1. Inside square $ ABCD$ with each side of length 1, there is a point $ P$, such that $ \triangle BPC$ is equilateral. Find the area of $ \triangle BPD$

2. $ P$ is a point inside equilateral $ \triangle ABC$. If $ \angle APB\equal{}116^0, \angle BPC\equal{}132^0$, find each angle of a triangle which sides are $ PA,PB$ and $ PC$

sorry for my bad wording... :blush:

[hide="numero uno"]
$ \angle CBD \equal{} 45^o, \angle CBP \equal{} 60^o\Rightarrow DBP \equal{} 60 \minus{} 45 \equal{} 15^o$

$ \overline{BP} \equal{} 1, \overline{BD} \equal{} \sqrt {2}$

$ \Rightarrow [BPD] \equal{} \frac {\sqrt {2}}{2} \sin(15) \equal{} \frac {\sqrt {3} \minus{} 1}{4}$[/hide]

[hide="number two"]

The origional angles are $ 116,132,112$

Angles of $ \bigtriangleup PA\minus{}PB\minus{}PC \equal{} 180\minus{}116, 180\minus{}132, 180\minus{}112 \equal{} 64^o,48^o,68^o$[/hide]

[hide="Bashing the heck out of #1"]
Draw in $ BD\equal{}\sqrt{2}$. $ PC\equal{}DC\equal{}1$, so $ \triangle PCD$ is isosceles. Because $ \angle PCB\equal{}\angle CBP\equal{}\angle BPC\equal{}60^\circ$, we have $ \angle PCD\equal{}90^\circ\minus{}60^\circ\equal{}30^\circ$. Therefore, $ \angle DPC\equal{}\angle PDC\equal{}75^\circ$. Draw in $ DP$.

Let $ \theta\equal{}\angle BDP$. We have that $ \angle DPB\equal{}75^\circ\plus{}60^\circ\equal{}135^\circ$. Thus, $ \frac{1}{\sin\theta}\equal{}\frac{\sqrt{2}}{\sin{135^\circ}}$

$ \sin{135^\circ}\equal{}\sin{45^\circ}\equal{}\frac{1}{\sqrt{2}}$.

So we have $ \sin\theta\equal{}\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\equal{}\frac{1}{2}\implies\theta\equal{}30^\circ$.

Now, $ [\triangle BPD]\equal{}\frac{1\cdot\sqrt{2}\cdot\sin15^\circ}{2}$.

$ \sin15^\circ\equal{}\sin(45^\circ\minus{}30^\circ)\equal{}\sin45^\circ\cos30^\circ\minus{}\cos45^\circ\sin30^\circ\equal{}\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}\minus{}1}{2}$.

Finally, $ [\triangle BPD]\equal{}\sqrt{2}\cdot\frac{1}{\sqrt{2}}\cdot\left(\frac{\sqrt{3}\minus{}1}{2}\right)\cdot\frac{1}{2}\equal{}\boxed{\frac{\sqrt{3}\minus{}1}{4}}$.
[/hide]

thx

can no 1 be done without trigonometry??

[hide="Yes, it can"]
Instead of calculating $ S_{BPD}$ directly, let's calculate $ S_{BPCD}$, then subtract $ S_{BCD}$.
Due to SAS, we have that $ \triangle CPD \simeq \triangle BPA$.

We have $ S_{BPC} \equal{} \frac {\sqrt {3}}{4}$

The height of $ \triangle DPA$ with relation to DA equals $ 1 \minus{} \frac {\sqrt {3}}{2} \equal{} \frac {2 \minus{} \sqrt {3}}4$.
Therefore, $ S_{DPA} \equal{} \frac {2 \minus{} \sqrt {3}}{4}$.

The total area must equal 1, so that $ 2S_{DPC} \equal{} 1 \minus{} \frac {\sqrt {3}}{4} \minus{} \frac {2 \minus{} \sqrt {3}}4$, or
$ S_{DPC} \equal{} \frac {1}{4}$

Finally, $ S_{BPD} \equal{} S_{BPCD} \minus{} S_{BCD} \equal{} \frac {\sqrt {3}}{4} \plus{} \frac {1}{4} \minus{} \frac {1}{2} \equal{} \boxed{\frac {\sqrt {3} \minus{} 1}4}$
[/hide]


algebra
polynomial
function
factoring polynomials

How the hell does one factor this?

(t^3 - 6t^2 +9t - 4)

I entered this into my ti-89 titanium and it got (t-4)(t-1)^2.  Thing is, how did it get an answer?! I've been stuck trying to figure this out!

Help please~!

The three most useful tools for factoring polynomials above the second degree are the factor theorem, the remainder theorem, and synthetic division. Look those up on Google or ask your teacher to go over them (they're part of the grade 11 functions curriculum in Canada, I'm not sure about the USA).

Factoring is really all about practice. You'll see things eventually. Until then, those tools I mentioned above will help you do them the rigorous way.

Hm, I don't think I've tried synthetic division and such yet.  They're usually my last resort since they can take quite a while at times.  Thanks for the reminder.  I'll give it a whirl.

Use the constant term over the leading coefficient to get possible integer values for the zeros of the equation:
+-4+-2+-1
-------------
+-1
so you get the possible values of +-4, +-2, +-1
Then just use synthetic division and try to see which values work:
4|1 -6  9  -4
  |    4 -8  4
  |______________
    1 -2 1 | 0
so we have (x^2-2x+1)(x-4) = t^3-6t^2=9t-4
Notice how the factor is x-4, that is because x = 4 for the zero...
Now you can continue trying for the other number... if they don't work then try the next one; if none work you either have complex/imaginary zeros or they are not integers...

Okay, thank you.  I've figured it out once I was reminded to use the remainder/factor theorem and such.  But, thanks for your time.  I'm sorry to trouble you so.

Basically, the best way is to find all possible rational roots from the leading and trailing coefficient (Rational Roots Theorem), then use Descartes' Rule of signs to narrow it down and finally use synthetic division to find the zeroes.

This is a good site for this topic

http://oakroadsystems.com/math/polysol.htm ;)

You can use Viete's.

Basically, let the polynomial in form of $(x-r)(x-s)(x-t)$ where $r,s,t$ are the roots that are not necessarily distinct (maybe, or maybe not). Multiplying out and matching the expression with your numbers in the polynomial, you should get 3 systems of equations. Then, if the answer is trivial, you should be able to find your roots.

Again, there isn't really "tip" for "easy" factoring on cubic or beyond. I mean, you can use roots of unity for some of them but not in all cases.


15.  Find the area of the largest equilateral triangle that can be inscribed in a rectangle with sides 10 and 11.

Hint: [hide] The side can be greater than 11, do you see why? Try drawing the figure and rotating the inscribed triangle. [/hide]

Solution: [hide] One side of the equilateral triangle must be the hypotenuse of a right triangle with leg 11. We then obtain the system 121 + x^2 = (10-x)^2 + (11-y)^2 = y^2 + 100. With some quick algebra, we get x = 20 - 11 :rt3:, area is 110 - 11x/2 - (10-x)(11-y)/2 - 10y/2 = 221 :rt3: - 330.[/hide]

mesamune, where do those x's and y's come from?...

Haha, thanks JBL. I'm not really good with pictures.


Find all composite integers M such that the product of 2005xM has exactly eight divisors

[hide]
The different ways that a number has exactly 8 factors are as follows (in prime factorization):
$a^1*b^1*c^1$, $a^3*b^1$, and $a^7$.
The prime factorization of $2005$ is $5*401$.
This means we can rule out $a^7$ since we know that $2005*M$ is going to have more than 1 prime factor.
We can also rule out $a^1*b^1*c^1$ because  $M$ must be composite and in order to get this factorization it would have to be prime.
This means that the factorization of the final number must be $a^3*b^1$.  Knowing the factorization of 2005 we can see that $a^3$ must be $5^3$ or $401^3$ so M must be either $5^2$ or $401^2$.
Therefore the only composite values for $M$ are  $25$ and $160,801$.
[/hide]


Ok here are some more for you

in the product 9*HATBOX=4*BOXHAT
find the six digit numbers, HATBOX and BOXHAT

Finally, A two inch cube (2*2*2) of Silver weighs 3 pounds and is worth 198.36. How much is a 3 inch cube of silver worth?

I know there are alot sorry but it's helpful to me.

Second problem:

The 3 inch cube should weigh 27/8 times as much as the 2 inch cube.

27/8 * 3 = 10.125

So, it weigh 10.125 lbs.

The ratio of the 2inch cube's price to the ratio of the 3inch cube's price is 10.125/3 = 3.375

3.375*198.36= 669.465 dollars!

[quote="uanaka"]in the product 9*HATBOX=4*BOXHAT
find the six digit numbers, HATBOX and BOXHAT[/quote]

[hide="*bump*"] *coughitcoughiscoughstillcoughunansweredcoughcough* [/hide]

$ 9(10^5)H\plus{}9(10^4)A\plus{}9(10^3)T\plus{}9(10^2)B\plus{}9(10^1)O\plus{}9(10^4)X\equal{}4(10^5)B\plus{}4(10^4)O\plus{}4(10^3)X\plus{}4(10^2)H\plus{}4(10^1)A\plus{}4(10^0)T$

That is as far as I can go.

[quote="uanaka"]in the product 9*HATBOX=4*BOXHAT
find the six digit numbers, HATBOX and BOXHAT[/quote]
Let $ HAT \equal{} x$ and $ BOX \equal{} y$.

Then, we have $ 9(1000x \plus{} y) \equal{} 4(1000y \plus{} x)\implies 8996x \equal{} 3991y\implies 692x \equal{} 307y$

Since $ \gcd(692, 307)\equal{}1$, $ x \equal{} 307$ and $ y \equal{} 692$.

Answer: $ HATBOX \equal{} 307692$, $ BOXHAT \equal{} 692307$


John has 10000 chances to get out of a maze. When a choice of direction can be made, John randomly chooses which way he will go. If he comes to a dead end, he looses a chance. Unfortuneately for John, he isn't too bright. Every time he starts over, he doesn't remember the correct path, and again randomly chooses his way. What is the probability that John will get out of the maze? Express your answer as a percentage to the nearest hundredth.

[img]http://img114.imageshack.us/img114/3443/mazexp0.png[/img]

$0.0007629$ %

Somebody please give him that map.

[hide]I counted $17$ places where there was a choice of 2 paths. The probability that he choose the correct path all 17 times is $\frac{1}{2^{17}}$.

The probability that he fails on any given try is $\frac{2^{17}-1}{2^{17}}$. The probability the he fails all 10000 times is $(\frac{2^{17}-1}{2^{17}})^{10000}$. The probability of him succeeding at least once is one minus that, which is approximately $7.35\%$.[/hide]

[quote="i_like_pie"]I see how you did it, randomdragoon, and I understand why that works, but I got a different answer.

[hide="My Solution"]On any one of his chances, he has a $\frac1{2^{17}}$ chance of getting out of the maze. That times $10000$ would be his chance of getting out on one of his tries.

$\approx7.63\%$

What's wrong with my solution? :maybe:[/hide][/quote]

What if there was $2^{17}$ chances? By your method, the probability would be $\frac{2^{17}}{2^{17}}=1$ which can't be right. Your method would work if he could remember the paths and never took the same path twice. However, you aren't accounting for the fact that he can repeat paths.


[quote="itazzirun"]do there exist functions f:R-->R and g:R-->R, g being periodic, such that x^3=f([x])+g(x) , for all real x?[/quote]
Again the same situation as in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114646[/url] and [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114637[/url]. This is question 1 from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114222[/url]. You just copied some other's problem without writing the source. Why?

for all real $\neq 0$
$h(x)=g(x+a)-g(x)=3ax(x+a)+a^{3}+f([x])-f([x+a])$
$h(0)=a^{3}+f(0)-f([a])$, et $h(-a)=a^{3}+f(0)-f([-a])$

if a is the period of g, we must have $h(x)=h(y)=0$ for all real $x,y$.
but we have for $p\in N$ , $h(p)\neq h(p+\frac{a+[a]+1}{2})$
(because $f([p])-f([p+a])=f([p+\frac{a+[a]+1}{2}])-f([p+\frac{a+[a]+1}{2}+a])$ and $m(p)\neq m(p+\frac{a+[a]+1}{2})$,$m(x)=3ax(x+a)+a^{3}=(x+a)^{3}-x^{3}$)

[size=142] there arn't functions f:R-->R and g:R-->R, g being periodic, such that x^3=f([x])+g(x) , for all real x[/size]


probability
counting
distinguishability
modular arithmetic
function
limit
vector

[quote="4everwise"]A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$[/quote]
[hide="solution with ComplexZetas method for these problems"]683[/hide]

Please post your solution, bpms. (I don't know what ``ComplexZeta's method'' is.)

[hide="Hint"]
Recursion
[/hide]

It reminds me about [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112379[/url]. In particular, the following theorem is true:
[i]We start from one vertex of a regular $n$-gon. In one move we can go to any of the two adjacent vertices with probability $\frac{1}{2}$. Then the probability that after $k$ moves we return to the starting vertex is
\[\frac{1}{2^{k}}\cdot\sum_{0\leqslant a\leqslant k\atop 2a\equiv k\pmod{n}}\binom{k}{a}.\]
[/i]
Te probability we want corresponds with $n=3$, $k=10$, so it equals
\[\frac{1}{2^{10}}\cdot \sum_{0\leqslant a\leqslant 10\atop 2a\equiv 10\pmod{3}}\binom{10}{a}= \frac{\binom{10}{2}+\binom{10}{5}+\binom{10}{8}}{2^{10}}= \frac{2\cdot 45+252}{2^{10}}= \frac{171}{512}.\]
Therefore the final result is $171+512=683$.

