This was originally posted at http://www.mathlinks.ro/viewtopic.php?t=252775, but wasn't solved. I have a strong hunch that m is at least close to 0.

Should be locked.

Japanese Communities Moderator

kunny


geometry
trapezoid
angle bisector
geometry proposed

[color=darkred]Let $ ABC$ be a triangle. The its $ B$- exincircle touches it at the points $ E\in AC$ and $ F'\in AB$. The its $ C$- exincircle touches it at the points $ F\in AB$ and $ E'\in AC$. Prove that the midpoints of the segments $ [BC]$, $ [EF]$ and $ [E'F']$ belong to a line which is parallelly with the bisector of the angle $ \widehat{BAC}$.
 [/color]

Not hard problem.But it can be very useful. :lol:
Solution.
Denote $ M,N,L$ be the midpoints of $ BC,FE,E'F'$ respectively. Let $ l_{a}$ be the angle bisector of $ \angle BAC$.So it is obviously that $ l_{a}\|F'E\|E'F$ and $ E'A\equal{}AF$,$ AF'\equal{}AE\Rightarrow E'F'EF$ is trapezoid with $ E'F'\equal{}FE\Rightarrow NL\| l_{a}$.
Now i'll prove that $ MN\|l_{a}$.Denote $ MN\cap AC\equal{}P,MN\cap AB\equal{}Q$.
Using Menelaus theorem to $ \triangle ABC$ and points $ M,P,Q$.:
$ \frac{BQ}{QA}\cdot\frac{AP}{PC}\equal{}1$
Using Menelaus theorem to $ \triangle AFE$ and points $ N,P,Q$.:
$ \frac{FQ}{QA}\cdot\frac{AP}{PE}\equal{}1\Rightarrow$
$ \frac{PC}{BQ}\equal{}\frac{PE}{QF}\Longleftrightarrow\frac{PC}{PE}\equal{}\frac{BQ}{QF}\Longleftrightarrow\frac{QF}{BF}\equal{}\frac{PE}{EC}$.It is obviously that $ EC\equal{}FB\Rightarrow QF\equal{}PE\Rightarrow QA\equal{}AP\Rightarrow PQ\|l_{a}\Rightarrow L,N,M$ are collinear.

[size=117][color=darkred][b]Nice Erken's proof ![/b] [/color][/size]

[color=darkblue][b]Remark.[/b] Since $ \{\begin{array}{c}F'\in (BA\ ,\ E'\in (CA\ :\ BF'=CE'=p\\\ F\in (BA\ ,\ E\in (CA\ :\ BF=CE=p-a\end{array}$, where $ 2p=a+b+c$
can apply the very usual lemma from my reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=161883 .[/color]


i need help with this problem:

when 4 standard six-sided dice are rolled, what is the most likely sum of the four numbers?

 :starwars:

[hide]Well, it's easy to see that the most common sum of 2 dice rolled is 7, because there are 6 ways to make 7 rolling 2 dice (1/6, 2/5, 3/4, 4/3, 5/2, 6/1), and there are less than 6 ways to make any other number.
So since it's 4 dice, we add 7 + 7 to get 14.a[/hide]

Harder question:

What's the probability that you roll a 14 with 4 dice?

[hide]There are $1296$ outcomes and $152$ of them make $14$ so $\frac{152}{1296}=\frac{1}{9}$[/hide]


topology
function
analytic geometry
calculus
derivative
geometry
ratio

This is from a "solved problems" physics book (for more problems, be sure to check out the suggestively named "Physics" sub-forum), but I'll put it here, since it hardly has anything to do with physics:

Prove that $ \left( \frac{\partial p}{\partial V} \right)_T \left( \frac{\partial V}{\partial T} \right)_p \left( \frac{\partial T}{\partial p} \right)_V \equal{} \minus{}1$, using properties of the Jacobians.

[hide="The solution starts like this..."]We have that $ \left( \frac{\partial p}{\partial V} \right)_T \equal{} \frac{[P,T]}{[V,T]}$ etc.
Now, I do know what a Jacobian is, but I can't understand what is being done...
The notation would the first of the problems on my list  :blush: (I mean those square brackets)
Can anybody clear it up?[/hide]

OK, it seems that we are talking about the following: let $ M$ be a 2-dimensional manifold and let $ f,g,h$ (or, if you prefer, $ P,V,T$) be any 3 functions such that each pair of them gives a coordinate system near a point. Let $ x,y$ be any coordinate system. Let $ [f,g]$ be the Jacobian of the mapping $ (x,y)\mapsto(f,g)$ at the point. Then the formulae you mentioned immediately follow from the chain rule and the multiplicativity of determinant. Perhaps, the intended interpretation was different, but this one works perfectly well. On the other hand, I do not see why we don't just consider the chain of mappings $ (f,g)\mapsto(g,h)\mapsto(h,f)\mapsto(f,g)$ and get the result immediately without introducing any $ [,]$. :?

Thanks! All of this seems pretty counter-intuitive, it took me about an hour to realize why the Jacobian of $ [g,h] \to [f,h]$ is that partial derivative, but I finally got it... Thanks once again and keep up the good work!

if I understand right we have :
$ x \equal{} x(y,z)$
$ y \equal{} y(x,z)$
$ z \equal{} z(x,y)$ with condition :
$ F(x,y,z) \equal{} 0$
then prove :
$ x'_{y}y'_{z}z'_{x} \equal{} \minus{} 1$
1. we can say that it's in exat  Menelia theorem .(from geometry)
2.we can express $ x'_{y},$ and $ y'_{z}$ from $ F(x,y,z) \equal{} 0$ and substitute . we are done .

[quote="Extremal"]
1. we can say that it's in exat  Menelia theorem .(from geometry)
[/quote]
Poor Menelaus of Alexandria, in addition to the Russian and Latin versions of his name, he now got a Georgian one :P. Anyway, that's not the point. What I was really going to ask is how the Menelaus theorem is related to the question under consideration :?. I tried to draw a few pictures interpreting the partial  derivatives but wasn't able to get one where they would be oriented ratios of some points on 3 extended sides of a triangle that lie on one line :(

[quote="fedja"][quote="Extremal"]
1. we can say that it's in exat  Menelia theorem .(from geometry)
[/quote]
Poor Menelaus of Alexandria, in addition to the Russian and Latin versions of his name, he now got a Georgian one :P. [/quote]
 :lol:
[quote="fedja"]
Anyway, that's not the point. What I was really going to ask is how the Menelaus theorem is related to the question under consideration :?. I tried to draw a few pictures interpreting the partial  derivatives but wasn't able to get one where they would be oriented ratios of some points on 3 extended sides of a triangle that lie on one line :([/quote]
I can't explain without image, but I can't draw image for this in $ R^{3}$ :huh: [b]I'm not sure[/b] but I think it's known about generalization of  this theorem  try to find here http://mathworld.wolfram.com/MenelausTheorem.html .
later I'll try to draw something  :) (esli poluchitsya)

This is called the Reciprocity Relation. Let $ z \equal{} z(x,y)$. The total differential of z is

$ dz \equal{} \left(\frac{\partial z}{\partial x}\right)_y dx \plus{} \left(\frac{\partial z}{\partial y}\right)_x dy$.

Now consider a process in which z stays constant and divide by $ dy_z$,

$ 0 \equal{} \left(\frac{\partial z}{\partial x}\right)_y \frac{dx_z}{dy_z} \plus{} \left(\frac{\partial z}{\partial y}\right)_x  \equal{} \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z \plus{} \left(\frac{\partial z}{\partial y}\right)_x$

$ \equal{}> \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z \equal{} \minus{}\left(\frac{\partial z}{\partial y}\right)_x \equal{} \minus{} \frac{1}{\left(\frac{\partial y}{\partial z}\right)_x}$

$ \equal{}> \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \equal{} \minus{}1$.


[hide="Hint"]Take the modulus of each side. What does this limit the solution set to? How can we parametrize our newly limited set in a way that will make it effective to find the solution set?[/hide]

[hide="solution"]$ a\minus{}bi\equal{}(a\plus{}bi)^3\equal{}a^3\plus{}3a^2bi\minus{}3ab^2\minus{}b^3i$

$ b^3\minus{}3a^2b\equal{}b$

If $ b\equal{}0$ then $ a\equal{}a^3$, or $ a\equal{}\minus{}1,0,1$. If $ b\neq 0$, then $ b^2\minus{}3a^2\equal{}1$, which is a Pell equation.

$ a^3\minus{}3ab^2\equal{}a$.

If $ a\equal{}0$, then $ \minus{}b\equal{}\minus{}b^3$, or $ b\equal{}\minus{}1,0,1$. If $ a\neq 0$, then $ a^2\minus{}3b^2\equal{}1$, which is another Pell equation.

We now have $ a^2\minus{}3b^2\equal{}b^2\minus{}3a^2$, or $ 4a^2\equal{}4b^2$, or $ a^2\equal{}b^2$. Thus $ a\equal{}b$ or $ a\equal{}\minus{}b$. We thus have the solutions $ (a,a)$, $ (a,\minus{}a)$, and $ (0,\pm 1)$ and $ (\pm 1,0)$.[/hide]

posons    z=[r;a] danc (conjugué de z)=[r;-a] et     z*z*z =[r*r*r, 3a]
danc si (conjugué de z)= z*z*z   alors ( so)  [r.-a]=[r*r*r;3a] d'où r=r*r*r  et
-a = 3a(2pi)  d'où r=1 ou r=-1 ou r=0
et a=0(pi/2)

Le langage primaire sur ce forum est l'anglais, et certains ne comprennent pas le français.

Translation: The primary language on this forum is English, and some people do not understand French.

[hide="Solution Using My Hint"]Taking the modulus gives $ |\overline{z}| \equal{} |z^3| \implies |z| \equal{} |z|^3 \implies |z| \in \{ 0,1 \}$.

[b]Case 1:[/b] $ |z| \equal{} 0 \implies z \equal{} 0$

[b]Case 2:[/b] $ |z| \equal{} 1 \implies \overline{z} \equal{} z^{ \minus{} 1} \implies z^3 \equal{} z^ {\minus{} 1} \implies z^4 \equal{} 1$[/hide]


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Prove that Z[i*sqrt(7)] is a noetherian ring and is not artenian[/code]

the ring is isomorphic to $\mathbb Z[X]/(X^2+7)\mathbb Z[X]$ therefore it is noetherian
the ring is an integral domain and not a field, hence it cannot be artinian


hello,i have found
$ x^2\plus{}y^2\plus{}z^2\plus{}w^2\equal{}\frac{3310893}{105733}$.
Is this right?
Sonnhard.

We may rewrite the system as follows:
$ A[X,Y,Z,W]^T\equal{}u$ where $ X\equal{}x^2,\cdots$, $ a_{i,j}\equal{}1/((2i)^2\minus{}(2j\minus{}1)^2)$ and $ u\equal{}[1,1,1,1]^T$.
$ X\plus{}Y\plus{}Z\plus{}W\equal{}u^T[X,Y,Z,W]^T\equal{}u^TA^{\minus{}1}u\equal{}S(A^{\minus{}1})$ where $ S(V)$ is the sum of the elements of $ V$.
Maple gives $ S(A^{\minus{}1})\equal{}36$ but there exists surely a prettier solution.