Another way to think of The Zuton Force's recursion is if we define $\ P_{n}$ the same way namely the probability that after $\ n$ moves the bug is at its starting point.  Therefore the probability the bug is at its starting point after $\ n$ moves is $\ 1/2$ times the probability that it is not on its starting point after $\ n-1$ moves.  So the recursion is $\ P_{n}=1/2(1-P_{n-1})$ repeating this recursion up to $\ n=10$ we get the answer.

$P_{n}=\frac{1}{2}(1-P_{n-1})$, you mean? And also, using this method you can find any other vertex after n moves easily cause it's symmetric..

[quote="The Zuton Force"][hide="Easy way to do"]
Let the starting point be point 1, and label the other two points 2 and 3.

The bug must move each turn, so
$P_{n+1}(1)=\frac{P_{n}(2)+P_{n}(3)}{2}$

Points 2 and 3 are indistinguishable, so:
$P_{k}(2)=P_{k}(3)\implies P_{n+1}(1)=P_{n}(2)$

Probabilities must add up to 1, so:
$P_{n+1}(2)=P_{n+1}(3)=\frac{1}{2}(1-P_{n+1}(1))$
$P_{n}(2)=\frac{(1-P_{n}(1))}{2}$
$P_{n+1}(1)=\frac{(1-P_{n}(1))}{2}$
We know $P_{0}(1)=1$

This is a very simple recursion, and we get $P_{10}(1)=\frac{171}{512}$
$m+n=171+512=\boxed{683}$
[/hide][/quote]

Could you explain how you get $P_{n+1}(2)=P_{n+1}(3)=\frac{1}{2}(1-P_{n+1}(1))$?

Thanks

woahhhhh

I think altheman found the smoothest solution

I mean the recursion solution lends itself to an easy generalization too, but that is a very clever solution.

[b]@mathgenius[/b]

Well, $P(2)$ and $P(3)$ are indistinguishable points. You could change the labels and it wouldnt change anything.

You also know that $P(1)+P(2)+P(3)=1$ because the probabilities must add up to 1.

So you get $P(2)=\frac{1}{2}(P(2)+P(2))=\frac{1}{2}(P(2)+P(3))=\frac{1}{2}(1-P(1))$

[quote="Altheman"][hide="solution"]lets draw a one to one correspondence with a binary string, consisting of A's and B's, let A be a move cw, and B be a move ccw, we must have that 3|(# of A's -#of B's)

now lets use the generating function $f(x)=(x+\frac{1}{x})^{10}$ then we want the sum of the coefficients of $x^{3k}$ for $k\in\mathbb{Z}$,

this is seen so many times, $\frac{1}{3}(f(1)+f(w)+f(w^{2}))$ where $w^{2}+w+1=0$, we get $\frac{1}{3}(2^{10}+2(w+w^{2})^{10}=342$
the total probability being $\frac{342}{2^{10}}=\frac{171}{512}\implies m+n=\boxed{683}$[/hide]

this leads to an easy generalization....[/quote]That's just brilliant. Nice, nice.  :)

Actually, I believe the answer to that is $6\underline{4}/128=1/2$.

On any given turn, there are only two possible locations for the bug - the pair of opposite corners containing the starting vertex after an even number of moves, and the other pair of corners if its after an odd number of moves. These are indistinguishable, so since 8 is an even number, the probability is 1/2.

EDIT: Did you mean for the bug to be able to use diagonals?
Then we have the recursion:
$P_{n+1}(\text{starting vertex})=\frac{1-P_{n}(\text{starting vertex})}{3}$
which gives $P(8)=\frac{547}{2187}$

Note that this equals $\frac{Ceiling[3^{7}/4]}{3^{7}}$, as this recursion converges on $1/4$

I don't think it's starting location and the point opposite it's starting is entirely symmetric...just off the top of my head 65/128 should probably be right

[quote="sapphyre571"]just so I know I have this right, if the bug were to start on a square, and make 8 moves as defined in the previous problem, the probability it returns to its starting point is 65/128?[/quote]
It is
\[\frac{1}{2^{8}}\cdot \sum_{0\leqslant a\leqslant 8\atop 2a\equiv 8\pmod{4}}\binom{k}{a}= \frac{1}{2^{8}}\cdot\sum_{0\leqslant a\leqslant k\atop 2\mid a}\binom{8}{a}= \frac{1}{2^{8}}\cdot\left(\frac{1}{2}\cdot 2^{8}\right)= \frac{1}{2}.\]
[quote="The Zuton Force"]Note that this equals $\frac{Ceiling[3^{7}/4]}{3^{7}}$, as this recursion converges on $1/4$[/quote]
The matter of convergence of the probability is very interesting. The limit of the probability of being in $j$-th vertex of $n$-gon after $k$ moves while $k\to\infty$ can be examined using a reasoning similar to $\mathbf{Altheman}$'s.
I showed in the topic mentioned in my post above that this probability equals $\frac{1}{2^{k}}\cdot\sum_{0\leqslant a\leqslant k\atop 2a\equiv k-j\pmod{n}}\binom{k}{a}$. Let $\omega: = \mathrm{e}^{\frac{2\pi i}{n}}$. Then using the fact $\frac{1}{n}\sum_{0\leqslant s\leqslant n-1}\omega^{sj}= \begin{cases}1\text{ for }n\mid j\\ 0\text{ for }n\nmid j\end{cases}$ we obtain
\begin{eqnarray*}\frac{1}{2^{k}}\cdot\sum_{0\leqslant a\leqslant k\atop 2a\equiv k-j\pmod{n}}\binom{k}{a}&=& \frac{1}{n\cdot 2^{k}}\cdot\sum_{0\leqslant a\leqslant k}\left(\binom{k}{a}\cdot\sum_{0\leqslant s\leqslant n-1}\omega^{s(2a-k+j)}\right)\\ &=& \frac{1}{n\cdot 2^{k}}\cdot\sum_{0\leqslant a\leqslant k\atop 0\leqslant s\leqslant n-1}\binom{k}{a}\cdot \omega^{s(2a-k+j)}\\ &=& \frac{1}{n\cdot 2^{k}}\cdot\sum_{0\leqslant s\leqslant n-1}\left(\omega^{s(j-k)}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}\cdot \omega^{2as}\right)\\ &=& \frac{1}{n}\cdot\sum_{0\leqslant s\leqslant n-1}\omega^{s(j-k)}\cdot \left(\frac{1+\omega^{2s}}{2}\right)^{k}\\ &=& \frac{1}{n}\cdot\sum_{0\leqslant s\leqslant n-1}\omega^{sj}\cdot \left(\frac{\omega^{s}+\omega^{-s}}{2}\right)^{k}.\end{eqnarray*}
Let us notice that if $\left|\frac{\omega^{-s}+\omega^{s}}{2}\right|<1$, then $\lim_{k\to\infty}\omega^{sj}\cdot \left(\frac{\omega^{s}+\omega^{-s}}{2}\right)^{k}=0$. But $\left|\frac{\omega^{-s}+\omega^{s}}{2}\right|\leqslant \frac{\left|\omega^{-s}\right|+\left|\omega^{s}\right|}{2}= \frac{1+1}{2}=1$ with equality iff the points $\omega^{s},\omega^{-s},0$ are collinear (of course the vectors $\omega^{s}$ and $\omega^{-s}$ must have the same direction). We have to consider two cases:
[b]1.[/b] $2\nmid n$
It is obvious that the only equality case of the triangle inequality occurs when $s=0$. Thus
\[\lim_{k\to\infty}\frac{1}{n}\sum_{0\leqslant s\leqslant n-1}\omega^{sj}\cdot \left(\frac{\omega^{s}+\omega^{-s}}{2}\right)^{k}= \frac{1}{n}\cdot\omega^{0}\cdot \left(\frac{\omega^{0}+\omega^{0}}{2}\right)^{k}= \frac{1}{n}.\]
[b]2.[/b] $2\mid n$
For even $n$ we have also a second equality case of the triangle inequality, i.e. $s=\frac{n}{2}$. Then
\[\lim_{k\to\infty}\frac{1}{n}\sum_{0\leqslant s\leqslant n-1}\omega^{sj}\cdot \left(\frac{\omega^{s}+\omega^{-s}}{2}\right)^{k}= \frac{1}{n}\left(1+\omega^{j\cdot\frac{n}{2}}\left(\frac{\omega^{\frac{n}{2}}+\omega^{-\frac{n}{2}}}{2}\right)^{k}\right)= \frac{1}{n}\cdot \left(1+(-1)^{j+k}\right).\]
We can consider a more general version of the problem: the vertices are numerated modulo $n$ and probability of moving from the $j$-th vertex to the $j+i$-th vertex is $p_{i}$ ($p_{0},\ldots,p_{n-1}\geqslant 0$, $\sum_{0\leqslant i\leqslant n-1}p_{i}=1$). Then the probability of being in the $j$-th vertex after $k$ moves is
\[\sum_{0\leqslant a_{1},\ldots,a_{k}\leqslant n-1\atop a_{1}+\ldots+a_{k}\equiv j\pmod{n}}p_{a_{1}}\cdot\ldots\cdot p_{a_{k}}.\]
By a method similar to the one used for the previous problem, it can be easily proved that if all $p_{i}$ are positive, this expression approaches to $\frac{1}{n}$ while $k$ approaches to infinity. The same result holds if there exists an $i$ for which $p_{i},p_{i+1}>0$, the proof of this statement is easy with cyclic matrices introduced in the topic [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112379[/url].

[quote="sapphyre571"]Sorry - I made a computation error (just like always)

let's try this again
pentagon, 9 moves

the probability that the bug ends at start is
$\frac{{9\choose2}+{9\choose7}}{2^{9}}=\frac{9}{64}$
correct?[/quote]
Correct. :)

okay, good. I'm learning.
what if, instead of a polygon, we had a cube or octahedron?

do any of you happen to know a formula (like the summation - it's easiest for me to remember) that work for polyhedrons as well?

I'm trying to do some of my own work on them by simplifying them to 2-D while keeping the vertex-travel probability equal, like an octahedron can be resembled by points A, B, and C, with A as the starting point, a path connecting A and B, a path connecting B and C, and two paths leading from B back to itself.

or something...

I don't think that combinatorical formula will work for an octahedron, since it is a more connected graph than the hexagon. However, the octahedron is easily done by recursion:

Label the starting vertex 1, the adjacent four 2, 3, 4, and 5, and the opposite vertex 6.

On the first move, there is a 0 probability the bug is on 1 or 6, and a 1/4 probability each that the bug is on 2, 3, 4, or 5. We notice now that from here on, verteces 2, 3, 4, and 5 are symmetric, and verteces 1 and 6 are symmetric.

Let $b_{n}$ be the probability that the bug is on vertex 2.
Note that $b_{n}=P_{n}(2)\implies b_{n}=P_{n}(2)=P_{n}(3)=P_{n}(4)=P_{n}(5)$ by symmetry.

Let $a_{n}$ be the probability that the bug is on vertex 1.
Note that $a_{n}=P_{n}(1)\implies b_{n}=P_{n}(1)=P_{n}(6)$.

$4b_{n}+2a_{n}=1$ since probabilities add up to 1, and $a_{n+1}=b_{n}$ since $P_{n+1}(1)=\frac14 (P_{n}(2)+P_{n}(3)+P_{n}(4)+P_{n}(5))$, giving:
$4a_{n+1}+2a_{n}=1 \implies \boxed{a_{n+1}=\frac{1-2a_{n}}{4}}$
with initial condition $a_{1}=0$ (note that we cannot use $a_{0}$ because the system is not yet symmetric).

For example, the probability that the bug returns to its starting vertex after $10$ moves is $a_{10}=\frac{171}{1024}$
Interestingly, this is half the probability it returns to its starting vertex on the triangle in the original question. This is in fact the case for all positive values of $n$ (can you see why from the recursion?)

The cube on the other hand is a very interesting situation.

We have 4 symmetric sets of points - starting vertex, the three verteces immediately adjacent to it, the vertex opposite the starting vertex, and the 3 verteces immediately adjacent to those. Call the probabilities $a$ $b$ $c$ and $d$ in that order. Note that at any given turn, either $a=d=0$ and $3b+c=1$ or $b=c=0$ and $3d+a=1$. The first is true after an odd number of moves and the second is true after an even number of moves. This is similar to the behaviour of probabilities in a polygon with an even number of verteces.

[quote="paladin8"][quote="Fraxia"][quote="Altheman"]
this is seen so many times, $\frac{1}{3}(f(1)+f(w)+f(w^{2}))$ where $w^{2}+w+1=0$, we get $\frac{1}{3}(2^{10}+2(w+w^{2})^{10}=342$
[/quote]

Could you explain how that works and is the sum of the coefficients of the $x^{3k}$ terms? :maybe:[/quote]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=binom&t=119683]here[/url].[/quote]

Oh wow, that's pretty neat  :)

[quote="The Zuton Force"]I don't think that combinatorical formula will work for an octahedron, since it is a more connected graph than the hexagon. [/quote]
But cyclic matrices also work easily.
Let us numerate the vertices of the graph of octahedron as shown in the attachment. Then the matrix of probabilities of moves of the bug is cyclic:
\[M= \frac{1}{4}\cdot \begin{bmatrix}0&1&1&0&1&1\\ 1&0&1&1&0&1\\ 1&1&0&1&1&0\\ 0&1&1&0&1&1\\ 1&0&1&1&0&1\\ 1&1&0&1&1&0\end{bmatrix}. \]
The probability of returning the bug to the $j$-th vertex after $k$ moves is $\left[M^{k}\right]_{0,j}$. Let us put $A=\frac{1}{2}\left(I+c_{0,3}(I)\right)$. Then it is easy to see that $M=\frac{1}{2}\left(c_{0,1}(A)+c_{0,-1}(A)\right)$ and since $A^{2}= \frac{1}{4}\left(I+c_{0,3}(I)+I+c_{0,3}\right)= A$ we have $A^{k}=A$. Furthermore, $\left[A\right]_{i,j}= \begin{cases}\frac{1}{2}\text{ for }i\equiv j\pmod{3}\\ 0\text{ for }i\not\equiv j\pmod{3}\end{cases}$. Thus
\[\left[M^{k}\right]_{0,j}= \left[\left(\frac{1}{2}\left(c_{0,1}(A)+c_{0,-1}(A) \right)\right)^{k}\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[\left(c_{0,-1}\left(A+c_{0,2}(A)\right)\right)^{k}\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[c_{0,-k}\left(\left(A+c_{0,2}(A)\right)^{k}\right)\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[c_{0,-k}\left(\sum_{a_{1},\ldots,a_{k}\in\{0,1\}}c_{0,2a_{1}}(A)\cdot\ldots\cdot c_{0,2a_{k}}(A)\right)\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[c_{0,-k}\left(\sum_{a_{1},\ldots,a_{k}\in\{0,1\}}c_{0,2\sum_{1\leqslant i\leqslant k}a_{i}}\left(A^{k}\right)\right)\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[c_{0,-k}\left(\sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{0,2a}(A)\right)\right]_{0,j}= \frac{1}{2^{k}}\cdot \left[\sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{k,2a}(A)\right]_{0,j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}\left[c_{k,2a}(A)\right]_{0,j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}\left[A\right]_{k,2a+j}= \frac{1}{2^{k+1}}\cdot \sum_{0\leqslant a\leqslant k\atop k\equiv 2a+j\pmod{3}}\binom{k}{a}= \frac{1}{2^{k+1}}\cdot \sum_{0\leqslant a\leqslant k\atop 2a\equiv k-j\pmod{3}}\binom{k}{a}. \]
It of course proves
[quote="The Zuton Force"]Interestingly, this is half the probability it returns to its starting vertex on the triangle in the original question. This is in fact the case for all positive values of $n$ (can you see why from the recursion?)[/quote]
for all values of $k$.

[quote="The Zuton Force"]The cube on the other hand is a very interesting situation. [/quote]
In this case we can use polidimensional cyclic matrices. We numerate the vertices of the cube using elements of $\mathbb{Z}_{2}^{3}$, i.e. triples $\left(i_{1},i_{2},i_{3}\right)$ where $i_{0},i_{1},i_{2}\in\{0,1\}$, such that these numbers are the same as cartesian coordinates of the vertices. Form the vertex $(a,b,c)$ we can go to any of $(1-a,b,c)$, $(a,1-b,c)$ and $(a,b,1-c)$. It is easy to see that differences of vertices connected in the graph are $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, so the fact that we can go from $i$-th vertex to the $j$-th is equivalent to the fact that we can go from the $0$-th to the $i-j$-th. Thus the matrix of probabilities of moves of the bug is cyclic. Its zeroth column is a $3$-vector of dimension $(2,2,2)$ whose only nonzero elements (all equal to $\frac{1}{3}$) are the $(1,0,0)$-th, the $(0,1,0)$-th and the $(0,0,1)$-th. Therefore this matrix equals $M=\frac{1}{3}\left(c_{(0,0,0),(1,0,0)}(I)+c_{(0,0,0),(0,1,0)}(I)+c_{(0,0,0),(0,0,1)}(I)\right)$. We want to find $\left[M^{k}\right]_{(0,0,0),\left(i_{0},i_{1},i_{2}\right)}$. Let us denote by $A_{j}$ the number of terms equal to $i$ in the sequence $\left(a_{1},\ldots,a_{k}\right)$. Then
\[\left[M^{k}\right]_{(0,0,0),\left(i_{0},i_{1},i_{2}\right)}= \frac{1}{3^{k}}\cdot \left[\left(c_{(0,0,0),(1,0,0)}(I)+c_{(0,0,0),(0,1,0)}(I)+c_{(0,0,0),(0,0,1)}(I)\right)^{k}\right]_{(0,0,0),(i_{0},i_{1},i_{2})}=\]
\[\frac{1}{3^{k}}\cdot \left[\sum_{a_{1},\ldots,a_{k}\in\{0,1,2\}}c_{(0,0,0),(A_{0},A_{1},A_{2})}(I)\right]_{(0,0,0),(i_{0},i_{1},i_{2})}= \frac{1}{3^{k}}\cdot \left[\sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}c_{(0,0,0),(A_{0},A_{1},A_{2})}(I)\right]_{(0,0,0),(i_{0},i_{1},i_{2})}= \]
\[\frac{1}{3^{k}}\cdot \sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}\left[c_{(0,0,0),(A_{0},A_{1},A_{2})}(I)\right]_{(0,0,0),(i_{0},i_{1},i_{2})}=\]
\[\frac{1}{3^{k}}\cdot \sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}\left[I\right]_{(0,0,0),(i_{0}+A_{0},i_{1}+A_{1},i_{2}+A_{2})}=\]
\[\frac{1}{3^{k}}\cdot \sum_{\substack{0\leqslant A_{0},A_{1},A_{2}\leqslant k\\ A_{0}+A_{1}+A_{2}=k\\ A_{0}\equiv i_{0},A_{1}\equiv i_{1},A_{2}\equiv i_{2}\pmod{2}}}\binom{k}{A_{0},A_{1},A_{2}}= \frac{1}{3^{k}}\cdot \sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}\prod_{0\leqslant j\leqslant 2}\frac{1+(-1)^{A_{j}+i_{j}}}{2}=\]
\[\frac{1}{3^{k}}\cdot \sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}\cdot \frac{1}{8}\sum_{j_{0},j_{1},j_{2}\in\{0,1\}}(-1)^{j_{0}(A_{0}+i_{0})+j_{1}(A_{1}+i_{1})+j_{2}(A_{2}+i_{2})}=\]
\[\frac{1}{3^{k}}\cdot \frac{1}{8}\cdot \sum_{j_{0},j_{1},j_{2}\in\{0,1\}}(-1)^{j_{0}i_{0}+j_{1}i_{1}+j_{2}i_{2}}\cdot \sum_{0\leqslant A_{0},A_{1},A_{2}\leqslant k\atop A_{0}+A_{1}+A_{2}=k}\binom{k}{A_{0},A_{1},A_{2}}(-1)^{j_{0}A_{0}+j_{1}A_{1}+j_{2}A_{2}}=\]
\[\frac{1}{3^{k}}\cdot \frac{1}{8}\cdot \sum_{j_{0},j_{1},j_{2}\in\{0,1\}}(-1)^{j_{0}i_{0}+j_{1}i_{1}+j_{2}i_{2}}\cdot \left((-1)^{j_{0}}+(-1)^{j_{1}}+(-1)^{j_{2}}\right)^{k}=\]
\[\frac{1}{8}\cdot \sum_{j_{0},j_{1},j_{2}\in\{0,1\}}(-1)^{j_{0}i_{0}+j_{1}i_{1}+j_{2}i_{2}}\cdot \left(\frac{(-1)^{j_{0}}+(-1)^{j_{1}}+(-1)^{j_{2}}}{3}\right)^{k}.\]
In particular, the probability of returning to the starting vertex is
\[\frac{1}{8}\cdot \sum_{j_{0},j_{1},j_{2}\in\{0,1\}}\left(\frac{(-1)^{j_{0}}+(-1)^{j_{1}}+(-1)^{j_{2}}}{3}\right)^{k}.\]

For the original problem, let the triangle be denoted as triangle $ABC.$ Denote $A_{n}$ the number of ways to get back to $A$ from $A$ after $n$ moves. Also, let $B_{n}$ be the number of ways to get to $A$ from $B$ in $n$ moves. Denote $C_{n}$ the same way. We have the recursions:

$A_{n}= B_{n-1}+C_{n-1}$
$B_{n}= A_{n-1}+C_{n-1}$
$C_{n}= A_{n-1}+B_{n-1}$

We have that $A_{1}= 0, B_{1}= 1, C_{1}= 1,$ so completing the recursion until the 10th case yields $A_{1}0 = 342.$ We divide by $2^{10}$ and add the resulting numerator and denominator after simplifying and get our answer to be $683.$

i want to try to get recursion for $ n\equal{}4$ (square). can somebody help me with the reasoning:
Label vertices $ 1,2,3,4$, and wlog suppose bug starts at $ 1$. To see probability that he returns to $ 1$ on $ n\plus{}1$st move, we know:
$ P_{n\plus{}1}(1) \equal{} \frac{P_{n}(2) \plus{} P_{n}(4)}{2}$
because vertices $ 2$ and $ 4$ are indistinguishable (just rotate figure)
$ P_{n\plus{}1}(1) \equal{} P_{n}(2)$

Now, how can I compute $ P_{n}(2)$?

also, is there a way to extend this idea to polyhedra?

[quote="shake9991"]i want to try to get recursion for $ n \equal{} 4$ (square)
...
Now, how can I compute $ P_{n}(2)$?[/quote]

Do the same thing over again, except use point 2 as your frame of reference:
$ P_{n \plus{} 1}(2) \equal{} \frac {P_{n}(1) \plus{} P_{n}(4)}{3} \equal{} P_n (1) \implies P_n (2) \equal{} P_{n\minus{}1} (1)$

Combining the two, you get:
$ P_{n\plus{}1} (1) \equal{} P_{n\minus{}1} (1) \implies P_{n} (1) \equal{} P_{n\minus{}2} (1) \equal{} P_{n\minus{}2k} (2)$

Using a similar argument, you can get $ P_{n} (2) \equal{} P_{n\minus{}2} (2) \equal{} P_{n\minus{}2k} (2)$

Using the above recursion, you get that for even $ n$, $ P_n (1) \equal{} P_2 (1) \equal{} \frac{1}{2}$ and $ P_n (2) \equal{} P_2 (2) \equal{} 0$, and these are also equal to $ P_n (3)$ and $ P_n (4)$, respectively.

For odd $ n$, we have $ P_n (1) \equal{} P_1 (1) \equal{} 0$ and $ P_n (2) \equal{} P_1 (2) \equal{} \frac{1}{2}$; again, these are equal to $ P_n (3)$ and $ P_n (4)$, respectively.

This makes sense intuitively; there is no way to be on points 1 or 3 after an odd number of moves, and no way to be on points 2 or 4 after an even number of moves, since each move, you either move from odd to even or from even to odd.

[quote="shake9991"]also, is there a way to extend this idea to polyhedra?[/quote]

The idea of using symmetric points in a recursion would have to be applied on a case-by-case basis. It is certainly always possible, but not necessarily always the best idea. While recursion may be more practical by hand for a very simple polyhedron or for a small value of n, if you were to write a general computer program or work generally, the matrix approach is superior.

Here is my solution from the Aops Wiki:
After n moves, there are $ 2^n$ possible paths for the ant. But the key is to realize that the number of paths that get back the the start after n moves is the same as the number of paths the do NOT get the ant to the start after n-1 moves. So after 1 move, there are 0 ways the get to the start. After 2 moves, there are $ 2^1\minus{}0$ ways to get to the start. After 3 moves, there are $ 2^2 \minus{} (2^1 \minus{} 0)$ ways to get to the start. Thus after 10 moves, there are $ 2^9 \minus{} (2^8 \minus{} (2^7 \minus{} (2^6 \minus{} (2^5 \minus{} (2^4 \minus{} (2^3 \minus{} (2^2 \minus{} 2))))))) \equal{} 342$ ways to get to the start, so the probability is $ 342/1024 \equal{} 171/512$ and the answer is 683.

Here are two problems I recently posted that involve a bug randomly moving on a 3D shape: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216473]The thread[/url]

[quote="Rzeszut"][i]We start from one vertex of a regular $ n$-gon. In one move we can go to any of the two adjacent vertices with probability $ \frac {1}{2}$. Then the probability that after $ k$ moves we return to the starting vertex is
\[ \frac {1}{2^{k}}\cdot\sum_{0\leqslant a\leqslant k\atop 2a\equiv k\pmod{n}}\binom{k}{a}.
\]
[/i][/quote]

Could someone please explain the above formula to me in simpler terms?  I'm finding the sigma and mod notation hard to understand.  Thanks in advance!