This exercise can be generalized; tobeno_1 has asked for the case $ n\equal{}4$ and the result is $ 36$; if we have $ n$ unknowns $ X_1,\cdots,X_n$ then, using Maple, I conjecture that $ X_1\plus{}\cdots\plus{}X_n\equal{}2n^2\plus{}n$. Unfortunately I have no proof.


Write $ \sin^{4}x$ in terms of $ \cos 2x$ and $ \cos 4x$

$ \sin ^ 4 x\equal{}(\sin ^ 2 x)^2\equal{}\left(\frac{1\minus{}\cos 2x}{2}\right)^2$.

I understood until that part, but the book went further. First it squared out the expression so:

$ \frac{1\minus{}2\cos2x \plus{} \cos^{2}2x}{4}$. I don't understand how they immediately jumped to the next step:

$ \equal{} \frac{1\minus{}2\cos2x}{4} \plus{} \frac{1\plus{}\cos 4x}{8}$

hello, i have $ \sin^4(x)\equal{}\frac{1}{8}\left(3\minus{}4\cos(2x)\plus{}\cos(4x)\right)$.
Sonnhard.

[quote="mathemagician1729"]
I understood until that part, but the book went further. First it squared out the expression so:

$ \frac {1 \minus{} 2\cos2x \plus{} \cos^{2}2x}{4}$. I don't understand how they immediately jumped to the next step:

$ \equal{} \frac {1 \minus{} 2\cos2x}{4} \plus{} \frac {1 \plus{} \cos 4x}{8}$[/quote]

$ \frac {1 \minus{} 2\cos2x \plus{} \cos^{2}2x}{4} \equal{} \frac{1\minus{}2\cos2x}{4} \plus{} \frac{\cos^2 2x}{4}$

$ \frac{\cos^2 2x}{4} \equal{} \frac{1}{4}\left(\frac{1\plus{}\cos4x}{2}\right) \equal{} \frac{1\plus{}\cos4x}{8}$

$ \Rightarrow \frac {1 \minus{} 2\cos2x \plus{} \cos^{2}2x}{4} \equal{} \frac {1 \minus{} 2\cos2x}{4} \plus{} \frac {1 \plus{} \cos 4x}{8}$

Sorry but how did you get $ \cos^{2}2x \equal{} \frac{1\plus{}\cos 4x}{2}$
Is there a way to derive that?

It's an identity:
$ \cos^2 x \equal{} \cos^2 x \minus{} \sin^2 x \plus{} \sin^2 x \equal{} \cos 2x \plus{} \sin^2x \equal{} \cos 2x \plus{} 1 \minus{} \cos^2x$

$ 2\cos^2x \equal{} 1 \plus{} \cos2x$

$ \cos^2x \equal{} \frac {1 \plus{} \cos2x}{2}$

And then put in $ 2x$ for $ x$.


function
calculus
derivative
search
calculus computations

TRUE or False, all continous functions are diffrentiable?

please help me with an example

One (and only one) of the following functions works as a counterexample:
1) $ f(x) \equal{} \frac {\cos(x^3\sin|x|)}{x^2 \plus{} 1}$
2) $ f(x) \equal{} \begin{cases}x^3 & x\ge 0 \\
x^2 & x < 0\end{cases}$
3) $ f(x) \equal{} \begin{cases}x^3 & x\ge 1 \\
x^2 & x < 1\end{cases}$
Can you figure out which one and why?

By the way, the correct spellings are D-I-F-F-[color=red]E[/color]-R-E-N-T-I-A-B-L-E and C-O-N-T-I-N-[color=red]U[/color]-O-U-S ;)

$ f(x) \equal{} |x|$ is an continuous function, but isn't differentiable at $ x \equal{} 0$.

$ f$ differentiable $ \Rightarrow$ $ f$ continuous $ \Rightarrow$ $ f$ antidifferentiable (indefinite integrable) $ \Rightarrow$ $ f$ has the Darboux property.
The converse does not hold in any of these cases.

When you say "integrable" which do you mean? That last implication not reversing is interesting. What is a counterexample to it?

Given a function $ f: I\rightarrow \mathbb{R}$, there exists $ F: I\rightarrow \mathbb{R}$ so that $ F'(x)\equal{}f(x)\forall x\in I$- this is what I mean.
No counterexample yet, but I think it is correct. :maybe:

You just need to find a function with the darboux property that isn't continuous in any point (a derivative is the pointwise limit of continuous functions and hence has a bunch of continuity points).

If you search for it, you will find examples posted on the site. I like the version with the image of every nontrivial interval equal to the whole real line.

A more pathological example:  the [url=http://en.wikipedia.org/wiki/Weierstrass_function]Weierstrass function[/url] is everywhere continuous but nowhere differentiable.

$ f(x)\equal{} \begin{cases}sin\frac{1}{x} & x\neq  0 \\  1/2 & x\equal{}0\end{cases}$ is a Darboux function but I do not think it is indefinite integrable.
Could anyone prove (or outline a proof) it is not indefinite integrable?

Wikipedia gives the [url=http://en.wikipedia.org/wiki/Conway_Base_13_function]Conway base-13 function[/url] as a Darboux function that is nowhere continuous.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31493]My version[/url]. Less explicit, but easier to understand.


geometry inequalities trigonometry LaTeX inradius circumcircle incenter

[quote="Sathej"]If $ x,\,y,\,z$ are positive real numbers, prove that:
\[ \left(x \plus{} y \plus{} z\right)^{2}\left(yz \plus{} zx \plus{} xy\right)^{2}\le 3\left(y^{2} \plus{} yz \plus{} z^{2}\right)\left(z^{2} \plus{} zx \plus{} x^{2}\right)\left(x^{2} \plus{} xy \plus{} y^{2}\right)
\]
[/quote]
[b] Solution.[/b] We have:
\[ 4\left(x^{2} \plus{} xy \plus{} y^{2}\right) \minus{} 3\left(x \plus{} y\right)^{2} \equal{} \left(x \minus{} y\right)^{2}\ge 0
\]
Multiplying the inequalities:
\[ 4\left(x^{2} \plus{} xy \plus{} y^{2}\right)\ge 3\left(x \plus{} y\right)^2
\]

\[ 4\left(y^{2} \plus{} yz \plus{} z^{2}\right)\ge 3\left(y \plus{} z\right)^2
\]

\[ 4\left(z^{2} \plus{} zx \plus{} x^{2}\right)\ge 3\left(z \plus{} x\right)^2
\]
we get:
\[ 64\left(y^{2} \plus{} yz \plus{} z^{2}\right)\left(z^{2} \plus{} zx \plus{} x^{2}\right)\left(x^{2} \plus{} xy \plus{} y^{2}\right)\ge 27\left(x \plus{} y\right)^{2}\left(y \plus{} z\right)^{2}\left(z \plus{} x\right)^{2}
\]
Thus, it suffices to show that:
\[ 81\left(x \plus{} y\right)^{2}\left(y \plus{} z\right)^{2}\left(z \plus{} x\right)^{2}\ge 64\left(x \plus{} y \plus{} z\right)^{2}\left(yz \plus{} zx \plus{} xy\right)^{2}
\]

\[ \Longleftrightarrow 9\left(x \plus{} y\right)\left(y \plus{} z\right)\left(z \plus{} x\right)\ge 8\left(x \plus{} y \plus{} z\right)\left(ab \plus{} bc \plus{} ca\right)
\]

\[ \Longleftrightarrow x\left(y \minus{} z\right)^{2} \plus{} y\left(z \minus{} x\right)^{2} \plus{} z\left(x \minus{} y\right)^{2}\ge 0\quad\mathbb{QED}
\]

[Moderator edit: Minor typo fixed, thanks to Conceive for the notice. - Darij]

Nice  :wink:
[quote="Adalbert"]where can we find all the problems from INMO 2007?[/quote]

PM to Sathej, please!

[quote="N.T.TUAN"][b]Problem.[/b]If $x, y,z$ are positive real numbers, prove that
 $(x+y+z)^{2}(yz+zx+xy)^{2}\leq 3(y^{2}+yz+z^{2})(z^{2}+zx+x^{2})(x^{2}+xy+y^{2})(*)$.

[b]Solution.[/b] Exist a triangle $ABC$ for $BC=a,CA=b,AB=c$ and point $M$ in it such that $MA=x,MB=y,MC=z$ and $\widehat{AMB}=\widehat{BMC}=\widehat{CMA}=120^{0}$.

We have $c^{2}=x^{2}+y^{2}+xy,b^{2}=z^{2}+x^{2}+zx,a^{2}=y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4}{\sqrt{3}}.S$ and $(x+y+z)^{2}=\frac{1}{2}.(a^{2}+b^{2}+c^{2}+4\sqrt{3}.S)$. Where $S$ is area of triangle $ABC$.

By $S=\frac{abc}{4R}$ we have
$(*)\Leftrightarrow a^{2}+b^{2}+c^{2}+4\sqrt{3}.S\leq 18R^{2}$. But this is TRUE because $a^{2}+b^{2}+c^{2}\leq 9R^{2}$ and $4\sqrt{3}.S\leq a^{2}+b^{2}+c^{2}$.[/quote]

problem is trivial using inherent properties of the point $M$ (Fermat's point)
using $c^{2}=x^{2}+y^{2}+xy,b^{2}=z^{2}+x^{2}+zx,a^{2}=y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4}{\sqrt{3}}.S$and $S=\frac{abc}{4R}$
the inequality gets reduced to $x+y+z \leq 3R$
i.e if O is the circumcentre then we have to prove that

$MA+MB+MC \leq OA+OB+OC$

which is true by original definition of the fermat's point(a point P inside triangle $ABC$ (with each angle less than $120^{0}$) such that $PA+PB+PC$ is minimum)

All the problems from INMO 2007 is here:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=2007+inmo&t=133114

[quote="rohitsingh0812"]

problem is trivial using inherent properties of the point $M$ (Fermat's point)
using $c^{2}=x^{2}+y^{2}+xy,b^{2}=z^{2}+x^{2}+zx,a^{2}=y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4}{\sqrt{3}}.S$and $S=\frac{abc}{4R}$
the inequality gets reduced to $x+y+z \leq 3R$
i.e if O is the circumcentre then we have to prove that

$MA+MB+MC \leq OA+OB+OC$

which is true by original definition of the fermat's point(a point P inside triangle $ABC$ (with each angle less than $120^{0}$) such that $PA+PB+PC$ is minimum)[/quote]
Nice,You are right  :)

Inmo 2007 /7
Prove for all positive reals x,y,z that
                                3(x2+y2+xy)(x2+z2+xz)(y2+z2+yz)>(x+y+z)2(xz+xy+yz)2
  solution: be  cause of homogeneity we normalize it as z=1:
                  thus we need to prove that
                                3(x2+y2+xz)(1+x2+x)(1+y2+y)>(1+x+y)2(x+y+xy)2
                                  put x+y=u and xy=v;
                                  on expanding it becomes
                                  3(u2+u4-2u2v+u3+u2v2+u3v-v+v2-uv-v3-
                                  uv2)>(u2+u4+2u3+v2+u2v2+2uv2+2uv+2u3v+4u2v)
                                  ó(2u4-8u2v)+(u3-4uv)+(2u2-8v)+(u3v-4uv2 ) +
                                       (2u2v2-8v3)-(2u2v-8v2)+(uv2+uv+5v+5v3-6v2)>0
                                   ó(u2-4v)(2(u2-4v)+6v+u+2+uv+2v2)+(uv2+uv)+
                                        5( v3/2-v1/2)2+4v2>0
                                       which is of course true.
          Anuj .
v6 means vto powr 6

[quote="hien"][quote="anuj kumar"]  be  cause of homogeneity we normalize it as z=1:
 [/quote]

I think it's wrong! :!:[/quote]
No, anuj kumar are true! By we can denote $a=\frac{x}{z},b=\frac{y}{z}$.