Suppose we make $ k$ moves, of which $ a$ are clockwise and $ k-a$ are counterclockwise. Then, in effect, we moved $ a-(k-a) = 2a-k$ moves clockwise. In order to be back on the starting vertex, these $ 2a-k$ moves must be a multiple of $ n$, the number of vertices. That is, if we in effect moved, say, $ 3n$ vertices clockwise, we clearly will end up on the starting vertex, since we made exactly 3 complete laps.

In other words, $ a$ must be such that $ 2a-k \equiv 0 \pmod{n} \iff \boxed{2a \equiv k \pmod{n}}$.

The other condition, that $ 0 \leq a \leq k$, follows from the fact that the number of clockwise moves cannot exceed the number of total moves.

The sum says that, for each possible $ a$ satisfying the two conditions listed above, (these are the conditions under the Sigma), include the probability that the bug will make $ a$ clockwise and $ k-a$ counterclockwise moves.

What is this value? There are $ \binom{k}{a}$ ways of choosing $ a$ of the $ k$ moves on which to move clockwise, and each such path has probability $ \frac{1}{2^k}$ (it was a 1/2 probability for each of the k moves).

Therefore, the total probability is $ \cdot\sum_{0\leq a \leq k \atop 2a\equiv k\pmod{n}} \frac {1}{2^{k}} \cdot \binom{k}{a}$, which is equivalent to what [b]Rzeszut[/b] wrote.


Given $2n-1$ integers, prove that one can choose exactly $n$ of them whose sum is divisible by $n$.

This is Erdös-Ziv-Ginzburg theorem, it has been discussed many times on this forum.
[b]ZetaX[/b] provided a proof of this result.

look here
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Given+2n-1+integers&t=31191]Integer sums[/url]
 :)


function calculus integration derivative LaTeX

1.	“f” is a differentiable function and f’’(x)>0.Prove that- (integration sign-with upper limit 2π to lower limit 0)f(x)cosxdx ≥ 0.
2.	“f” is a differentiable function, satisfying the condition xf(x) =ln x .Now prove that f (n) (1) = (-1) n+1(1+1/2+1/3+1/4+…….+1/n) n! where f (k) x denotes the k-th derivative
pls. solve them up :)  :) .extremely sorry for not having LATEX and hence the integratrion sign calamity :blush:  :blush:

I have got this paper from somewhere.
For the first integral use integration by parts two times.
I have forgot how I have done the second one,but probably you can find out the pattern in derivative and you can prove it by induction.

Oh!God,forgot about induction.very very thank u :blush: .But would u kindly slove the first one rigorously.I've stuck after two steps due to f(x). :D

@math12061992
Just do integration by parts to the left term twice.then it is obvious.It is quite hard for me to write the full solution here in latex.
@nuclear_alchemist
There is no fixed number of seats.But generally 30-55 students get through both in bstat and bmath.


solve in N the equation
x^5+y^4+z^3=(x+y+z)^2

hello! sorry but this isn't the right emplacment for your diophantinne equation, you must post it on number theory forum,

I think it 's so easy.We can denote max(x,y,z)= t and...

:roll: ok .can you post to me the solution please.because i want to see the full solution ......

Let $ t \equal{} max(x,y,z)$, then if $ t \equal{} x$
$ x^5 \plus{} y^4 \plus{} z^3\ge t^5 \plus{} 2$, and $ (x \plus{} y \plus{} z)^2\le 9t^2$, but $ t^5 \plus{} 2 > 9t^2$ if $ t > 2$.
$ x \equal{} 2, y \equal{} 1 \equal{} z$ is not solution.
If $ t \equal{} y$, then $ t^4 \plus{} 2 > 9t^2$ for $ t > 2$. $ y \equal{} 2,x,z\le 2$ give not solution.
Therefore $ t \equal{} z$ and from $ x^5 \plus{} y^4 \plus{} z^3\ge t^3 \plus{} 2 > 9t^2\ge (x \plus{} y \plus{} z)^2$ for $ t\ge 9$ we get $ t\le 8$.
$ x \equal{} y \equal{} 1$ $ z^3 \plus{} 2 \equal{} (z \plus{} 2)^2$ - no solution.
$ x \equal{} 2,y \equal{} 1$ $ z^3 \plus{} 33 \equal{} (z \plus{} 3)^2$ - no solution.
$ x \equal{} 1,y \equal{} 2$ $ z^3 \plus{} 17 \equal{} (z \plus{} 3)^2 \to (x,y,z) \equal{} (1,2,2).$


See if you can solve it!

$ \frac{3}{2} \cdot \frac{2}{1} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \dots \cdot \frac{k}{k\plus{}1} \geq \frac{1}{8}$

Find the largest value of $ k$.

[hide="Source"]
UNCC
[/hide]

[quote="isabella2296"]See if you can solve it!

$ \frac {3}{2} \cdot \frac {2}{1} \cdot \frac {1}{2} \cdot \frac {2}{3} \cdot \dots \cdot \frac {k}{k \plus{} 1} \geq \frac {1}{8}$

Find the largest value of $ k$.

[hide="Source"]
UNCC
[/hide][/quote]

did you get that from intermediate algebra?!

Actually, Introduction to Algebra took that from UNCC(see source).


Have look at [url=http://imo2006.dmfa.si/participants/MDA.html]Alex[/url] and [url=http://imo2006.dmfa.si/participants/USA.html]Zach.[/url] They look quite similar! Yeah and I guess the same goes for performance. They are both members of the ML/AoPS forum.

Not really that similar :?
But I must admit that their hairstyles look exactly the same ;)

deleted. i apologize if anyone find this offensive.

[quote="mozilla"]Marija Jelić of Serbia and Montenegro, guy or girl?

[/quote]
[b]lemma 1[/b]
If a female contestant forgets to shave herself ,do not confuse her with a male .
[b]lemma 2[/b]
If a male contestant forgets to have a proper haircut,do not confuse him with  a female.
 ;)

[quote="mozilla"]Marija Jelić of Serbia and Montenegro, guy or girl? :D

http://imo2006.dmfa.si/participants/SCG.html[/quote]

Not sure, but I'd say girl too... look the name of the 4º participant, it's quite similar

I believe Mozilla understand that [b]Marija[/b] is female name

There is a fun similarity between them. I find that amusing. Marija is definitely a girl...I think.

A funny thing: how are the USA contestants been treated in MOP? Because the pictures show contestants that look 'sleepy' while the leaders and observers are having upbeat portraits :D

Yes, Marija is a female name. Girl ...  :)

What about Marius and Sorin Radulescu? :D

[quote="10000th User"]
A funny thing: how are the USA contestants been treated in MOP? Because the pictures show contestants that look 'sleepy' while the leaders and observers are having upbeat portraits :D[/quote]

Keen observations like that can also help you solve math problems. You're onto something, in case you couldn't already tell :P

[quote="silouan"]How many twins do you know in the IMO history ?? :D[/quote]
Mexico had ones in the IMO of Japan. Though only one of them made it on the team. The other went as observer. They used to call them "El buena y el mala" (Something like the "good one" and the "bad one")

Sorin and Marius Radulescu were twins but only one of them,Sorin went to IMO and got silver.[/quote]

(To respond to the trivia challenge) I think there were Greek twins in 2004, and (secondhand) Indian twins in 2002.

I remember that Slovenia had two twin contestants in Greece. I remember deputy leader Irena telling me she took a course to know how to tell one form the other.

[quote="Makaveli"]
I would like to kindly require from the moderators to delete this topic as quickly as possible because it is pointless and it might be offensive to some people.
Thank you in advance[/quote]

I'm sorry if I offend anybody. I don't mean to offend anyone, I am just simply asking whether the contestant is a girl or a guy.

[quote="orl"][quote="Makaveli"][quote="mozilla"]Marija Jelić of Serbia and Montenegro, guy or girl? :D

http://imo2006.dmfa.si/participants/SCG.html[/quote]
I would like to kindly require from the moderators to delete this topic as quickly as possible because it is pointless and it might be offensive to some people.
Thank you in advance[/quote]

No, I won't. There is no issue of getting offended here. That are just some random observations on pictures to be found on the IMO web site. Nothing more, nothing less.  :)[/quote]
The question of getting offended is more of personal character, but anyway, you are the one who has the power of deciding which topics should go on existing, though this one shows no concern of feelings of other people.

[quote="mozilla"]Marija Jelić of Serbia and Montenegro, guy or girl?

[/quote]
-.-
Shes a girl, ofcourse.


Are $ a,b,c$ rational or real. Please be more accurate.

[quote="poonprang"][size=150]
If a√3+b / b√3+c[/size] is rational number

Pf a^2+b^2+c^2/a+b+c is integer[/quote]

If the problem is "Let $ a,b,c\in\mathbb Z$ such that $ \frac {a\sqrt 3 \plus{} b}{b\sqrt 3 \plus{} c}\in\mathbb Q$. Prove that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\in\mathbb Z$", then :

Let $ p,q\in\mathbb Z$ with $ \gcd(|p|,|q|) \equal{} 1$ such that $ \frac {a\sqrt 3 \plus{} b}{b\sqrt 3 \plus{} c} \equal{} \frac pq$

If $ p \equal{} 0$, then $ a \equal{} b \equal{} 0$ and $ c\neq 0$ and so $ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c} \equal{} c\in\mathbb Z$
If $ p\neq 0$ :

Then $ (aq \minus{} pb)\sqrt 3 \equal{} pc \minus{} bq$ and so $ aq \equal{} pb$ and $ pc \equal{} bq$

$ aq \equal{} pb$ and $ \gcd(p,q) \equal{} 1$ $ \implies$ $ a \equal{} up$ and $ b \equal{} uq$

$ pc \equal{} bq$ becomes $ pc \equal{} uq^2$ and $ \gcd(p,q) \equal{} 1$ $ \implies$ $ u \equal{} vp$ and $ c \equal{} vq^2$

So : $ a \equal{} vp^2$ and $ b \equal{} vpq$ and $ c \equal{} vq^2$ and so $ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}$ $ \equal{} v\frac {p^4 \plus{} p^2q^2 \plus{} q^4}{p^2 \plus{} pq \plus{} q^2}$ $ \equal{} v(p^2 \minus{} pq \plus{} q^2)$ $ \in\mathbb Z$

Q.E.D.


do you have PDF files of these theorems or proofs?? i would really appreciate if someone put one together or post it here at least so i can TeX it myself for reference. if you have one, please send it to es7abl1sh3d@gmail.com. Thanks in Advance.

Well, I don't know if the administrators would have such files or not, but I don't think so. Nevertheless, why don't YOU do such a compilation and publish it on the forum. It shouldn't be that difficult, I mean, you can have the text surce from the forum directly (put "quote", and it will give you the source), and after that, well you will need to edit a little, but most of the job will be already done. Just an idea.

Best,


geometry
3D geometry
number theory
prime factorization

6.Three segments can divide a vircular region into at most seven regions. What is the maximum number of regions that a circular region can be divided into by 100 segments?

8.The product of integers 240 and k is a perfect cube. What is the smallest possible positive value of k?

I know the answers to them...just not HOW you got them.Please show how you got it, not just the answer.

Thanks-a-billion!! :D

[hide="hint for 6"] Experiment with smaller cases; what's the maximum regions you can divide the circle into with 1 line? 2? 3? etc. You should find a pattern that gives you the answer.[/hide]

[hide="solution for 8"] $ 240\equal{}10\cdot24\equal{}2\cdot5\cdot2\cdot2\cdot2\cdot3\equal{}2^4\cdot3\cdot5$. In a perfect cube, the exponents of every number are multiples of 3. So, we'll need to add two to the exponent of 2, two to the exponent of 3, and two to the exponent of 5. That means multiplying by $ 2^2\cdot3^2\cdot5^2\equal{}900$. Our answer is 900.[/hide]

[hide="1"] Where there are 0 lines, there is 1 region. When there is 1 line, there are 2 regions. Next, the third line will have to cut through the past 2 lines. This creates 3 more lines. The fourth one will create 4 more, etc. Finally it can be deduced that largest number of possible regions is, with $ n$ being the number of lines:
\[ \frac {n(n \plus{} 1)}{2} \plus{} 1
\]

\[ \frac {100(101)}{2} \plus{} 1 \equal{} \boxed{5051}
\]
[/hide]
[hide="2"] When $ x$ is the number and $ p$ is its prime factorization and $ a,b,c$ any numbers:
\[ x^3 \equal{} p_1^{3n_1} \cdot p_2^{3n_2} \cdot p_3^{3n_3} \cdots
\]
Thus, we prime factorize 240:
\[ 2^4 \cdot 3 \cdot 5
\]
Thus, $ x$ must equal $ 2^2 \cdot 3^2 \cdot 5^2 \equal{} \boxed{900}$.[/hide]

---EDIT---
Thanks for pointing that out  :P !

[quote="gatorlover124"]6.Three segments can divide a vircular region into at most seven regions. What is the maximum number of regions that a circular region can be divided into by 100 segments?[/quote]

[hide=" Click for Solution"]

Experimenting with smaller numbers, you can quickly come up with that with $ n$ lines, the maximum number of regions will be $ \frac {n(n \plus{} 1)}{2} \plus{} 1$.  Plugging in for $ x$, we get:
\[ \frac {100(101)}{2} \plus{} 1
\]

\[ 50(101) \plus{} 1
\]

\[ \boxed{5051}
\]
[/hide]

@akhil: Don't forget to add the $ 1$!