I'm sorry about my carfuless thinking that event

[quote="April"]
\[4\left(x^{2}+xy+y^{2}\right)-3\left(x+y\right)^{2}=\left(x-y\right)^{2}\ge 0 \]
[/quote]
 well anyways

since $x$ $y$ $z$ are real no.s so is there algebraic combination but we know $p^{2}$ $\geq0$ for all $p$ belonging to real number.that can be proved using some logic.....

[i]beautiful proofs [/i]surely can have some [i]managable typographic erro[/i]rs

[quote="N.T.TUAN"]$c^{2}=x^{2}+y^{2}+xy,b^{2}=z^{2}+x^{2}+zx,a^{2}=y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4}{\sqrt{3}}.S$[/quote]

This is very NICE! But i like to knows proof of this points of Ferma. Can you help me?

Well, this is my solution!  :D

[b]Problem[/b]

$3(x^{2}+xy+y^{2})(y^{2}+yz+z^{2})(z^{2}+zx+x^{2}) \geq (x+y+z)^{2}(xy+yz+zx)^{2}$

[u]Solution[/u]

[b]Lemma[/b] 1,

  $9(x+y)(y+z)(x+z) \geq 8(x+y+z)(xy+yz+zx)$ For all x, y, z are positive.

  [hide="Prove"]
  Identity;$(x+y)(y+z)(x+z)=(x+y+z)(xy+yz+zx)-xyz$

  So it's equivalent to;

            $(x+y+z)(xy+yz+zx) \geq 9xyz$

  Which trivial by AM - GM
  [/hide]

 From Lemma 1, we have that.

  $\frac{81}{64}(x+y)^{2}(y+z)^{2}(x+z)^{2}\geq (x+y+z)^{2}(xy+yz+zx)^{2}$

 So our problem is now equivalent to ;

   $(x^{2}+xy+y^{2})(y^{2}+yz+z^{2})(z^{2}+zx+x^{2}) \geq \frac{27}{64}(x+y)^{2}(y+z)^{2}(x+z)^{2}$

But we have ;

   $x^{2}+xy+y^{2}\geq \frac{3}{4}(x^{2}+2xy+y^{2})$

   $y^{2}+yz+z^{2}\geq \frac{3}{4}(y^{2}+2yz+z^{2})$

   $z^{2}+zx+x^{2}\geq \frac{3}{4}(z^{2}+2zx+x^{2})$

     (These can be proved easily by AM - GM)

  By Multiplying these three inequalities, we will have the inequality as required.

-------------------------------------------------------------
Wps

thank u very much me@home .when i posted it i did not know latex.but now i have learnt a bit .

[quote="ANA123"][quote="April"]
$4(x^{2}+xy+y^{2})\ge 3(x+y)$
[/quote]

Why? :blush:  :wink:[/quote]
Just a type error
$4(x^{2}+xy+y^{2})\ge 3(x+y)^{2}$

Muirhead does kill the inequality but in a very ugly way. here is my proof:
Expanding the $ LHS$ and $ RHS$  :oops:   :oops:   :oops:
We have to prove the following:
$ \displaystyle\sum_{sym} x^4y^2 \plus{} \displaystyle\sum_{sym} x^3y^3 \plus{} \displaystyle\sum_{sym} x^4yz \plus{} \frac{5}{2} \displaystyle\sum_{sym} x^2y^2z^2 \plus{} 8\displaystyle\sum_{sym} x^2y^3z \le \frac{3}{2} \displaystyle\sum_{sym} x^2y^2z^2 \plus{} 3\displaystyle\sum_{sym} x^4y^2 \plus{} \frac{3}{2} \displaystyle\sum_{sym} x^3y^3\plus{} 6\displaystyle\sum_{sym} x^2y^3z \plus{} \frac{3}{2}\displaystyle\sum_{sym} x^4yz$.

This simplifies to :
$ 2\displaystyle\sum_{sym} x^4y^2 \plus{} \frac{1}{2} \displaystyle\sum_{sym} x^3y^3\plus{} \frac{1}{2}\displaystyle\sum_{sym} x^4yz \ge 2\displaystyle\sum_{sym} x^2y^3z \plus{} \displaystyle\sum_{sym} x^2y^2z^2$.

By the Muirhead's Theorem/Inequality,
$ 2\displaystyle\sum_{sym} x^4y^2 \ge 2\displaystyle\sum_{sym} x^2y^3z$
$ \displaystyle\sum_{sym} x^3y^3 \ge \displaystyle\sum_{sym} x^2y^2z^2$
$ \displaystyle\sum_{sym} x^4yz \ge \displaystyle\sum_{sym} x^2y^2z^2$.
Adding up the above 3 inequalities, we get the inequality.
The proof is very ugly though. :oops:

i am TERRIBLY sorry, but i don't really know latex. so this is in the non-latex format (whatever THAT means)

my soln.

(x + y + z)^2 * (xy + yz + zx)^2 [u]>[/u] 3* (x^2 + y^2 +xy)(z^2 + y^2 +zy)(x^2 + z^2 +xz)

now.

each product term on the right hand side is equal to

(x - y)^2 + 3xy [u]>[/u] 3xy        (1)

so, the question is sufficient to show that

(x + y + z)^2 * (xy + yz + zx)^2 [u]>[/u] 3^4 * (xyz)^2

(because of (1) )

which is the same as

(x + y + z)^2 * ( (1/x) + (1/y) + (1/z))^2 [u]>[/u] 3^4

i.e.

((x + y + z)^2) / 3^2 [u]>[/u]  3^2 / ( (1/x) + (1/y) + (1/z))^2

which is the AM-HM inequality squared. it's ok to take the root on both sides as x, y, z have been previously declared as positive reals.

Q.E.D.

[quote="kuing"][quote="Sathej"]If $ x$, $ y$, $ z$ are positive real numbers, prove that
\[ \left(x \plus{} y \plus{} z\right)^2 \left(yz \plus{} zx \plus{} xy\right)^2 \leq 3\left(y^2 \plus{} yz \plus{} z^2\right)\left(z^2 \plus{} zx \plus{} x^2\right)\left(x^2 \plus{} xy \plus{} y^2\right) .\]
[/quote]

this ineq holds for any [b]reals[/b] $ x,y,z$. :maybe:[/quote]

Well, I think April's proof will still hold(With changes to the last two steps...)...isn't it??
Or are you referring to the proof with the fermat's point with negatives being seen as taken in the opposite direction?

If you have your own proof....Please bring it to light :)

[quote="Ronald Widjojo"]No, accuracy, your proof is still incorrect..  :wink:[/quote]

yeah.. oops..  :blush:

[quote="Sathej"]If $ x$, $ y$, $ z$ are positive real numbers, prove that
\[ \left(x \plus{} y \plus{} z\right)^2 \left(yz \plus{} zx \plus{} xy\right)^2 \leq 3\left(y^2 \plus{} yz \plus{} z^2\right)\left(z^2 \plus{} zx \plus{} x^2\right)\left(x^2 \plus{} xy \plus{} y^2\right) .\][/quote]
See here a stronger thing:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=467847

Using Ravi Substitution, let $x+y=c, y+z=a, z+x=b$. Then $a, b, c$ are the sides of a triangle. We then have $x=s-a, y=s-b, ,z=s-c$ and $x+y+z=s$, where $s=\frac{a+b+c}{2}$. We then have \begin{align*}x^2+xy+y^2&=(x+y)^2-xy \\ &=c^2-\frac{1}{4}(c+a-b)(c+b-a)\\ &=\frac{3}{4}c^2+\frac{1}{4}(a-b)^2 \\ &\ge\frac{3}{4}c^2\end{align*} Moreover, $xy+yz+zx=(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a)$. thus it suffices to show that \[s\sum(s-a)(s-b)\le\frac{9}{8}abc.\] But, $\sum (s-a)(s-b)=r(4R+r)$, where $r, R$ denote inradius and circumradius respectively, as usual. Since we have $abc=4Rrs$ (by $[ABC]=rs$) the inequality reduces to \[r(4R+r)\le\frac{9}{2}Rr.\] this is simply $2r\le R$. This follows from $IO^2=R(R-2r$. where $I$ is incenter and $O$ is circumcenter. We are done. $\Box$

Uff how do you people notice such geometric things?
Here's algebraic solution
$3\prod(x^2+xy+y^2)\geq \frac{81}{64}\prod (x+y)^2$
So, we are left  to prove
$9(x+y)(y+z)(z+x)\geq 8(x+y+z)(xy+yz+zx)\Leftrightarrow \sum_{sym}xy^2\geq 6xyz$ which is true by AM-GM

@above, \(x^2 + y^2 + xy = x^2 + y^2 - 2xy\cos\left(\frac{2\pi}{3}\right)\). Never ever ever ever forget this.


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Mafia

Hello,

I had actually expected that there would already have been a topic about this.  For those of you who have not been following this up :
last week( 12th of july) Israeli soldiers were attacked by Hezbollah near the Lebanese border and two of them were abducted.  In order to free them, Israel started carrying out air raids, to which Hezbollah responded by firing missiles into northern Israel, as far as Haifa.

First of all, I wanted to acknowledge my horror concerning this violence, on BOTH sides.

Do we actually have Israelis or Lebanese among us, who could be suffering as we speak  :o ?.

What do you think?  Did Hezbollah plan this?  Did Israel allow its soldiers to be kidnapped in order to gain control in Lebanon again (and made the mistake of underestimating the response)?  Is the timing (abduction in Gaza, G8 summit and recent North Korean missile test) coincidence?

And how will this play out?  Another lengthy occupation?  A stronger Hezbollah position within Lebanon?  A weaker position (even being disarmed)?  Interventions by Syria or Iran?  Another civil war even seems to be possible in Lebanon.

Please keep this topic civil, I don't want it to get locked.

I just thought that this event could not be ignored here considering the suffering  :(

Hezbollah enters Israeli territory with armed troupes (some say this could be seen as a declaration of war, and this is what it is per definition). They kidnap Israeli soldiers. Israel attacks back without proportions. Hezbollah knew that, and so did Israel.

So I think both sides knew this situation would come. Unfortunately, the civil population is getting caught in-between.

I don't think Iran will do something "in public" (except financing and stuff like that), but I'm not sure about Syria.

Like Fredbel6, I definitely acknowledge the suffering on both sides.

As for the blame, clearly Hezbollah started it, Hezbollah is a terrorist organization, Hezbollah is purposely targeting civilians while Israel is trying not to target civilians, and Israel has a right to defend itself. On the other hand, it does appear that Israel has gone a bit overboard on the aggression.

The potential outcomes are pretty straightforward: national and individual damage, weakening of Hezbollah, etc. One important issue I would like to note though, for Israel, is that its attack on Lebanon may cause a growing sense of hatred against Israel among the Lebanese people (as well as other Arab nations). Various news sources I've watched have indicated that many of the Lebanese are angry at Israel. I mean, if it's Israel whom you're frightened of and who's been terrorizing your people, wouldn't you feel something against Israel? I think Israel has to make sure that the Lebanese people truly understand what's really behind this conflict. As well, I think Israel might be able to clean up some of this damage by giving great financial aid to help rebuild Lebanon once this is over.

As for how it will end, I would guess that either some international force will help bring a cease-fire, or that Israel will decide that it has destroyed as much of Hezbollah as it can with airstrikes and will then change its gameplan.

good that you brought up this topic, fredbel. well personally i think hizbollah might have started this whole thing but israel has gone too far. i was just watching the news 2 mins back and it was really horrible. lots of innocent ppl have been killed and the channel actly claimed that more women and children have been, killed than men. i saw a 4 yr old boy with heavy injuries on his leg and shoulder.

  sure what the hizbollah did was not correct, but this is no way to act and i'm surprised that more ppl are not getting angry abt this attack by israel.i'm no expert in analysing history but i wonder why israel is caught in so many strifes- palestine, lebanon etc. :|

Well, we are still not sure whether or not the Israeli soldiers were attacked/abducted on the Israeli side or the Lebanese side of the border, at first Lebanon claimed it was on their side.  But now Hezbollah themselves has admitted they have been planning this for at least five months...

Israel often stated that they have completely withdrawn from Gaza and southern Lebanon, but Hezbollah claims they are still not even as Israel still holds Hezbollah members in prison...

Hezbollah probably are terrorist, but I always wonder : it sure must be nice to decide who is a terrorist and who is not.  It is true that Hezbollah is probably intentionally hiding among the normal population, but why are Israelis bombing christian parts of the city (trying to provoke a civil war again?).
Question : is an Israeli settler throwing rocks at foreign workers trying to restore water supplies in the West Bank a terrorist?  You probably wouldn't hear that on the news. :huh:  On the other hand Palestinian boys have been shot for throwing rocks at tanks...

I think there is no point in arguing who started this, it will always come down to the same debate about whether or not Israel is a legitimate state, something of which most Israelis are convinced and most Arabs certainly not.

I am quite certain that Israel wants to eradicate Hezbollah, but knows how difficult that is, and is therefore resorting to a collective punishment.  This could, maybe it will, result in firm action of the Lebanese government against Hezbollah, or ...it could work the other way around and boost their status....

Edit : because of kouboy's very recent post :

well the news will not show you what REALLY goes on (as Robert Fisk once said : truth is always the first victim of war).
There is actually a website (Angry Arab) collecting images (this site is obviously choosing a side), do not proceed if you are quickly shocked, or a minor :
[url]http://angryarab.blogspot.com/2006/07/hanady-salman-friend-and-editor-at-as.html[/url]
This is a collective punishment.  It is trying to obtain a goal by harming civilians and yes, making them afraid : terrorizing them.
Suppose a bunch of Iraqis, whose family died during rocket attacks in Baghdad in 2003, went to New York to do this?  How would the world react to THAT :huh: ?

[quote]this attack by israel.i'm no expert in analysing history but i wonder why israel is caught in so many strifes- palestine, lebanon etc. [/quote]

This is hard to answer without getting biased.  Essentially the Israelis (jews) claim the land based on historic arguments and Palestinians ((sunni) muslims) see them as oppressors.  Israelis see the land is theirs and consider Palestinian violence terrorism.
Lebanon is harder (I am still trying to solve the puzzle and I will probably never completely succeed) :Hezbollah is a Shia militia formed in 1982 because of the Israeli invasion of Lebanon during the Lebanese civil war.  That Lebanese civil war is a consequence of the already complicated ethnicv make up of Lebanon (shia muslims, christians, druzes,...) together with the burden of Palestinian refugees from Israel.

Now already there are probably gonna be people disputing the accuracy of what I said (and they will probably be right) , and this was just a very very incomplete explanation.

Absolutely No one among civilians take advantage of all that but let us be realistic a bit :The Israeli responce forces lebanon to be back to 1990 wich is the year when the civil war ended, it's the infrastructure of Lebanon which is destroyed .
A part from that the suffering of civilians from both sides isn't the same in southern lebanon they don't even have shelters and till the moment there are more than 400 lebanese killed, so saying that both israelis and lebanese suffer might be real but admiting that they suffer the same way is extremely wrong.
At last i'd say something about the role of Media in all that because i noticed that there is a big gap between the information received by people living in the Middle east and the one that the rest of world gets, I say that because i watch western arab and even IBA(israeli channel) news
an other clear example of that gap is for instance the difference of CNN's coverage of events between the cable edition (the one watched by americans) and the international edition.Personally i try to get the information from all these sources in order to have a clear image about what actually happens.
And here are two blogs about the current events :
[url]fromisraeltolebanon.info[/url]
[url]ontheface.blogware.com[/url]

A bit off topic :
[hide]Well coverage is of the most crucial importance.  I think the Flemish media isn't doing a good job and the Wallonian isn't either (they seem to spend more time on those few Belgians in Lebanon), the Dutch have correspondents on both sides and get to interview important Israelis about everyday.  I also receive BBC, and I try to compare it with ZDF, Arte and France 2

CNN isn't that bad compared to FOX!  Unfortunately those greedy cable distributors here kicked CNN of, so I can only visit their website.

al jazeera has a pretty decent website actually![/hide]

Fredbel, I don't believe that Israel is purposely trying to punish or target the civilians of Lebanon; instead I think it's doing quite the opposite: it's trying to minimize civilian damage. I think the problem here is that Israel has not gone anywhere near close enough to the end of protecting civilians over destroying terrorist infrastructure.

As for your mention of throwing rocks, neither the Israel settler nor the Palestinian boys are terrorists.

I think what the U.S. Ambassador to the UN, John Bolton, has said perfectly echoes my sentiments:

"What Hezbollah has done is kidnap Israeli soldiers and rain rockets and mortar shells on innocent Israeli civilians. What Israel has done in response is act in self-defense. And I don't quite know what the argument about proportionate force means here. Is Israel entitled only to kidnap two Hezbollah operatives and fire a couple of rockets aimlessly into Lebanon?

The situation is that Israel has lived under the terrorist threat of Hezbollah for years, and these most recent attacks have given it the legitimate right, the same right America would have if we were attacked, to deal with the problem. And that's what they're doing."

Sure, Israel has caused lots of civilian casualties and its horrible...but the fact is, if Hezbollah actually knew HOW to fire their rockets properly, they would have killed a lot more civilians. Israel's problem is that Hezbollah hides out in densely-populated suburban areas, effectively using civilians as human shields--both physically, and to deflect media attention.

But really:

 "And I don't quite know what the argument about proportionate force means here. Is Israel entitled only to kidnap two Hezbollah operatives and fire a couple of rockets aimlessly into Lebanon?"

is the clincher for me.

[quote="kouboy"]

  sure what the hizbollah did was not correct, but this is no way to act and i'm surprised that more ppl are not getting angry abt this attack by israel.i'm no expert in analysing history but i wonder why israel is caught in so many strifes- palestine, lebanon etc. :|[/quote]

One inference deserves another...maybe you should ask instead why Arabs everywhere live in war, under dictatorships, or in poverty--regardless of whether or not they control the country and/or vast reserves of oil.

Here's a simple analogy, and even better, I'll make it so that everyone can relate on a personal level (because really, almost nobody here can understand what it's like to be an Israeli): two men break into your house, kill your sibling and kidnap your parents. One of the intruders is arrested and taken to jail, but the other escapes and holds your father and mother hostage. The demands? He wants his friend back and the two of them can go free in return for your family to remain only somewhat intact (they are, after all murderers).

I really think this example speaks for itself.

[quote="kouboy"]@jpps- what you say also makes sense. sure israel is entitled to defend itself, but at what cost? hundreds of innocent ppl in return for the 2 who were captured? also just because they cannot find the hizbollah, they cant just fire into civilian areas.israel must get its intelligence agency to find out where they are,they cant keep killing the ordinary lebanese guy.

 also i post this question to everyone who is interested, do you think that we can make a difference to all thats happening now-violence,war,increasing hatred among nations? i hope this stays as a good natured discussion without hurting anyone.

p.s- if theres any israeli here or lebanese for that matter i'd like to hear what you have to say[/quote]

very well said.
i think this something which shouldn't be discussed here.
probably it will be useless to discuss here.

2 solider don't cost 100s of life.
however their war is  something milestone.it will not end up so fast.
it will be carried on for some years like history tells.

[b]fredbel6[/b][quote]what do you think? Did Hezbollah plan this? Did Israel allow its soldiers to be kidnapped in order to gain control in Lebanon again (and made the mistake of underestimating the response)?[/quote]
i think their demands were to set free those captive people from last wars.

imo international forces are needed to bring peace.

Kidnapping the israeli soldiers isn't justified if and only if we forget about lebanese and palestinian prisoners in israeli jails, among them there are those who stayed there for more than 30 years.Hezbollah justification of what it has done is that it aims to free those prisoners ,i am not supportive of Hezbollah myself but what i am sure of is that resistance of occupation is a legitimate right and resistants were always called terrorists by the occupiers. Peace is possible if everyone has what he owns back.
By the way i should remind you of war victims till the moment :
37 israelis/460 lebanese.
Peace

[quote="jpps1"]Here's a simple analogy, and even better, I'll make it so that everyone can relate on a personal level (because really, almost nobody here can understand what it's like to be an Israeli): two men break into your house, kill your sibling and kidnap your parents. One of the intruders is arrested and taken to jail, but the other escapes and holds your father and mother hostage. The demands? He wants his friend back and the two of them can go free in return for your family to remain only somewhat intact (they are, after all murderers).

I really think this example speaks for itself.[/quote]
Analogies are as usual a method which turns twisted thoughts into logic.