Oh, sorry for that: the right inequality is 2(a^2+1)(b^2+1)(c^2+1)>=(a+1)(b+1)(c+1)(abc+1). I apologise for my ignorance. Sorry! Now, solve it!

Congratulations, I spent a lot of time on this one and I couldn't solve it. It is an extraordinary solution.

I agree with Harazi. It is really an extraordinary and beautiful solution.


Let $a,b,c>0$ and $abc(a+b+c)=1.$Prove that $a^5+b^5+c^5 leq 3 a^2 b^2 c^2 sqrt3.$

[quote="cezar lupu"]sorry,it's geq :oops:[/quote]
are you sure about the right side? (If i'm not mistaken, you mean $3 \sqrt{3}$? I get $3 \sqrt{\sqrt{3}}$ instead.

I'm sorry again. Yes, what you have obtained is correct.

[quote="cezar lupu"]Can you post your solution? I'm interested if there's another one different from my solution.[/quote]
from the given condition we can obtain that $a+b+c \ge \sqrt{3\sqrt 3}$
Then, $(a^5+b^5+c^5)(a+b+c)^2 \ge \frac{(a+b+c)^7}{81} \ge \frac{(\sqrt{3\sqrt3})^7}{81}=3\sqrt{\sqrt3}$

[quote="mahbub"]can u plz tell me which inequalities have u used for ur each steps?[/quote]
amgm, power mean

[quote="mahbub"]well,

$a+b+c \geq %Error. "cbroot" is a bad command.
{abc}$

then how:

$a+b+c \geq \sqrt{3\sqrt{3}}$?[/quote]
I think the condition is $abc(a+b+c)=1$ ;)

Sorry i forgot to reply u! that i have understood the AM-GM part. i think u did this:
(a+b+c) >= 3cbroot(abc)
(a+b+c)*cbroot(a+b+c) >= 3
(a+b+c)^(4/3) >=3
(a+b+c) >= sqrt(3sqrt(3))

 :)

Ok. Now the power mean part! :(

tell me if i am wrong:

fifthroot( (a^5+b^5+c^5) / 3) >= (a+b+c)/3

now take power five and gather 3 at RHS.  :)

I think I have done it! Thanks to my creator and to siuchong :lol:

Siuhochung,I proved this is inequality in thesame way.
  We have more general: $a^n+b^n+c^n \geq (a+b+c)^n/3^(n-1).$

[quote="cezar lupu"]Siuhochung,I proved this is inequality in thesame way.
  We have more general: $a^n+b^n+c^n geq (a+b+c)^n/3^(n-1).$[/quote]
this is the power mean inequality.


a,b,c>0 prove that
  (a+b+c)(a/b+b/c+c/a)  >= 3 :sqrt: 3(a 2 +b 2 +c 2 )

[/quote][/list]

suppose that a >= c >= b then ac >= ab >= bc.
Use twice :( Inegalité de reordenement: I hope its means the same in English
( Otherwise ask Pierre to translate   :P )
ca 2 +ab 2 +bc 2  >= 3abc.
Or a+b+c >=  :sqrt: (a 2 +b 2 +c 2 )
that's all :D

[quote="Vasc"]To my mind, the following inequality is hard enough.

Let $a,b,c$ be positive numbers such that $a^2+b^2+c^2=3$. Prove that

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})(a+b+c) \ge 9$.  ;)[/quote]
My ugly proof:

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})(a+b+c) \ge 9\Leftrightarrow$
$\Leftrightarrow(a^2c+b^2a+c^2b)^2(a+b+c)^2\geq27a^2b^2c^2(a^2+b^2+c^2)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^6c^2+2a^5c^3+2a^5b^2c+2a^5c^2b+$
$+\frac{1}{2}a^4b^4+\frac{1}{2}b^4c^4+4a^4b^3c+4a^4c^3b+6a^3b^3c^2)\geq22\sum_{cyc} a^4b^2c^2.$
But the last inequality is true by Cauchy. :)

I think it's nicer:
let $c=\max\{a,b,c\}.$ Hence,
$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)(a+b+c) \ge 9\Leftrightarrow$
$\Leftrightarrow(a-b)^{2}\left(\frac{a+b+c}{ab}-\frac{6}{3+a+b+c}\right)+$
$+(c-a)(c-b)\left(\frac{a+b+c}{ac}-\frac{6}{3+a+b+c}\right)\geq0,$ which triviality.

The following inequality is still true:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{k\sqrt{3(a^{2}+b^{2}+c^{2})}}{a+b+c}$ with $k=2+\sqrt{6}$ and the solution is the same as arqady's solution.

How about this one:
$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge \frac{13(a^{2}+b^{2}+c^{2})}{ab+bc+ca}-4$

[quote="10maths_tp0609"]The following inequality is still true:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{k\sqrt{3(a^{2}+b^{2}+c^{2})}}{a+b+c}$ with $k=2+\sqrt{6}$
[/quote]
You mean that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k-3\ge \frac{k\sqrt{3(a^{2}+b^{2}+c^{2})}}{a+b+c}$ :wink:
[quote="10maths_tp0609"]... and the solution is the same as arqady's solution.
[/quote]
It's your method, 10maths_tp0609 :lol:

[quote=Vasc]To my mind, the following inequality is hard enough.

Let $a,b,c$ be positive numbers such that $a^2+b^2+c^2=3$. Prove that

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})(a+b+c) \ge 9$.  ;)[/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy76sbMKMaf6b&690[/img]


vector
analytic geometry
inequalities
combinatorics proposed
combinatorics

Let $n$ be a positive integer.  Let $S_{1}, S_{2}, \ldots, S_{k}$ be a collection of $2n$-element subsets of $\{1, 2, 3, 4, \ldots, 4n-1, 4n\}$ so that $S_{i}\cap S_{j}$ contains at most $n$ elements for all $1 \leq i < j \leq k$.  Prove that $k \leq 8n-2$.

Let $x \in \{1,2,...,4n\}$ and $A_{1},A_{2},...,A_{m}$ those sets among the given $k$ which do not contain $x$. I prove that $m \leq 4n-1$. Let $v_{1},v_{2},...,v_{m}\in \{0,1\}^{4n-1}$ be the characteristic vectors of these $m$ sets. We know that $v_{i}^{2}=2n$ and $v_{i}\cdot v_{j}\leq n$. On one hand, \[(v_{1}+...+v_{m})^{2}\leq 2mn+2{m \choose 2}n=nm(m+1) \eqno(1)\] while on the other hand, \[(v_{1}+...+v_{m})^{2}= \sum_{j=1}^{4n-1}{\text{(the sum of the j-th coordinates)}^{2}}\geq \frac{(2nm)^{2}}{4n-1}\eqno(2)\] The last inequality by Cauchy. By (1) and (2) we get $m \leq 4n-1$. Therefore, each number is contained in at least $k-(4n-1)$ sets. Now, $\mid S_{1}\mid +\mid S_{2}\mid +...+\mid S_{k}\mid $ is $2kn$, but it is also at least $4n(k-(4n-1))$, hence $k \leq 8n-2$.


superior algebra
superior algebra unsolved

Here is the problem I'm tryng face, but due to the lack of knowledge from algebra I can' deal with it. I'll be grateful for any help and time spent on it.

In infinite dimensional space a notion of 'flag' is replaced by the 'chain of subspace' (ordered by inclusion). Using Zorn theorem prove that  every chain is the subchain of the maximal one.

By 'flag' we understand strictly increasing sequences of subspaces $ L_0 \subset L_1 \subset ... \subset L_n$ of the space $ L$. Number $ n$ is called the length of the flag. Flag $ L_0 \subset L_1 \subset ... \subset L_n \subset ...$ is called maximal if $ L_0 \equal{} \{0\}$, $ \cup{L_i} \equal{} L$ and for all $ i$ between $ L_i$ and $ L_{i\plus{}1}$ we cannot include a subspace.

Nobody1111

so where is the problem? apply zorn to the subchains with the obvious order of refinement.


inequalities
geometry proposed
geometry
U sviti matematyky

Let $ G$ be the centroid of triangle $ ABC.$ Denote by $ r,$ $ r_1,$ $ r_2$ and $ r_3$ the inradii of triangles $ ABC,$ $ GBC,$ $ GAC$ and $ GAB$ respectively and by $ p$ the semiperimeter of triangle $ ABC$. Prove that $ \frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\frac{1}{r_3}\ge\frac{3}{r}\plus{}\frac{18}{p}.$

Centroid $ G$ divides $ \triangle ABC$ into three equivalent triangles $\triangle GBC ,\triangle GCA,\triangle GAB$

$ \frac {_1}{^3}[\triangle ABC] \equal{} r_1[\frac {_a}{^2} \plus{} \frac {_1}{^3}(m_b \plus{} m_c)]$

$ \frac {_1}{^3}[ \triangle ABC] \equal{} r_2[\frac {_b}{^2} \plus{} \frac {_1}{^3}(m_a \plus{} m_c)]$

$ \frac {_1}{^3}[\triangle ABC] \equal{} r_3[\frac {_c}{^2} \plus{} \frac {_1}{^3}(m_a \plus{} m_b)]$

$ \frac {1}{r_1} \plus{} \frac {1}{r_2} \plus{} \frac {1}{r_3} \equal{} \frac {3p \plus{} 2(m_a \plus{} m_b \plus{} m_c)}{[\triangle ABC]} \equal{} \frac {3}{r} \plus{} \frac {2(m_a \plus{} m_b \plus{} m_c)}{[\triangle ABC]}$

Using $ h_a \plus{} h_b \plus{} h_c \ge 9r$ and the fact that medians are greater-equal to altitudes

$ m_a \plus{} m_b \plus{} m_c \ge h_a \plus{} h_b \plus{} h_c \ge 9r \ \Longrightarrow \ \frac {2(m_a \plus{} m_b \plus{} m_c)}{[\triangle ABC]} \ge \frac {18}{p}$

$\Longrightarrow \  \frac {1}{r_1} \plus{} \frac {1}{r_2} \plus{} \frac {1}{r_3} \ge \frac {3}{r} \plus{} \frac {18}{p}$


[quote="Orfikoss"]Ψάχνοντας βρήκα αυτό, νομίζω βοηθάει: https://nrich.maths.org/discus/messages/2069/6484.html?1059044863
Δες το τρίτο ποστ
btw είναι στα αγγλικά.
Αν θες να σου το μεταφράσω πες μου[/quote]

O.K mate thanx for your help, but what i asked it is not the proof but the laws of logic INVOLVED in the proof.

Now the post that you referred me to it describes equivalence between statements.

After one of my student ask me the question i open couple of books of logic and i started to learn the laws of logic.

To my surprise i knew none of the laws of logic by name. For example i learned  that one of the laws of logic is called M.Ponens (κανονας αποσπασεως νομιζω στα ελληνικα),another one addition introduction,Universal Elimination e.t.c ..e.tc.

BUT i cannot figure yet HOW AND WHICH are the laws of logic INVOLVED in  the above proof

φιλε καμμια φορα μια ερωτηση σε βγαζει εξω απο τα ρουχα σου .Ειμαι πολυ ΠΕΡΙΕΡΓΟΣ να δω ποιος θα απαντηση στην ερωτηση

[quote="Γιώργος"][quote="τρικλινο"]γεια σας φιλοι ξανα.

Επροσπαθουσα λοιπον να αποδειξω στους μαθητες μου .αβ=0 <===> α=0 η β=0,οταν καποιος ειπε:

Οταν λεμε οτι καθε μαθηματικη αποδειξη στηριζεται στους κανονες της λογικης στην προκειμενη αποδειξη ποιοι ειναι αυτοι οι κανονες??

Mπορειτε να βοηθησετε???[/quote]

Έστω $ a\cdot b \equal{} 0$ και $ a\neq 0$. Θα αποδείξουμε ότι $ b \equal{} 0$. Έχουμε ότι $ b \equal{} 1\cdot b \equal{} (a^{ \minus{} 1}\cdot a)\cdot b \equal{} a^{ \minus{} 1}\cdot (a\cdot b) \equal{} a^{ \minus{} 1}\cdot 0 \equal{} 0$
P.S Ελπίζω να βοήθησα έστω και λίγο... :)[/quote]

Ευχαριστω Γιωργο αλλα οπως απηντησα και στον Οrficco δεν ενδιαφερομαστε για το πως αποδεικνυεται αβ=0 <===>α=0 η β=0,αλλα για τους ΚΑΝΟΝΕΣ ΤΗΣ ΛΟΓΙΚΗΣ πανω στους οποιους στηριζεται η αποδειξη.

Ευχαριστω για την αποδειξη Anyway

@moderators: nomizo prepei na metaferthei to topic sta genika i toulaxiston sta sxolika.

[quote="Demetres"]Οι δύο κανόνες λογικής που επιτρέπεται να χρησιμοποιηθούν ειναι οι

γ1) Modus Ponens: Αν έχω ήδη γράψει τις προτάσεις p και $ p \Rightarrow q$, τότε επιτρέπεται να γράψω και την πρόταση q.
γ2) Generalization: Αν έχω γράψει την πρόταση p και η μεταβλητή x δεν υπάρχει στην p (αν υπάρχει να είναι bound στην p και όχι free) τότε επιτρέπεται να γράψω την πρόταση $ (\forall x)p$.