Well, not all are murderers. I've heard of prisoners with their <1year old children in the same cell. I'm having a hard time seeing a child who can barely talk as a terrorist.

Here is another simple analogy.

A few mini-ants kill some beetles. Now the mini-ants flee and hide inside an ant colony, safely in the middle. The beetle gets pissed of and start bombing the colony, of course they [i]try [/i] not to hurt the other ants. It's not hard to understand that the bombs will kill civil ants.

Do you see why analogies don’t always work?

Since you like math. Find the number of civil casualties and the number of Hezbollah soldiers that have been killed. Also look for number of civil injuries.
Now, try to find some data on how many Hezbollah members there are. Use the data from the previous calculations and count how many civil casualties it would take to kill all the Hezbollah members. Also count injuries that it would lead too.
Last time I checked (according to AFP a few hours ago) the numbers were 363 casualties where 43 were "not civilians" (don't know how many of them where actually hezbollah members).
So I'm not sure they (both parts) really care too much about the civil population.

Mohamed, your point is interesting, as I have seen it discussed in an very recent (after the beginning of this war) interview with former Israeli prime minister Netanyahu .

He says that death toll does not justify the actions of one side, and he referred to the huge number of German casualties in the second World War.  I do not completely disagree- one doesn't know what Hezbollah would do if they had more and better rockets - but I do think this reaction is way over the top.

He also claimed that Israel does not deserve any of this, as they have completely withdrawn from Gaza and southern Lebanon.  He does have a point here, especially about Lebanon.  How can Hezbollah see themselves as occupied.  They are not Palestinians?  They have been firing missiles, AFTER the withdrawal as well if I remember correctly.  That I find hard to understand.

Greetings

[quote="fredbel6"]

He says that death toll does not justify the actions of one side, and he referred to the huge number of German casualties in the second World War.  I do not completely disagree- one doesn't know what Hezbollah would do if they had more and better rockets - but I do think this reaction is way over the top.
[/quote]

Certainly, death toll doesn't justifies Hezbollah's actions but completely discredits in my opinion israel's. And actually it is not the fact of the number of casualties as Netanyahu reminds us with his comparison. It is the way by which this desproportionate death toll is reached. Israel is not attacking Hezbollah. I find hard to understand an attack on a terrorist organization with a dubious number of activists directly related with armed confrontation and which by the way some people seem to think that many of them are hiding in tunnels, while others including Israel claim they hide guess where!, suburban neighborhoods, as I say, I find hard to understand such an attack when they are bombing densely populated areas in Beirut. How many Hezbollahs have been killed in the air bombings? We know nevertheless that families with children running away from copters have indeed been bombed, this is news from yesterday so it should be easy for you to check it. Also may be it's me but I've noticed this tendency to call it right away a war crime if Hezbollah fires a missile in Haifa, and much more reluctancy when it comes to judge bombings on civilians with no Hezbollah connections in Beirut.

Israel in my opinion is attacking Lebanon and besides in a very calculated fashion. Israel knows pretty well Lebanon is in any case is poor country with a very diminished army with very old equipment. If Lebanon would have wanted to fully support Hezbollah's actions and back them up it woud have been suicidal...

Nevertheless and this is probably my strongest argument, for what I've heard, this has not been Lebanon's response, and this is CNN, Fouad Siniora speaking, Lebanese prime minister to the people

"SINIORA (translator): The right way and the only way to face that we only save the real estate that will protect everybody and preserve the rights and preserve their safety and the government has declared from the very beginning of the explosive situation that it was not aware of what happened, and did not adopt the operation that Hezbollah has captured two Israeli soldiers, and that did not -- did not cause this.

Hezbollah has no right to destroy Lebanon. And destroy its infrastructure, and this experience pushes us more to withstand and to hold on to the Lebanese state, to protect the Lebanese against any threats and aggression. Lebanon cannot stand if the government is the last one to know and the first one to be asked, the state alone is the only one who has the legitimate decision for peace or war because it represents the will of the Lebanese people, that who wanted to live together. We are facing the aggression today, unified and we will continue to be unified from behind the state, from behind the constitutional institutions ".

This is july 15th, in what seems to me a clear message of not taking sides with Hezbollah. And if these are the declarations of lebanese prime minister I feel inclined to think that at that moment, probably, a diplomatic solution could have been reached.

But no one is interested in diplomatic solutions nowadays it seems. War and war if it suits you. Hezbollah is doing as expected, also targeting civilian areas, rocketing every city in Israel. Of course they are not saints, but while they can be labelled a terrorist organisation and treated so, Israel is one full country with full support from the US. The good guys. I know Hezbollah has political power in Lebanon with many seats in the parliament. Still why do I feel that there is no intention to fix things with the islamic world?

Nobody seems to remember either that this has already happened before. Hezbollah, in the year 2005 if I remember well, abducted israel's soldiers and it didn't ended up in an armed conflict. It seems that they were after the release of some lebanese who had been in israel's prisons for twenty years or more, since the lebanon war. I admit I'm not pretty sure of the precise facts, I could look them up, I'm just not in the mood right now. But I'm certain this has happened before and it was resolved with the exchange of prisoners, I think Germany mediated, ergo, diplomaticly. It also seems that this time this was again the intention of Hezbollah since it appears that there are still lebanese prisoners in Israel.

[\quote ]He also claimed that Israel does not deserve any of this, as they have completely withdrawn from Gaza and southern Lebanon.  He does have a point here, especially about Lebanon.  How can Hezbollah see themselves as occupied.  They are not Palestinians?  They have been firing missiles, AFTER the withdrawal as well if I remember correctly.  That I find hard to understand.
[/quote]

As I understand, Hezbollah is not seeing itself as occupied. The intention of the abductions was as I said, to exchange prisoners with Israel. I don't recall Hezbollah launching missiles recently to Israel, before all this of course, but it's very possible. I dont' know, I will definitely have to look on that because it's an important fact. I also find kind of funny Netanyahu saying 'we don't deserve this' because it seems to me like a sort of pleading phrase in order to convince the world of his truth. Yeah well, I'm not lebanese nor palestinian, if I were, I would probably had the right to ask, what is it that you don't deserve? Our terrorism in your country? (assuming I'm a palestinian who has already accepted israel as a country of course) To have us kidnapping your soldiers? Oh, indeed you don't. You know there might be some things that we don't deserve either... not that I'm complaining but you see we might be able to make our lives a bit better if we instead of crying loud to the outside world and staring fiercely at each other in our homes and why not, sending a missile or two from time to time, we did bla bla...
Of course I'm a very positive palestinian, in the real world I would have to say such things as an israeli too, so excuse me, it's just that it was Netanyahu who said it.

Oh I almosf forgot, wasn't it Netanyahu who opposed the withdrawal of Israel forces from Gaza??  Another funny thing to have him using it as an argument.

Enfin, what can I say? I was optimistic a few years ago and entered these discussions every time, now I do it with less and less frequency since I don't see solutions and it depresses me.

Daniel

[color=red]A few edits to improve the english[/color]

Daniel , about the word 'deserve', I don't actually remember him saying that, I was trying to make a summary of his claims.   The interview was a video conference with a Dutch reporter in the Netherlands and Netanyahu, and they both spoke English.
(Well, I just wanted to straighten that out).

One question : WHERE is the Lebanese Army?  Hezbollah doesn't really listen to the Lebanese government, they are not the official army.  So where is the army?  I haven't seen any (non Israeli tank) in Lebanon since all this started?  I know they are a small army that probably is inferior to the Israeli army, but still....

[u]About Hezbollah since 2000 :[/u]
in may 2000 Israel withdrew from Lebanon.  However Hezbollah considers the Shebaa Farms still as occupied Lebanese territory by the Israelis, while the UN seems to consider it part of the Golan heights, Syrian territory captured in 1967 by Israel.

Also in 2000, they captured three soldiers during an attack on those farms.  In 2004 they traded their bodies (they were dead) along with one Israeli man, for 30 Lebanese and more than 400 Palestinian prisoners (alive) , along with bodies and the location of [b]Israeli mines in South Lebanon[/b].

They have killed soldiers and wounded civilians with their rockets since the withdrawal, but claim this was retalliation for illegal entry into Lebanese airspace by the Israelis, and they have also sent an Iranian unmanned machine over Israel apparently

This description however is probably not complete.

@kouboy: This conflict isn't solely about the Israeli soldiers who were kidnapped; it's about the mere existence of Hezbollah (a terrorist organization) and Hezbollah's rocket fire into Israel.

@kahsi: Just because many Israeli soldiers have targeted innocent Palestinian civilians as if they were terrorists, does that mean that Hezbollah is not terrorist?

@MoHaMeD and Fredbel6: I certainly believe that the death toll is not right to judge whether each side has more or less right to do what it's doing. First, let's look at each sides' military tactics:

- Hezbollah has been actively targeting civilians.
- Israel has been trying and doing a very bad job at not targeting civilians.

Even though Israel has been doing a bad job at not targeting civilians, Hezbollah still must have a greater proportion of civilian injuries to military capability than Israel. Clearly, the reason that Lebanon has suffered more damage is that Israel has much greater military capability: It is a full country with a strong military *and* help from the United States. Hezbollah, on the other hand is simply a foreign terrorist organization in Lebanon.

Thus, I think Israel is justified in taking military action against Hezbollah; on the other hand, Israel's fault is that it has not done so properly.

ps: yay $7^{3}$ posts

@calc rulz: I never said that. My feelings for Hezbollah are not cheerful ones. We all know that what Hezbollah is doing is wrong, so I am not going to talk about that. It’s trivial. I am going to focus on how Israel is handling this.
Just because Hezbollah are bad guys doesn’t justify all the actions of Israel, now does it?
Israel has the right to act since they were attacked by a foreign power. But, does this mean that they do everything in their power to spare civil lives?
They are now fighting Hezbollah and not Palestinians with primitive weapons etc. Many say that Hezbollah are more organized than Al-Qaeda. They probably have more advanced weapons too. This is a hard nut to crack for Israel. A nut they won’t be able to crack with airplanes. You see, this is where most experts agree. There is no way to strike Hezbollah by dropping bombs from the sky. Israel will have to enter with troops. They know that, and so does everyone else. But yet they continue doing this (the bombings). It’s not hard to understand the intentions…
But, they have started with troops now (after demolishing the place (this was the intention)), so hopefully this will lead to less civil casualties. And they are [i]considering[/i] NATO help.

To be honest, I think Israel cares about Lebanese civilians just a bit more than Hezbollah cares about Israelis. Just a little bit and not more than that… And blaming on a "bad job" is ridiculous considering the bombing from the airplanes. No one actually thought they would succeed with that. The intention was simple: bomb everything and make chaos so that it will be easier for our troops when they enter Lebanon.

PS: Israeli ministers have said that Hezbollah have their base in Syria. So why wont they attack Syria?

[quote="fredbel6"]
One question : WHERE is the Lebanese Army?  Hezbollah doesn't really listen to the Lebanese government, they are not the official army.  So where is the army?  I haven't seen any (non Israeli tank) in Lebanon since all this started?  I know they are a small army that probably is inferior to the Israeli army, but still....[/quote]

Hezbollah seems to be exactly that, the unofficial army of south Lebanon. Of course, it's something unusual to have an army disconnected of the government ruling in a portion of your country. Anyway here, detailed account of the capacities of lebanon army. Notice no air force [url=http://www.globalsecurity.org/military/world/lebanon/airforce-equipment.htm]here[/url]. Also wikipedia informs us that in 2005 Lebanon announced that it would end conscription in two years. With the twist of events I find that unlikely. In contrast conscription in Israel is goes up to three years for almost every male above 18 and for many females. I think even after conscription every year the men must go for a period into a military base and the army takes care of their salaries during that time. This again I can not confirm cause I don't remembe where I took this knowledege from. Israel has nukes and doesn't sign any world treaty concerning the non-use of weapons of chemical or biological sort.

[quote="fredbel6"]
[u]About Hezbollah since 2000 :[/u]
in may 2000 Israel withdrew from Lebanon.  However Hezbollah considers the Shebaa Farms still as occupied Lebanese territory by the Israelis, while the UN seems to consider it part of the Golan heights, Syrian territory captured in 1967 by Israel.
[/quote]

This is also the position of the lebanese government about the farms. So it's not like Hezbollah is feeling occupied, it's Lebanon who feels still a bit occupied. And Lebanon doesn't second Hezbollah...or didn't, in the last days I'm not aware of how the acceptance of Hezbolla among the lebanese has evolved. Besides, Hezbollah leader Sheik Hassan declared on july 12th that [b]only[b] the exchange for three lebanese held in Israel, would set free the israeli soldiers.

[quote="fredbel6"]

Also in 2000, they captured three soldiers during an attack on those farms.  In 2004 they traded their bodies (they were dead) along with one Israeli man, for 30 Lebanese and more than 400 Palestinian prisoners (alive) , along with bodies and the location of [b]Israeli mines in South Lebanon[/b].
[/quote]

You're right, this was what I remembered. Didn't remember that they returned the soldiers dead though. I find curious that Israel gave so much apparently.

[quote="fredbel6"]
They have killed soldiers and wounded civilians with their rockets since the withdrawal, but claim this was retalliation for illegal entry into Lebanese airspace by the Israelis, and they have also sent an Iranian unmanned machine over Israel apparently
[/quote]

I still can not find this documented. Could you please say where are you taking it from?

Daniel[/b]

[quote="dhernandez"][quote="fredbel6"]

He says that death toll does not justify the actions of one side, and he referred to the huge number of German casualties in the second World War.  I do not completely disagree- one doesn't know what Hezbollah would do if they had more and better rockets - but I do think this reaction is way over the top.
[/quote]

Certainly, death toll doesn't justifies Hezbollah's actions but completely discredits in my opinion israel's. And actually it is not the fact of the number of casualties as Netanyahu reminds us with his comparison. It is the way by which this desproportionate death toll is reached. Israel is not attacking Hezbollah. I find hard to understand an attack on a terrorist organization with a dubious number of activists directly related with armed confrontation and which by the way some people seem to think that many of them are hiding in tunnels, while others including Israel claim they hide guess where!, suburban neighborhoods, as I say, I find hard to understand such an attack when they are bombing densely populated areas in Beirut. How many Hezbollahs have been killed in the air bombings? We know nevertheless that families with children running away from copters have indeed been bombed, this is news from yesterday so it should be easy for you to check it. Also may be it's me but I've noticed this tendency to call it right away a war crime if Hezbollah fires a missile in Haifa, and much more reluctancy when it comes to judge bombings on civilians with no Hezbollah connections in Beirut.

Israel in my opinion is attacking Lebanon and besides in a very calculated fashion. Israel knows pretty well Lebanon is in any case is poor country with a very diminished army with very old equipment. If Lebanon would have wanted to fully support Hezbollah's actions and back them up it woud have been suicidal...

Nevertheless and this is probably my strongest argument, for what I've heard, this has not been Lebanon's response, and this is CNN, Fouad Siniora speaking, Lebanese prime minister to the people

"SINIORA (translator): The right way and the only way to face that we only save the real estate that will protect everybody and preserve the rights and preserve their safety and the government has declared from the very beginning of the explosive situation that it was not aware of what happened, and did not adopt the operation that Hezbollah has captured two Israeli soldiers, and that did not -- did not cause this.

Hezbollah has no right to destroy Lebanon. And destroy its infrastructure, and this experience pushes us more to withstand and to hold on to the Lebanese state, to protect the Lebanese against any threats and aggression. Lebanon cannot stand if the government is the last one to know and the first one to be asked, the state alone is the only one who has the legitimate decision for peace or war because it represents the will of the Lebanese people, that who wanted to live together. We are facing the aggression today, unified and we will continue to be unified from behind the state, from behind the constitutional institutions ".

This is july 15th, in what seems to me a clear message of not taking sides with Hezbollah. And if these are the declarations of lebanese prime minister I feel inclined to think that at that moment, probably, a diplomatic solution could have been reached.

But no one is interested in diplomatic solutions nowadays it seems. War and war if it suits you. Hezbollah is doing as expected, also targeting civilian areas, rocketing every city in Israel. Of course they are not saints, but while they can be labelled a terrorist organisation and treated so, Israel is one full country with full support from the US. The good guys. I know Hezbollah has political power in Lebanon with many seats in the parliament. Still why do I feel that there is no intention to fix things with the islamic world?

Nobody seems to remember either that this has already happened before. Hezbollah, in the year 2005 if I remember well, abducted israel's soldiers and it didn't ended up in an armed conflict. It seems that they were after the release of some lebanese who had been in israel's prisons for twenty years or more, since the lebanon war. I admit I'm not pretty sure of the precise facts, I could look them up, I'm just not in the mood right now. But I'm certain this has happened before and it was resolved with the exchange of prisoners, I think Germany mediated, ergo, diplomaticly. It also seems that this time this was again the intention of Hezbollah since it appears that there are still lebanese prisoners in Israel.

Daniel

[color=red]A few edits to improve the english[/color][/quote]
Bravo!
Very well written! These days it’s hard to find unbiased opinions. I often find people very pro-Israel or pro-Arab. And these people seem to think their side is flawless and that everything they do is justified no matter what actions is taken. "The ends justify the means" seems to be correct for all means…

Daniel, in your last post you asked for sources, and I will admit it : Wikipedia :

But :

[url]http://news.bbc.co.uk/2/hi/middle_east/1279456.stm[/url] about an attack by Hezbollah in 2001
[url]http://news.bbc.co.uk/2/hi/middle_east/3990773.stm[/url] about that unmanned drone from Iran

[url]http://www.un.org/apps/sg/sgstats.asp?nid=449[/url] about anti aircraft artillery from Hezbollah hitting northern Israel

[u]Question : [/u] even a long time ago, I was trying to understand the Lebanese civil war (you never completely "get" a conflict but understanding the major parties is always a start) and I didn't really succeed.  I tried doing this again since this conflict started, and got confused again :
I read that at one time Amal (shia group supported by Syria) was fighting Hezbollah (shia group supported by Iran) and Palestinian armed groups.
Isn't it quite weird that now everybody points to both Syria and Iran as the supporters in Hezbollah? :huh:

khasi, is that a rhetorical question?  Attacking Syria is a lot harder than Lebanon.  It has a much larger population, and enjoys good ties with Iran for the moment.  It's not like Lebanon, a small country with almost no defenses.  And just like the USA, I think they don't like to hit someone who can really hit back HARD.
On the other hand, they have hit Syria quite recently [url]http://news.bbc.co.uk/2/hi/middle_east/3183788.stm[/url]

Regarding the lebanese civil war it started when palestinians that were kicked out by Jordanian King Husain after the black september moved to Lebanon. It's a long story but let's say that at that time some lebanese didn't want to be a part of the arab-israeli conflict these guys were mainly christians but not all of them actually the war started when a palestinian bus was bombed thus the war was between those who supported the palestinian cause in Lebanon and those who did not and who were allies of israelis and were financed by Israel and guess what?? at first they were backed by syrians too so it's a long story.Moreover a lot of intelligence services agencies were involved in that "durty" war, personally i wouldn't be surprised if i knew that it was an american/Israeli network that gave weapons to so many militias even those that are against israel, as they have given weapons to Iran in its war against Iraq and as the Israeli mafia sells weapons to palestinians, unless it's about classic weapons and it brings benefits.By the way there are those who think that the US this time wants Hezbollah's story over because of what is supposed to be Hezbollah's attacks againt american commandos during the civil war.
Let's be back to the syrian case well according to New York Times american administration might tend to stop their pressure on it for a while so that they make it further from Iran and thus Iran will remain without allies in the region. Personally i don't think Syria is strong enough to face an attack let's remember the israeli attack in 2005 against Syria and how weak the syrian reaction was.
Actually the only military power in the region is Israel that had more than 200 nuclear bombs in the 80's according to Mordachai Vaanono (originally moroccan jewish who converted to christianity in Jail in Israel) and Israel wants to make sure of this supremacy by preventing any other country in the region from having a scientific advancement that may have a military application :This is what most arabs think.

[u]About Mordechai Vanunu :[/u]=

[url]http://www.vanunu.freeserve.co.uk/[/url] this is a site (biased of course!) that strives for his freedom.
He was abducted on Italian soil by the Mossad.  That is the same organisation accused of shooting Gerard Bull in my country

Yes, it is weird that USA cries on all the time about other nations (than the big five) developing WMD while it is the only nation having actually used them in a war, and their 'buddy' Israel is definitely number six.

But to get back to my question : how did Hezbollah transform from an amal fighting (and thus Syrian interference opposing) militia to a very pro Syrian organization?

Precisely Vaanono has met an american jewish woman who pretented being a divorced american tourist and told him to come with her to Italy, there is an interesting book (i read it in french) that talks about Mossad operations.
Regarding the Hezbollah/Amal fightings it's due to the war of camps:
http://en.wikipedia.org/wiki/War_of_the_Camps
Syria has tolerated the creation of Hezbollah by Iranians at first to help Amal fighting israelis in Central and southern Lebanon.
in lebanese politics there are a lot of odd things of politicians changing their position completely due to the change of circumstances.
Back again to Israel/Hezbollah war.Israel has targeted from 2000 till today so many Hezbollah's members whom it has killed in Mossad's operations (last monthes there have been 3 Mossad cells discovered in Lebanon those who work in these cells are lebanese financed who receive the orders and the financial support directly from the Mossad).

For what it's worth, I heard on the BBC/NPR the other day that America partly funded Hezbollah.