Τρίκλινο κάτι τέτοιο ψάχνεις; Αν ναι πες μου να αρχίσω να γράφω τα αξιώματα λογικής (του κατηγορηματικού λογισμού) καθώς και την απόδειξη. (Δεν την έχω γράψει ακόμη αλλά της 100 προτάσεις δεν τις γλυτώνουμε με τίποτα. (Λόγω του ότι τα αξιώματα λογικής είναι ελάχιστα και θα χρειαστεί να αποδείξουμε και διάφορα θεωρήματα.)[/quote]

Σύμφωνω με τον Demetres, κάτι τέτοιο θέλει πάρα πάρα πολύ (και αρκετά άχαρο μετά από ένα σημείο) γράψιμο (btw, Dem, πραγματικά θαυμάζω το κουράγιο σου που είσαι έτοιμος να γράψεις 100 γραμμές formal logic!). Πάντως, πιστεύω ότι η απόδειξη σε formal logic πρέπει να υπάρχει ακόμα και από κάποιο πρόγραμμα αυτοματισμού αποδείξεων (το CADE ή όπως αλλιώς το λένε). Τρίκλινο, αν σ'ενδιαφέρει απλώς να δεις την απόδειξη, βάλε στο google automated reasoning και ψάξε τα αντίστοιχα transcripts.

Cheers,

Durandal 1707

[quote="Demetres"]Νομίζω πως πρέπει να μείνει εδώ μιας και έχει σχέση με μάθημα λογικής που διδάσκεται σε φοιτητές. Ο φίλος μάλλον αναφέρεται στους κανόνες λογικής που χρησιμοποιούνται στον κατηγορηματικό λογισμό.

Λοιπόν, έχουμε την γλώσσα της θεωρίας σωμάτων με όλα τα αξιώματα (π.χ. $ \forall x (1x \equal{} x)$) κτλ. Σε αυτή την γλώσσα θέλουμε να αποδείξουμε το θεώρημα

$ ab \equal{} 0 \Leftrightarrow (a \equal{} 0) \vee (b \equal{} 0)$

Μια τέτοια απόδειξη στον κατηγορηματικό λογισμο ειναι μια ακολουθία από προτάσεις. Η πρώτη πρόταση είναι η
$ ab \equal{} 0$, η τελευταία πρόταση είναι η $ (a \equal{} 0) \vee (b \equal{} 0)$ και όλες οι ενδιάμεσες προτάσεις είναι

α) Αξιώματα της θεωρίας σωμάτων.
β) Αξιώματα λογικής.
γ) Προκύπτουν από κανόνες λογικής οι οποίοι εφαρμόζονται σε προηγούμενες προτάσεις.

Οι δύο κανόνες λογικής που επιτρέπεται να χρησιμοποιηθούν ειναι οι

γ1) Modus Ponens: Αν έχω ήδη γράψει τις προτάσεις p και $ p \Rightarrow q$, τότε επιτρέπεται να γράψω και την πρόταση q.
γ2) Generalization: Αν έχω γράψει την πρόταση p και η μεταβλητή x δεν υπάρχει στην p (αν υπάρχει να είναι bound στην p και όχι free) τότε επιτρέπεται να γράψω την πρόταση $ (\forall x)p$.

Τρίκλινο κάτι τέτοιο ψάχνεις; Αν ναι πες μου να αρχίσω να γράφω τα αξιώματα λογικής (του κατηγορηματικού λογισμού) καθώς και την απόδειξη. (Δεν την έχω γράψει ακόμη αλλά της 100 προτάσεις δεν τις γλυτώνουμε με τίποτα. (Λόγω του ότι τα αξιώματα λογικής είναι ελάχιστα και θα χρειαστεί να αποδείξουμε και διάφορα θεωρήματα.)

Επίσης τι δίδασκες στους μαθητές σου και σε ποιο επιπεδό;

ΥΓ: Συγνώμη για μερικά αγγλικά αλλά δεν γνωρίζω την ελληνική ορολογία. Θα ψάξω αργότερα να την βρω.[/quote]

O.K DEMETRES start shooting in ENGLISH or GREEK ,it is no problem for me.But not 100 sentences.FROM the books that i read the laws of logic usually involved in a proof are : addition introduction,addition elimination, Disjunction introduction Disjunction elimination ,M.Ponens,De Morgan e.t.c in propositional logic and Universal Introduction,Universal Elimination,Existential Introduction and Existential Elimination in predicate calculus.
ΤΗΕ law of conditional proof and the proof by contradiction

Βεβαια οι παραπανω νομοι ειναι θεωρηματα των αξιωματων της λογικης.Επισης εχουν διαφορετικα ονοματα σε διαφορετικα βιβλια,αλλα καταλαβαινεις γενικα τι θελω να πω,δεν θα χρειαστη να αποδειξης τους παραπανω νομους (ΘΕΩΡΗΜΑΤΑ).

Επισης η κανονικη φορμουλα που θελουμε να αποδειξουμε ειναι:

$ \forall a\forall b$( ab=0 <===> a=0vb=0) που πολυ καλλα εγαψες αλλα δεν εβαλες μπροστα τους ποσοδεικτες

Κανω ιδιαιτερα σε group απο μαθητες γυμνασιου λυκειου οχι ολο το χρονο και αναλυση σε διερευνητικη μορφη

Τρίκλινο, δίκιο έχεις για τους ποσοδείκτες. Αρκεί όμως να αποδείξω την πρόταση που έγραψα και μετά να κάνω δυο χρήσεις του Universal Generalisation.

Παρεμπιπτόντως, στο πιο πάνω post μου, έχω γράψει λανθασμένα το Universal Generalisation. Αυτό που έπρεπε να γράψω είναι:

Universal Generalisation: Αν έχω γράψει την πρόταση $ p$ και η μεταβλητή $ x$ είναι δεσμευμένη (bound) σε όλες  τις συνθήκες που χρησιμοποιήθηκαν κατά την απόδειξη της $ p$, τότε μπορώ να γράψω την πρόταση $ \forall x p$.

[H μεταβλητή $ x$ ασφαλώς και θα βρίσκεται ελεύθερη (free) στην $ p$ αλλιώς δεν έχει και νόημα να χρησιμοποιήσω Universal Generalisation.]

Για αξιώματα λογικής θα χρησιμοποιήσω αυτά που βρίσκονται στο βιβλίο ''Notes on Logic and Set Theory'' του P. T. Johnstone. (Από βιβλίο σε βιβλίο ίσως διαφέρουν.) Αυτά είναι

1) $ p \Rightarrow (q \Rightarrow p)$
2) $ (p \Rightarrow (q \Rightarrow r)) \Rightarrow ((p \Rightarrow q) \Rightarrow (q \Rightarrow r))$
3) $ \neg \neg P \Rightarrow p$
4) $ (\forall x p) \Rightarrow p[t/x]$ (Εδώ η $ x$ είναι ελεύθερη μεταβλητή της $ p$ και η $ t$ είναι term (όρος) με όλες τις μεταβλητές τις δεσμευμένες στην $ p$.
5) $ (\forall x (p \Rightarrow q)) \Rightarrow (p \Rightarrow (\forall x) q)$ (Εδώ η $ x$ δεν πρέπει να είναι ελεύθερη μεταβλητή της $ p$.)
6) $ \forall x (x\equal{}x)$
7) $ (\forall x \forall y)((x\equal{}y) \Rightarrow (p \Rightarrow p[y/x]))$  (Εδώ η $ x$ είναι ελεύθερη μεταβλητή της $ p$ και η $ y$ δεν είναι δεσμευμένη στην $ p$.

Εκτός απο αυτά τα αξιώματα επιτρέπεται να χρησιμοποιήσω τα Modus Ponens (MP) και Univesal Generalisation (UG) σαν κανόνες λογικής συμπερασματολογίας.

Κανονικά οτιδήποτε άλλο θα πρέπει να το αποδεικνύω αλλά για να γλυτώσουμε χώρο θα τα ονομάζω θεωρήματα λογικής. (Για παράδειγμα το Disjunction Introduction αποδεικνύεται εύκολα από (1) και MP:

1) $ A$
2) $ A \Rightarrow (\neg B \Rightarrow A)$
-----------------------------------------
3) $ \neg B \Rightarrow A$ (που εξ'ορισμού είναι το $ A \vee B$)

Θέλω λοιπόν να αποδείξω $ (ab \equal{} 0) \Rightarrow ((a\equal{}0) \vee (b\equal{}0))$ δηλαδή $ (ab \equal{} 0) \Rightarrow (a \neq 0 \Rightarrow b\equal{}0)$. Αρχικά όμως δεν θα δείξω αυτό. Θα πάρω ώς δωσμένη συνθήκη το $ ab \equal{} 0$ και θα αποδείξω το $ a \neq 0 \Rightarrow b\equal{}0$. (Το Deduction Theorem (συμπερασματολογικό θεώρημα) μου λέει ότι αυτό είναι αρκετό. Εν πάσει περιπτώσει, θα δείξω μετά πως μπορεί να μεταφραστεί η απόδειξη σε αυτό που θέλουμε.) Χρησιμοποιώντας ακόμη μια φορά το συμπερασματολογικό θεώρημα, μπορώ να πάρω και το $ a \neq 0$ ως δεδομένο και θέλω να αποδείξω ότι (με αυτά τα δύο δεδομένα) $ b \equal{} 0$.

Έχουμε λοιπόν:

1) $ (\forall x)(x \neq 0 \Rightarrow (\exists c) cx \equal{} 1)$ (Αξίωμα Θεωρίας Σωμάτων)
2) $ a \neq 0 \Rightarrow (\exists c) ca \equal{} 1$ (Αξ. Λ. 4 στο (1))
3) $ a \neq 0$ (Δεδομένο)
4) $ (\exists c) ca \equal{} 1$ (MP στα (2),(3))
5) $ (\forall x) x0 \equal{} 0$ (Αξίωμα Θεωρίας Σωμάτων)
6) $ (ab \equal{} 0) \Rightarrow (((\forall x) x0 \equal{} 0) \Rightarrow ((\forall x) x(ab) \equal{} ab))$ (Εφαρμόζοντας 2 φορές το Αξ. Λ. 4 στο Αξ. Λ. 7)
7) $ ab\equal{}0$ (Δεδομένο)
8) $ ((\forall x) x0 \equal{} 0) \Rightarrow ((\forall x) x(ab) \equal{} ab)$ (MP στα (6),(7))
9) $ (\forall x) x(ab) \equal{} ab$ (MP στα (5),(8))
10) $ c(ab) \equal{} ab$ (Αξ. Λ. 4 )
11) $ c(ab) \equal{} 0$. (Θεώρημα Λογικής στα 6 και 10. Θέλει απόδειξη!)
12) $ (ca)b \equal{} c(ab)$ (Αξίωμα Θεωρίας Σωμάτων και Αξ. Λ. 4 τρεις φορές)
13) $ (ca)b \equal{} 0$. (Θεώρημα Λογικής στα 11 και 12)
14) $ 1b \equal{} 0$. (Από 4,13 και Existential Elimination το οποίο θέλει απόδειξη)
15) $ b\equal{}0$. (Αξίωμα Θεωρίας Σωμάτων + Θεώρημα Λογικής στο 14)

Μέχρι τώρα έχω δείξει ότι $ \{ab \equal{} 0 , a \neq 0 \} \vdash b\equal{}0$. Μένει τώρα να δικαιολογηθούν μερικές γραμμές στην απόδειξη, (προς το τέλος άφησα αρκετές γραμμές πίσω) και να τροποποιήσω την απόδειξη αρχικά σε απόδειξη του  $ \{ab \equal{} 0\} \vdash (a \neq 0) \Rightarrow (b\equal{}0)$ και μετέπειτα σε απόδειξη του $ \emptyset \vdash (ab\equal{}0) \Rightarrow ((a \neq 0) \Rightarrow (b\equal{}0))$

Πες μου αν είσαι ευχαριστημένος από τα πιο πάνω για να συνεχίσω ή αν κάτι χρειάζεται περισσότερη επεξήγηση.

[quote="Demetres"]Για αξιώματα λογικής θα χρησιμοποιήσω αυτά που βρίσκονται στο βιβλίο ''Notes on Logic and Set Theory'' του P. T. Johnstone. (Από βιβλίο σε βιβλίο ίσως διαφέρουν.)
[...]
Μέχρι τώρα έχω δείξει ότι $ \{ab \equal{} 0 , a \neq 0 \} \vdash b \equal{} 0$.
[/quote]

:o

Dude... respek στο κουράγιο σου! :omighty:

Durandal 1707

[quote="Durandal"]
:o

Dude... respek στο κουράγιο σου! :omighty:

Durandal 1707[/quote]

Μπα, από ένα σημείο και μετά (κάπου στις γραμμές 9-10) είχα βαρεθεί και τα έγραφα λίγο μαζικά.  :D  Με λίγους πρόχειρους υπολογισμούς που έκανα αν τα έγραφα όλα και μετά μετέτραπα όλες τις χρήσεις του deduction theorem θα έβγαιναν καμιά 300αριά γραμμές.  :)

[quote="Demetres"]Τρίκλινο, δίκιο έχεις για τους ποσοδείκτες. Αρκεί όμως να αποδείξω την πρόταση που έγραψα και μετά να κάνω δυο χρήσεις του Universal Generalisation.