[quote="fredbel6"][u]About Hezbollah since 2000 :[/u]
in may 2000 Israel withdrew from Lebanon.  However Hezbollah considers the Shebaa Farms still as occupied Lebanese territory by the Israelis, [/quote]

Hizbollah and its Iranian paymasters have been saying loudly for years that any Israel-controlled territory is an illegitimate Zionist occupation.   Iran's president is, of course, famous for his more recent threats to annihilate the Zionist enemy in a fiery (i.e. nuclear) holocaust.   It is a bit ridiculous to focus on whether some particular slice of territory might justify the Hizbullah position or to discuss Hizbullah as some sort of nation-entity whose opinions are relevant.

I have not read any complaints about "collective punishment" when the Hizbullahis routinely bombed civilian targets in Israel for many years, first with Katyusha rockets (as did the PLO before them) and now with Qassam missiles.  Nor have I heard an explanation of how the Lebanese population, which separated itself so clearly from the Syrian military, somehow managed to overlook the Hizbullah dominion in Southern Lebanon (much like Al Qaeda in Afghanistan, "owned and operated by Osama bin Laden" according to a famous CIA report). The passive-aggressive indignation of the Lebanese, when bombing is finally used to control the mad dogs that they and their various militias declined to control for 20 years, rings false.

Fleeting guest, your comments in the other post did convince me I was going off topic, and therefore I decided to bump this thead back up, instead of writing everything I think of Israel in "Israel bombed UN observers" thread.

Iran claims to be working on a peaceful nuclear program.  They have done press conferences with (in English!) banners saying "The world without zionism".  But do you have a source for claims they want to launch an unprovoked nuclear attack against Israel (It would be hard to not hit Israel too hard, it is so small any such actions would most likely result in harming Lebanon, Syria, Jordan, Palestinians,...)

What I was doing in that thing you quoted was trying to piece together Hezbollah's actions between the withdrawal and this conflict.  Some sources claim they have not fought over land except the Shebaa farms.  But I do not know, so I will gladly accept sources saying completely different things.

Hezbollah's hiding between civilians in Beirut or launching rockets very close to UN refugee camps (and even seeking shelter inside the camp directly afterwards) is often explained by the "forced asymmetric warfare" argument, they cannot win in an open war against the overwhelming Israelian military force, so they have to resort to all sorts of different techniques.  I've seen reporters on Dutch television (which is weird, they shouldn't give an opinion) saying that that still doesn't give Israel license to target civilian areas which is a crime.  But it cannot be denied : Hezbollah is commiting a crime themselves.   Neither party should use incapability as an excuse.
(It should be pointed out that many people, usually Hezbollah members themselves, deny that they hide between civilians in Beirut)

Israel often drops leaflets before bombing, ordering them to leave certain areas or stay inside.  But leaving has already been made practically impossible in some part of the country.
But it is true : Hezbollah does not drop any leaflets and aim rather wildly than target military installations.

Technical questions : how does the alarm system work in Haifa?  I mean do they radar or visual confirmation of rockets flying over?  How fast do these rockets fly?  And can their trajectory still be changed after launched (either by being programmed to do so or by remote control)?
I am just wondering why it is so hard for the technologically superior Israelian forces to immediately strike back at the exact location a rocket was just launched.


calculus
integration
combinatorics solved
combinatorics

We have 12 Coins.One of them (we dont know which one?)
has different weight from the others.With three time scaling (the scale has two pans) find the differrent coin. :lol:

Well known, from some IMO longlist, if I remember correctly.

What is the point of posting this in the "Solved Problems" section??

i am so sorry i realy dont know!? because this is my friends problem!but i think it is very nice! :D

can anybody sove this nice problem?come on lets try it! :)

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=balance++coins&t=11629

Let's consider the problem solved and the topic closed.

thank you very much Arne,you realy helped me. :D

Thank you Arne but in thet link there is no solution for this problem,there is only their ideas,anyway than you again. ;)

There's more than one page of discussion - the solution is on one of the later pages.

i have already solved it and here is my solution:
first of all we divide the coins in three group such that in each group there is four coins.
\[ \omega_1=[a_1,a_2,a_3,a_4] \setminus   \omega_2={[b_1,b_2,b_3,b_4]} \setminus \omega_3={[c_1,c_2,c_3,c_4]} \]in this writting $ (a,b)$ means that we put $a $ in one of the pans and we put $b$ in the other one and then we scale it.
now lets start to solve this problem(cool problem):step1)if $ (\omega_1,\omega_2) $ then:a)if two pans were equal then:
   we understand that all coins in $ \omega_1$ and $\omega_2$ are normal and the different coin is in $\omega_3$.   then $ ({[c_1,c_2]},{[c_3,c_4]}) $ and after it $ ({[c_3,c_2]},{[c_4,a_1]})$ and from it we can find the different coin. :D b)if they were not equal then:   we can suppose that $ \omega_1>\omega_2$ (because there is no different(?)). then $({[a_1,b_1,c_1]]},{[a_2,b_2,a_3]}) $then :1)if  $ {[a_1,b_1,c_1]}>{[a_2,b_2,a_3]} $ then $(a_1,c_1)$ and from it we can find the different coin.(?)
2)if  $ {[a_1.b_1,c_1]}<{[a_2,b_2,a_3]} $ then $ (a_2,a_3) $ and from it we can find the different coin.(?)
3)if  \[ {a_1,b_1,c_1}={a_2,b_2,a_2} \] then $ (b_3,b_4) $ and from it we can find the different coin.(?)
so we solved the problem!!..

thanks a lot .i think it is nice too. :D

do i know you ali_ms2004 ?because one of my friend's name is ali ?are you ali ?(i mean ali e khodemoon?)

Do you know that the least number k of weighing is egual to log3(2n+3) in case the number 2n+3 is an integral power of 3 and [log3(2n+3)+1] in case 2n+3 isn't egual to an integral power of 3?

It's very difficult to prove but the result is very useful!

[quote="Lilian"]What about if we have 13 coins instead of 12?[/quote]
i think it will be different.
and another thing about what you said about the general answer,are you sure?

Yes! I've found that in a very good book!

can you tell me which book is this and what does it say,please? :?:

It's an old Russian book written by Yaglom.

Are you sure that is impossible for 13 coins?

[quote="Gibbenergy"]Hey guys.
  I have strugle with this one too.
   In fact, there is impossible for 13 coins with only 3 tries. But sorry! I don't know how to solve it. My friend have done it by using probability. But.. I totally get lost as the second of his page! :D[/quote]

I have posted the same problem here :)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65414[/url]

I know a problem in this range:
we have $2n$ coins which have regular weights and $2n$ coins with less weights
but we know that the weights of the regular coins are equal and the different coins have the same weight
proove that we can find the unregular coins with less than $3n$scales
does it help u?? :?:

[quote="Gibbenergy"]Hey guys.
  I have strugle with this one too.
   In fact, there is impossible for 13 coins with only 3 tries. But sorry! I don't know how to solve it. My friend have done it by using probability. But.. I totally get lost as the second of his page! :D[/quote]

It actually is possible for 13.  Probably not 14 though.

how is it possible with 13? didn't the formula given early show that it had to be 4?

[quote="jensen"]i have already solved it and here is my solution:
first of all we divide the coins in three group such that in each group there is four coins.
\[ \omega_1 \equal{} [a_1,a_2,a_3,a_4] \setminus \omega_2 \equal{} {[b_1,b_2,b_3,b_4]} \setminus \omega_3 \equal{} {[c_1,c_2,c_3,c_4]}
\]
in this writting $ (a,b)$ means that we put $ a$ in one of the pans and we put $ b$ in the other one and then we scale it.
now lets start to solve this problem(cool problem):step1)if $ (\omega_1,\omega_2)$ then:a)if two pans were equal then:
   we understand that all coins in $ \omega_1$ and $ \omega_2$ are normal and the different coin is in $ \omega_3$.   [b]then $ ({[c_1,c_2]},{[c_3,c_4]})$ and after it $ ({[c_3,c_2]},{[c_4,a_1]})$ and from it we can find the different coin.[/b] :D b)if they were not equal then:   we can suppose that $ \omega_1 > \omega_2$ (because there is no different(?)). then $ ({[a_1,b_1,c_1]]},{[a_2,b_2,a_3]})$then :1)if  $ {[a_1,b_1,c_1]} > {[a_2,b_2,a_3]}$ then $ (a_1,c_1)$ and from it we can find the different coin.(?)
2)if  $ {[a_1.b_1,c_1]} < {[a_2,b_2,a_3]}$ then $ (a_2,a_3)$ and from it we can find the different coin.(?)
3)if
\[ {a_1,b_1,c_1} \equal{} {a_2,b_2,a_2}
\]
then $ (b_3,b_4)$ and from it we can find the different coin.(?)
so we solved the problem!!..[/quote]
i have a little problem with the part that is bold.
because we don't know if the special coin is heavier or lighter.

I have $ 13$ little metal balls which with same appearance,
there is a special ball with unrealized different weight , I can use
a balance without any weights to find out it. How can I find the
special ball by weighting the balls three times only?

[b]Solution:[/b] mark the balls with number $ 1,2,\cdots,13$ respectively.

(1)compare A={1,2,3,4} with B={5,6,7,8}

(2)compare C={1,2,5,9} with D={3,6,10,11}

we can make the 3rd compare according to the result of (1) and (2) to find out the special ball.

How do I get the first and the second compare? $ A, A \cap C, B \cap D, A \cap D, B \cap C, and \ E \equal{} \overline{A\cup B\cup C\cup D}$,
their sizes need some kind of balance.

I think,we can consider it like this.if $n$ is the number of weighings,THEN,summation of 3^0 + 3^1 + .........+3^(n-1) coins can be found out.I think it is somewhat similar to da problem,where they tell you to divide some weight into some parte,so that any integral weight can be measured from 1 to that weight.It is also related to power of 3.

Among 12 coins, it's possible to find the counterfeit coin and whether it's heavier or lighter. Among 13 coins, it's possible to find the counterfeit coin, but not whether it's heavier or lighter.

13 coins:
Let the coins be A,B,C,D,E,F,G,H,I,J,K,L,M.

(1) First, compare A,B,C,D to E,F,G,H.
If (1) balance, the counterfeit is among I,J,K,L,M. Then: (2) Compare A,B,C against I,J,K.

If (2) balance, the counterfeit is among L or M. Compare A and L to get the counterfeit coin; if it balances, it's M, otherwise it's L. Note that we can't decide whether M is heavier or lighter if it's the counterfeit coin, but if the counterfeit coin is among the other twelve, we can also decide whether the counterfeit's heavier or lighter.

If (2) don't balance, the counterfeit is among I,J,K, and we know whether it's heavier or lighter. WLOG assume the counterfeit coin is heavier (thus I,J,K is heavier than A,B,C). Compare I and J. The heavier one is the counterfeit; if both balance, K is the counterfeit.

If (1) don't balance, WLOG A,B,C,D is heavier. Then the counterfeit is among A,B,C,D which is heavier, or among E,F,G,H which is lighter. Then: (3) Compare A,B,E and C,D,F.

If (3) balance, the counterfeit is either G or H which is lighter. Compare G and H to get the counterfeit coin; the lighter one is the counterfeit.