Παρεμπιπτόντως, στο πιο πάνω post μου, έχω γράψει λανθασμένα το Universal Generalisation. Αυτό που έπρεπε να γράψω είναι:

Universal Generalisation: Αν έχω γράψει την πρόταση $ p$ και η μεταβλητή $ x$ είναι δεσμευμένη (bound) σε όλες  τις συνθήκες που χρησιμοποιήθηκαν κατά την απόδειξη της $ p$, τότε μπορώ να γράψω την πρόταση $ \forall x p$.

[H μεταβλητή $ x$ ασφαλώς και θα βρίσκεται ελεύθερη (free) στην $ p$ αλλιώς δεν έχει και νόημα να χρησιμοποιήσω Universal Generalisation.]

Για αξιώματα λογικής θα χρησιμοποιήσω αυτά που βρίσκονται στο βιβλίο ''Notes on Logic and Set Theory'' του P. T. Johnstone. (Από βιβλίο σε βιβλίο ίσως διαφέρουν.) Αυτά είναι

1) $ p \Rightarrow (q \Rightarrow p)$
2) $ (p \Rightarrow (q \Rightarrow r)) \Rightarrow ((p \Rightarrow q) \Rightarrow (q \Rightarrow r))$
3) $ \neg \neg P \Rightarrow p$
4) $ (\forall x p) \Rightarrow p[t/x]$ (Εδώ η $ x$ είναι ελεύθερη μεταβλητή της $ p$ και η $ t$ είναι term (όρος) με όλες τις μεταβλητές τις δεσμευμένες στην $ p$.
5) $ (\forall x (p \Rightarrow q)) \Rightarrow (p \Rightarrow (\forall x) q)$ (Εδώ η $ x$ δεν πρέπει να είναι ελεύθερη μεταβλητή της $ p$.)
6) $ \forall x (x \equal{} x)$
7) $ (\forall x \forall y)((x \equal{} y) \Rightarrow (p \Rightarrow p[y/x]))$  (Εδώ η $ x$ είναι ελεύθερη μεταβλητή της $ p$ και η $ y$ δεν είναι δεσμευμένη στην $ p$.

Εκτός απο αυτά τα αξιώματα επιτρέπεται να χρησιμοποιήσω τα Modus Ponens (MP) και Univesal Generalisation (UG) σαν κανόνες λογικής συμπερασματολογίας.

Κανονικά οτιδήποτε άλλο θα πρέπει να το αποδεικνύω αλλά για να γλυτώσουμε χώρο θα τα ονομάζω θεωρήματα λογικής. (Για παράδειγμα το Disjunction Introduction αποδεικνύεται εύκολα από (1) και MP:

1) $ A$
2) $ A \Rightarrow (\neg B \Rightarrow A)$
-----------------------------------------
3) $ \neg B \Rightarrow A$ (που εξ'ορισμού είναι το $ A \vee B$)

Θέλω λοιπόν να αποδείξω $ (ab \equal{} 0) \Rightarrow ((a \equal{} 0) \vee (b \equal{} 0))$ δηλαδή $ (ab \equal{} 0) \Rightarrow (a \neq 0 \Rightarrow b \equal{} 0)$. Αρχικά όμως δεν θα δείξω αυτό. Θα πάρω ώς δωσμένη συνθήκη το $ ab \equal{} 0$ και θα αποδείξω το $ a \neq 0 \Rightarrow b \equal{} 0$. (Το Deduction Theorem (συμπερασματολογικό θεώρημα) μου λέει ότι αυτό είναι αρκετό. Εν πάσει περιπτώσει, θα δείξω μετά πως μπορεί να μεταφραστεί η απόδειξη σε αυτό που θέλουμε.) Χρησιμοποιώντας ακόμη μια φορά το συμπερασματολογικό θεώρημα, μπορώ να πάρω και το $ a \neq 0$ ως δεδομένο και θέλω να αποδείξω ότι (με αυτά τα δύο δεδομένα) $ b \equal{} 0$.

Έχουμε λοιπόν:

1) $ (\forall x)(x \neq 0 \Rightarrow (\exists c) cx \equal{} 1)$ (Αξίωμα Θεωρίας Σωμάτων)
2) $ a \neq 0 \Rightarrow (\exists c) ca \equal{} 1$ (Αξ. Λ. 4 στο (1))
3) $ a \neq 0$ (Δεδομένο)
4) $ (\exists c) ca \equal{} 1$ (MP στα (2),(3))
5) $ (\forall x) x0 \equal{} 0$ (Αξίωμα Θεωρίας Σωμάτων)
6) $ (ab \equal{} 0) \Rightarrow (((\forall x) x0 \equal{} 0) \Rightarrow ((\forall x) x(ab) \equal{} ab))$ (Εφαρμόζοντας 2 φορές το Αξ. Λ. 4 στο Αξ. Λ. 7)
7) $ ab \equal{} 0$ (Δεδομένο)
8) $ ((\forall x) x0 \equal{} 0) \Rightarrow ((\forall x) x(ab) \equal{} ab)$ (MP στα (6),(7))
9) $ (\forall x) x(ab) \equal{} ab$ (MP στα (5),(8))
10) $ c(ab) \equal{} ab$ (Αξ. Λ. 4 )
11) $ c(ab) \equal{} 0$. (Θεώρημα Λογικής στα 6 και 10. Θέλει απόδειξη!)
12) $ (ca)b \equal{} c(ab)$ (Αξίωμα Θεωρίας Σωμάτων και Αξ. Λ. 4 τρεις φορές)
13) $ (ca)b \equal{} 0$. (Θεώρημα Λογικής στα 11 και 12)
14) $ 1b \equal{} 0$. (Από 4,13 και Existential Elimination το οποίο θέλει απόδειξη)
15) $ b \equal{} 0$. (Αξίωμα Θεωρίας Σωμάτων + Θεώρημα Λογικής στο 14)

Μέχρι τώρα έχω δείξει ότι $ \{ab \equal{} 0 , a \neq 0 \} \vdash b \equal{} 0$. Μένει τώρα να δικαιολογηθούν μερικές γραμμές στην απόδειξη, (προς το τέλος άφησα αρκετές γραμμές πίσω) και να τροποποιήσω την απόδειξη αρχικά σε απόδειξη του  $ \{ab \equal{} 0\} \vdash (a \neq 0) \Rightarrow (b \equal{} 0)$ και μετέπειτα σε απόδειξη του $ \emptyset \vdash (ab \equal{} 0) \Rightarrow ((a \neq 0) \Rightarrow (b \equal{} 0))$

Πες μου αν είσαι ευχαριστημένος από τα πιο πάνω για να συνεχίσω ή αν κάτι χρειάζεται περισσότερη επεξήγηση.[/quote]

ΦΙΛΕ δημητρη εισαι μοναδικος.Βεβαια η παραπανω αποδειξη δεν ειναι 300 γραμμες ,αν χρησιμοποιησης θεωρηματα αντι για αξιωματα στον pridicate calculus.

Εχεις προσπαθησει να κανεις formal proof σε καμμια αποδειξη στην αναλυση.βεβαια οχι στο style της παραπανω αποδειξις.

Ξερεις ποσες αποδειξεις στην αναλυση ειναι λαθος .?? Υπαρχει αλλος τροπος να αποδειξης οτι μια αποδειξη ειναι λαθος ,εαν δεν την γραψεις  γραμμη γραμμη  δικαιλογοντας καθε γραμμη???

Δεν μου ειπες τι σπουδαζεις.

Δωσε μου λιγο χρονο να μελετησω την αποδειξη σου και θα σου απαντησω

[quote="τρικλινο"]
Ξερεις ποσες αποδειξεις στην αναλυση ειναι λαθος .?? Υπαρχει αλλος τροπος να αποδειξης οτι μια αποδειξη ειναι λαθος ,εαν δεν την γραψεις  γραμμη γραμμη  δικαιλογοντας καθε γραμμη???
[/quote]

Δεν καταλαβαίνω. Αν την δικαιολογήσεις γραμμή, γραμμη πώς μπορεί να είναι λάθος; Δεν έχει πέσει στην αντίληψη μου πάντως καμιά λανθασμένη απόδειξη. Μήπως προβληματίζεσαι με το second order logic;

Επιτελους σταθηκα τυχερος και με την βοηθεια ενος φιλου βρηκα στο βιβλιο του Angelo Margaris ,1st order Mathematical Logic στην σελλιδα 126 μια πληρη formal proof του θεωρηματος:.... $ \vdash$  .$ \forall x\forall y$( xy=0 <===>x=0 v y=0)

Eδω ο συγγραφεας δινει πρωτα μια κανονικη αποδειξη και κατοπην μια φορμαλιστικη.

Και¨..... στα Αγγλικα ( ακολουθει αποδειξη και στα Ελληνικα) ,εχουμε.

Suppose xy=0 and y $ \neq$ 0.Then there is a z such that yz=1.Then x= 1.x=x(yz) = (xy)z =0z =0.

και τωρα η formal proof.

1) xy=0 & y $ \neq$ 0..........................................................................assumption ,to apply the deduction theorem,or rule of conditional proof

2) xy =0  .............................................................................from (1) and using addition elimination ( p&q ====>p)

3) y $ \neq$ 0.........................................................................from (1) and using again addition elimination or named differently conjunction elimination

4) $ \forall x$ [ x $ \neq$ 0 ===> $ \exists z$( xz=1)] .................................................................................................field axiom ,in real Nos too

5)   y $ \neq$ 0 ===> $ \exists z$( yz=1)...............................................from (4) and using Universal Elimination where we put x=y,differently called specialization,or Universal Instantiation .It is the theorem according to which we can drop the Universal quantifier substituting the general variable x by a specific one y provided this can be done.

Example: $ \forall x\exists y$ ( x<y). here if we drop the universal quantifier and we put x=y in the formula we get :

.........................................$ \exists y$ ( y<y )....y smaller than itself,which is not cerrect.....................................
................that explains the phrase " provided this can be done"...............................................................................

6) $ \exists z$ ( yz=1).............................................................from (5) and (3) and using M.Ponens [( p & p==.>q)===>q]

7) yz =1................................................................................from (6) and using Existential Elimination,or the C rule,and at the same time we start a hypothesis keeping z under certain restrictions,for example we cannot generalize it,until we get a formula where z is bounded.Then we discharge the hypothesis like we do with the deduction theorem.

8) $ \forall x ( x1\equal{}x)$..........................................................................................axiom.

9) x1 = x.................................................................................from (8) and using U.Elimination where we put x=x

10) x(yz) = x............................................................................by substituting (7) into (9) ,a theorem in predicate calculus.It works like substitution in ordinary mathematical proofs.

11) $ \forall x\forall y\forall z ( (xy)z \equal{} x(yz))$........................................................axiom........

12) (xy)z = x(yz)......................................................................from (11) and using U.Elimination 3 times where we put x=x,y=y,z=z.

13) 0z = x(yz)..........................................................................by substituting (2) into (12).............................

14) $ \forall x\forall y\forall z ( xy \equal{} yx )$...................................................................axiom............................

15) 0z = z0 ................................................................................from (14) and using U.Elimination where we put x=0 and y=z.

16) $ \forall x ( x0 \equal{} 0)$.............................................................................................theorem in fields,in real Nos too.......

17) x0 = 0...................................................................................from (16) and using U.Elimination where we put x=x...

18)  x= 0.....................................................................................from (10), (13) ,(15) ,(17) and using the theorem of equality..............................$ \forall x\forall y\forall z$ [ ( x=y & y= z) ====>x= z] repeatedly.

19)   x=0 ....................................................................................here we end the hypothesis we started in line (7),because the formula x=0 has no free z

20) xy = 0 & y$ \neq$ 0 =====> x=0.............................................from (1) to (19) and using the deduction theorem.......

21) xy = 0 ====> x = 0 v y = 0.....................................................from (20) and using the theorem in propositional logic ,where the formula : P&Q===>R is equivalent P===>(~Q v R).So if we put , (xy = 0)=P ,( y$ \neq$ 0)=Q, ( x= 0) = R.
from line 20 we get line 21

22) $ \forall x\forall y\forall z$ ( xy=0 ====> x=0 v y=0).................from (21) and using the theorem of generalization,or Universal introduction 3 times.

The generalization is done again under certain restrictions.

Example: if we generalized the variable x in lines (19) or (18) we would have got:  $ \forall x$ ( x=0) a false statement.

Usually in mathematical proofs comes AFTER the application of the deduction theorem.

In the proof above i have added o more detailed justification for each line of that offered by the Author.

I also disagree with the way that the Author tackles the proof.For example i would have used the deduction theorem twice and so  there would be no need to use the complicated law in line 21,but  a more simple and well known law,like addition introduction disjunction elimination M.Ponens e.t.c  e.t.c.

And in our case material implication :  p vq <====> ~p--->q

I COPIED this proof from the book to show that we can have formal proofs for mathematical problems not 300 lines LONG,even for difficult problems in analysis too.