If (3) don't balance, WLOG A,B,E is heavier. Then the counterfeit coin is either A or B which is heavier, or F which is lighter. Compare A and B to get the counterfeit coin; the heavier one is the counterfeit, or if they balance, F is the counterfeit.

If we are given one coin known to be genuine and 13 coins where it is suspected that one of them is counterfeit (either heavier or lighter), we can state whether these 13 coins are genuine or not; if not, we can also decide which coin is counterfeit and whether it's heavier or lighter.

Reason that 13 coins can't be decided completely:
There are 26 cases come initially pairs (this coin is heavier, this coin is lighter). A balancing operation can separate these cases into three groups. Since we have three operations, we can separate these cases into 27 groups. However, since they initially come in pairs, and the first operation cannot break any of these pairs, the first operation can at best separate the cases into 8-8-10 group. The 10 cases group can't be separated into one cases by two operations; at best, this will separate them into eight groups of one case and one group of two cases. To ensure that we can determine the counterfeit coin, this group of two cases must be composed of the case where a certain coin is heavier and the case where the same coin is lighter, so even if this group is chosen, we can still determine the counterfeit coin.

By the above reasoning, prove that we can't determine a counterfeit coin among 14 suspected coins. Is it possible to determine a counterfeit coin among 14 suspected coins (it is ensured to be exactly one counterfeit coin among the fourteen) if given one extra coin known to be genuine?


Putnam
combinatorics unsolved
combinatorics

Given any distinct $mn+1$ postive integer $a_1,a_2,\cdots ,a_{mn+1}$ .Prove that one can find either

(i) $m+1$ integer such that neither two among them is divisible by each other

OR

(ii) $n+1$ integer , such that when they are arranged in increasing order , the $i^{th}$ integer can always divide the $(i+1)^{th}$ integer where $i=1,2,3...n$

i think it is a simple result from Dilworth's theorem

Yes!It's a old problem from Putnam MO 1966...Here is offical solution which I knew:
For each $a_i$ define $b_i$ be maximal length of the sequence which starting from $a_i$ such that each term is divisible by the next term.If there exist $j$ such that $b_j\geq n+1$ ,we are done
   If not according Pigeon Hole Principle we shall get at least $m+1$ the numbers $b_i$ have the same value,but now $m+1$ corresponding value of $a_i$ do not satisfy the first condition.Contradiction!!
   And What is Dilworth's theorem,Zorro?

http://www.mathlinks.ro/Forum/viewtopic.php?t=31425 could be of interest for you ;)


Are upper bounds restricted to positive numbers?

[color=cyan]A quick example:
the upper bound of the set of real numbers less than or equal to
-1 is -1.[/color]

Let's be a bit more precise than that. In Joel's example, -1 is [i]an[/i] upper bound (in fact the least upper bound), but it is not [i]the[/i] upper bound. 0 is an upper bound also since all the terms must be less than or equal to 0.

posting under a friend's account cuz im at school and dont remember my password... he does.

Basically, I understand ur responses, but I was asking this question in reference to polynomials, specifically page 79 bottom part of aops vol 2, where it talks about finding the uppper bound of a polynomial. I knew about this technique before, but it doesnt make sense here. For x^2 + 6x + 8.... which has roots -2 and -4, inputting -5 into synthetic brings out an anwer with the quotient and the remainder positive, thereby declaring that there are no roots greater than -5, obviously not true and contradicting what the book says. I am thinking here that in reference to this case, only positive numbers can be upper bounds. Is this true? Ill respond more in depth later, but bell about to ring.

Yeah, you're right - we should have specified here that c>0 (in the context of the book).  (Also, c<0 for the least upper bound.)  See if you can prove why these work.

Curiously, in the 10 years this book has been out, you're the first one to catch this.  Nice job.

(To the others on this thread - he's referring to the upper bound on the roots of a polynomial, not on the polynomial itself.)

Um... By least upper bound, are u meaning lower bound?

Because I believe that the lower bound has to be negative as well, considering that in x^2 - 5x + 6, when I applied synthetic with 4, it comes to an alternating sign, so wouldnt that mean there are no roots less than it? But of course, 2 and 3 are the roots.

And thanks for the props in seeing this... There are a few other mistakes i ahve seen in each volume, but I havent said them unless Im absolutely sure im right.

Yeah, I meant lower bound - fingers went faster than brain in typing.

It seems to me that if all the signs are positive including the remainder, then that's an upper bound for the roots. After that point, if you go up, the coefficients will just be increasing, so you can't get any more roots. If you get alternating signs, that's probably a lower bound, since then you could replace x with -x everywhere, and you'd have all the same signs. I hope I'm not doing anything stupid.

After that point, if you go up, the coefficients will just be increasing, so you can't get any more roots. If you get alternating signs, that's probably a lower bound, since then you could replace x with -x everywhere, and you'd have all the same signs.

Excuse me when I say, "huh?" Please explain this.

Oh sorry about not explaining it very well. Suppose you put in some really big value and perform polynomial division. Then you will get all positive coefficients for the resulting quotient and also a positive coefficient for the remainder. Then if you put in an even bigger number, the coefficients of the quotient and the remainder will just increase (try it!). You can get a lower bound too if there are alternating signs because that's the same as p(-x) rather than p(x), and that's equivalent to an upper bound of the roots of p(-x), or the lower bound of the roots of p(x). I hope that makes a bit more sense than my last post.


Prove the following inequality:

$ \dfrac{1}{2}*\dfrac{3}{4}*\dfrac{5}{6}*\dfrac{7}{8}*\cdots*\dfrac{99}{100} < \dfrac{1}{\sqrt {101}}$

[hide="hint"]
Consider the number:
$ \dfrac{2}{3}*\dfrac{4}{5}*\dfrac{6}{7}*\dfrac{8}{9}*\cdots*\dfrac{100}{101}$
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[hide="solution"]
funtwo is correct.
Here is a little more detailed solution.

Let $ x \equal{} \dfrac{1}{2}*\dfrac{3}{4}*\dfrac{5}{6}*\dfrac{7}{8}*\cdots*\dfrac{99}{100}$
and let $ y \equal{} \dfrac{2}{3}*\dfrac{4}{5}*\dfrac{6}{7}*\dfrac{8}{9}*\cdots*\dfrac{100}{101}$

We wish to prove that $ x<\dfrac{1}{\sqrt{101}}$. Observe that $ xy \equal{} \dfrac{1}{101}$.

But $ x<y$ because $ \dfrac{1}{2}<\dfrac{2}{3}, \dfrac{3}{4}<\dfrac{4}{5}...$ (in general $ \dfrac{n\minus{}1}{n}<\dfrac{n}{n\plus{}1}$ and $ (n\minus{}1)(n\plus{}1) \equal{} n^2\minus{}1<n^2$). So that $ x^2<xy \equal{} \dfrac{1}{101}$, from which $ x< \dfrac{1}{\sqrt{101}}$, Just what we wanted to prove.
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$ \dfrac{1}{2}*\dfrac{3}{4}*\dfrac{5}{6}*\dfrac{7}{8}*\cdots*\dfrac{99}{100}$ is less than $ \dfrac{2}{3}*\dfrac{4}{5}*\dfrac{6}{7}*\dfrac{8}{9}*\cdots*\dfrac{100}{101}$ (Consider each corresponding factor) and multiplying these two numbers yields 1/101
QED

This is slightly off topic, but i noticed that if we multiple the numerator by the denominator and the denominator by the denominator, we can express the above as

$ \frac {100!}{2^{100}(50!)^2} \le \frac {1}{\sqrt {101}}$

$ \sqrt {101}\binom{100}{50} \le 2^{100}$

$ \sqrt {101} \le \dfrac{\binom{100}{0} \plus{} \binom{100}{1} \plus{} \binom{100}{2} \plus{} ... \plus{} \binom{100}{100}}{\binom{100}{50}}$

Using funtwo's solution, we can generalises to

$ \sum_{k \equal{} 0}^{2n} \frac {\binom{2n}{k}}{\binom{2n}{n}} \ge \sqrt {2n \plus{} 1}$ :)

EDIT: thanks tonypr, i fixed it

Sorry if I'm wrong but do you mean:
$ \sum_{k\equal{}0}^{2n} \frac{\binom{2n}{k}}{\binom{2n}{n}} \ge \sqrt{2n\plus{}1}$


Jack, who always drives at the same constant velocity, leaves home daily in his car to pick up his wife Jill at her office. He reaches her office at 5pm, and immediately drives her home. One exceptional day, Jill left her office at 4pm and began walking towards home. When she met Jack, who was on the way to pick her up, he turned around and immediately drove her home, arriving 20 minutes earlier than usual. How long did Jill walk?

[hide]

i assumed the problem is asking for the amount of time, instead of amount of distance
i don't know if it's safe to assume it takes more than one hour for jack to arrive at jill's office, but i think it worked

so my variables were:
$d$ = distance between home and office, mi
$r_1$ = speed of car, mph
$r_2$ = speed of walking, mph (Jill)
$t$ = normal round trip time, hr

basic travel formula:$\frac{d}{r} = t$
time between car's departure and Jill's departure: $\frac{d}{r_1}-1$ hours
distance traveled until Jill departs: $d-r_1$ miles
distance traveled by the car in one hour: $r_1$ miles

for clarity, compare the following two equations
${(\text{time between departure times})+\frac{(d - r_1(\text{time between departure times}))}{r_1+r_2}}+\frac{r_1(\text{time required for them to meet, since Jill's departure})+d-r_1}{r_1} = 2t-\frac{1}{3}\\ \frac{d}{r_1}-1+\frac{d-r_1(\frac{d}{r_1}-1)}{r_1+r_2}+\frac{r_1(\frac{r_1}{r_1+r_2})+d-r_1}{r_1}=2t-\frac{1}{3}$

$\frac{d}{r_1}-1+\frac{r_1}{r_1+r_2}+\frac{r_1}{r_1+r_2}+\frac{d}{r_1}-1=2t-\frac{1}{3}\\ 2(\frac{d}{r_1}+\frac{r_1}{r_1+r_2}-1)=2t-\frac{1}{3}\\ \frac{d}{r_1}+\frac{r_1}{r_1+r_2}-1=t-\frac{1}{6}\\ \frac{d}{r_1}+\frac{r_1}{r_1+r_2}=t+\frac{5}{6}\\ \frac{r_1}{r_1+r_2}=\frac{5}{6}$

\[\begin{eqnarray*} \text{time required for them to meet, since Jill's departure} &=& \text{Jill's travel time}\\ &=& \frac{r_1}{r_1+r_2}\\ &=& \frac{5}{6} \end{eqnarray}\]
Jill travels for 50 minutes (?)

EDIT: checking, Jack travels toward Jill for 50 minutes and Jill travels toward Jack for 50 minutes, and when the trip back takes 50 minutes, so the round trip takes 100 minutes. solving the last equation, Jack travels 5 times as fast as Jill, so the normal trip would have taken 120 minutes. right?...

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