 ΑΚΟΛΟΥΘΕΙ η αποδειξη στα Ελλινικα στο επομενο ποστ

και τωρα η formal proof στα Ελληνικα

1) xy=0 &$ y\neq  0$..........................................................................υποθεση για να εφαρμοσουμε το θεωρημα παραγωγης,η κανων υποθετικης αποδειξης

2) xy =0 .............................................................................απο (1) και χρησιμοπ αφαιρετικη συζευξη ( p&q ====>p) ,η κανων απλουστευτικος

3) $ y\neq 0$...............................................................................απο (1) και χρησιμοπ.ξανα αφαιρετικη συζευξη

4)$ \forall x[ x \neq0 \rightarrow \exists z( xz\equal{}1)]$ ............................................................αξιωμα  στα (fields) η στους πραγματικους αριθμους

5)$ y \neq0 \rightarrow \exists z( yz\equal{}1)$.............................................απο (4) και χρησιμοπ.  Ειδικευση του καθολικου η εξαλειψη του γενικου η εξαλειψη του καθολικου η αποβολη του γενικου Οπου βαζουμε x=y .  Ειναι το θεωρημα συμφωνα με το οποιο αποβαλλουμε  το γενικο ποσοδεικτη αντικαθιστοντας την γενικη μεταβλητη x  με τη ειδικη μεταβλητη η σταθερη η ελευθερη μεταβλητη y υποτιθεμενου οτι αυτο μπορει να γινει .

παραδειγμα:  $ \forall x\exists y( x<y)$. Εδω εαν αποβαλλουμε τον γενικο ποσοδεικτη και οπου x βαλλουμε y  τοτε εχουμε:

.........................................$ \exists y ( y<y )$....δηλαδη οτι υπαρχει  y μικροτερο του εαυτου του,το οποιον ειναι αδυνατο. ....................................
................αυτο εξηγει την φραση «υποτιθεμενου οτι αυτο μπορει να γινει»...............................................................................

6) $ \exists z( yz\equal{}1)$.............................................................Απο (5) και (3) και χρησιμοπ. Τον κανονα αποσπασεως. [( p & p==.>q)===>q]

7) yz =1................................................................................Απο (6) και χρησιμοπ. Ειδικευση του υπαρξιακου η εξαλειψη του υπαρξιακου η αποβολη του υπαρξιακου η κανονας διαλογης και συντοχρονως αρχιζουμε μια υποθεση διατηροντας το z κατω υπο ορισμενας συνθηκες,π.χ δεν μπορουμε να το γενικευσουμε ( δηλαδη να γραψουμε $ \forall z$ P) μεχρι να φτασουμε σε μια φορμουλα οπου το z δεν ειναι ελευθερο.Τοτε αποβαλλουμε την υποθεση οπως κανουμε με το θεωρημα παραγωγης.

8)$ \forall x( x1\equal{}x)$..........................................................................................Αξιωμα.

9) x1 = x.................................................................................Απο (8) και χρησιμοπ. Εξαλειψη του γενικου οπου βαζουμε x=x.

10) x(yz) = x............................................................................Αντικαθιστοντας  την (7) εις το (9)και χρησιμοπ. Ενα θεωρημα στον predicate calculus.Aυτη η αντικατασταση γινεται οπως η αντικατασταση στις συνηθεις μαθηματικες αποδειξεis.

11)$ \forall x\forall y\forall z[ (xy)z\equal{}x(yz)]$........................................................Aξιωμα........

12) (xy)z = x(yz)......................................................................Απο (11) και χρησιμοπ. Εξαλειψη του γενικου 3 φορες οπου βαζουμε x=x,y=y,z=z.

13) 0z = x(yz)..........................................................................Aντικαθιστοντας την (2) στην (12)..............................

14)$ \forall x\forall y( xy \equal{} yx )$...................................................................Αξιωμα............................

15) 0z = z0 ................................................................................Απο (14) και χρησιμοπ. Εξαλειψη του γενικου οπου βαζουμε x=0,y=z..

16) $ \forall x( x0 \equal{} 0 )$.............................................................................................Θεωρημα στα (fields )η στους πραγματικους αριθμους.......

17) z0 = 0...................................................................................Απο (16) και χρησιμοπ. Εξαλειψη του γενικου οπου βαζουμε      x=z.

18) x= 0.....................................................................................Aπο (10), (13), (15), (17) και χρησιμοπ. Το θεωρημα της ισοτητας ..............................  $ \forall x\forall y\forall z[ ( x\equal{}y \wedge y\equal{} z) \equal{}\equal{}\equal{}\equal{}>x\equal{} z]$ επανειλημμενα.

19) x=0 ....................................................................................Εδω αποβαλλουμε την υποθεση που αρχισαμε εις την γραμμη (7) επειδη η φορμουλα x=0 δεν εχει ελευθερο z.

20) xy = 0 & $ y\neq 0$ =====> x=0.............................................Απο (1) μεχρι την γραμμη (19) και χρησιμοπ.το θεωρημα της παραγωγης η κανονα της υποθετικης αποδειξεως..................

21) xy = 0 ====> x = 0 v y = 0.....................................................Απο (20) και χρησιμοπ. To θεωρημα της προτασιακης λογικης,οπου η φορμουλα  : P&Q===>R  ειναι ισοδυναμη με  P===>(R v ~Q).Ετσι εαν βαλλουμε , (xy = 0)=P ,( $ y\neq 0$)=Q, ( x= 0) = R.
Απο την γραμμη (20) παιρνουμε την γραμμη (21) .

22) $ \forall x\forall y ( xy\equal{}0 \equal{}\equal{}\equal{}\equal{}> x\equal{}0 \vee y\equal{}0)$.................Απο (21) και χρησιμοπ.Το θεωρημα Της γενικευσεως του ειδικου η της καθολικης γενικευσεως η της επαναφορας του γενικου, 3 φορες.

Η ΓΕΝΙΚΕΥΣΗ γινεται  υπο ορισμενες συνθηκες..

Παραδειγμα : εαν γενικευσουμε την μεταβλητη x στις γραμμες (19) η (18) θα παρουμε:  $ \forall x( x\equal{}0)$ ,λαθος φορμουλα.

Συνηθως στις μαθηματικες αποδειξεις η γενικευση γινεται  ΚΑΤΟΠΙΝ της εφαρμογης του θεωρηματος της παραγγωγης.

 Στην ανωτερω αποδειξη εχω προσθεσει ποιο λεπτομερης δικαιολογιες απο αυτες του συγγραφεως.

Eπισης διαφωνω με τον τροπο αποδειξεως του συγγραφεως.Π.Χ θα ειχα χρησιμοπ.το θεωρημα παραγωγης δυο φορες και ετσι δεν θα υπηρχε αναγκη να χρησιμοπ. το λιγο περιπλοκο θεωρημα της γραμμης (21 ) ,αλλα ποιο απλοι κανονες λογικης οπως της αφαιρετικης συζευξης, της αποσπασεως ,της διαζευξεως κ.τ.λ  κ.τ.λ

 Και εις την περιπτωση μας τον κανονα αντικαταστασεως : p vq <====> ~p--->q

 Aντεγραψα αυτη την αποδειξη για να δειξω οτι μπορει να εχουμε φορμαλιστικες αποδειξεις οχι 300 γραμμες αλλα μερικες γραμμες ,ακομα και σε δυσκολα προβληματα στην αναλυση.

ΕΠΙΣΗΣ αυτος ο τροπος αποδειξης ,η και λιγοτερο λεπτομερης ,ειναι ο ΜΟΝΟΣ ΤΡΟΠΟΣ ευρεσεως τυχων λαθων στις μαθηματικες αποδειξεις και ιδιως αυτων στην αναλυση.

Ετσι στην ανωτερω αποδειξη εχουν χρησιμοποιηθει οι εξης κανονες λογικης:

1)	p&q ===> p : κανων αφαιρετικης συζευξεως
2)	[p & (p---->q)]====> q : κανων αποσπασεως
3)	Κανων εξαλειψεως του γενικου.
4)	Κανων επαναφορας του γενικου
5)	Κανων αντικαταστασεωw : Εαν μια αληθης (αντιστοιχως ψευδης) συνθετη προταση περιεχεη μιαν προταση p και p<===>q ειναι αληθης,τοτε αντικαθιστωντες εις την συνθετη αυτη προταση την p με την q προκυπτει μια νεα αληθης ( αντστοιχως ψευδης) συνθετη προταση
6)	Kανων εξαλειψεως του υπαρξιακου
7)	[P&q---->] r<===> [p------>(~q v r)]: κανων στην προτασιακη λογικη ,δεν εχει ειδικο ονομα
8) Kαι το θεωρημα της παραγωγης η κανονας υποθετικης αποδειξεως

Ακολουθει η αποδειξη   $ \forall x\forall y ( x\equal{}0 \vee y\equal{}0\rightarrow xy\equal{}0 )$


Let $x_1, x_2, \ldots, x_m$ be positive real numbers.

Prove that if
\[ \left\lfloor n \sum_{j = 1}^m x_j \right\rfloor = \sum_{j = 1}^m \left\lfloor n \cdot x_j \right\rfloor  \]
Then at most one $x_j$ is non integral.

Are we assuming that $n$ is a positive integer?

If so, then note that $\lfloor a_1 + a_2 + ... + a_k \rfloor \geq \lfloor a_1 \rfloor + \lfloor a_2 \rfloor + ... + \lfloor a_k \rfloor$, where $a_1, a_2, ..., a_k$ are positive real numbers, and equality holds iff the sum of the fractional parts of $a_1, a_2, ..., a_k$ is less than 1.  If you can prove that for some choice of $n$ the sum of the fractional parts of $nx_1,nx_2,...,nx_m$ is greater than one, you have the proof.  I thought I had it, but I'd have to refine my method more, so I'll let somebody else do it.

[quote="Altheman"]what about x=0.1,0.2 and n=2

we have floor(.6)=floor(.2)+floor(.4)=0[/quote]
I think it means "for arbitrary n"; so, you might be able to find a specific n value that gives equality, but not all n will give equality.

Yes indeed, $n$ is an positive integer, and we suppose the condition holds for all positive integer $n$


1995-1996 Chapter Competition Sprint #13:

How many different, positive three-digit numbers can be made using any three of the following five digits: 1, 2, 2, 3, and 3?

[hide=" I think the answer is:"]
18
[/hide]

I know that you can do cases, and I've figured it out with cases..... and i know that you can just brute force it and just count it and hope you got all of them....but is there a way that you can do it, kinda like where you just simply multiply a few things together, like when you choose 4 ways to pick a hat, or something like that, and it's 4*3*2*1?

Thanks,
LHM

It's 18.
You can make the problem simpler, but still with some casework:
There are a few unique combinations you can get with 1,2,2,3 and 3.
122
123
133
223
233

Then, you can use the formula for the number of ways to arrange n letters/numbers with n repeated digits/letters w/e.

So 3!/2!+3!+3!/2!+3!/2!+3!/2!=3+6+3+3+3=18. This is slightly better than brute force, but I'm not sure of a better method.

[hide]
You can have 1, 2, 2 ; 1, 3, 3 ; 1, 2, 3
There are 3 ways to arrange them for both of the first ones and 6 for the last one.
You can also have 2, 2, 3 ; 2, 3, 3
These each have 3 ways to be arranged.
So 3+3+6+3+3 = 18. [/hide]

i don't know how to do the reveal hidden thing or latex, but here it is..

5 choices for the first one 4 choices for the second and so it turns out like this:
5X4X3
but we have to divide by 2! * 2! because the two threes and two twos are indistinguishable.
so we have
(5X4X3)/4 = 60/4 = 15 hmmmmmmm well you get the idea if not the answer

I actually posted that solution at first. But realizing it was wrong...anyway, in competition, I would use brute force, as time permits.

[hide]I don't think there is a super fast method of doing it in 1 second, but you could save time by:
assuming that the 1 didn't exist.. then you would have 4*3*2/4=6 possibilites
Then add the 1 in and make it so that 1 has to be one of the digits.
It can go in 3 different spots and for each spot, there are 2*2=4 different ways of choosing the two digits (you can do 2*2 because there are 2 digits for 2 and 3 so it doesn't matter)
3*4=12.
12+6=18. This method should save you maybe 10 seconds :)[/hide]


If $ a$ and $ b$ are integers such that $ x^2 \minus{} x \minus{} 1$ is a factor of $ ax^3 \plus{} bx^2 \plus{} 1$, then $ b$ is

$ \textbf{(A)}\ \minus{}2 \qquad \textbf{(B)}\ \minus{}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

hello, we have $ ax^3\plus{}bx^2\plus{}1\equal{}(xa\plus{}a\plus{}b)(x^2\minus{}x\minus{}1)\plus{}x(2a\plus{}b)\plus{}a\plus{}b\plus{}1$, the remainder must be zero, so we get the conditions
$ 2a\plus{}b\equal{}0$
and
$ a\plus{}b\equal{}\minus{}1$
solving this system we get $ a\equal{}1$ and $ b\equal{}\minus{}2$, so the answer is $ \text{A}$.
Sonnhard.


modular arithmetic
number theory unsolved
number theory

Prove $ 2^{2.3^{n\minus{}1}} \equiv 1 \plus{} 3^n mod \, 3^{n\plus{}1}$   for $ n \in \mathbb{N}$

we have $ 3^n \mid \mid 2^{2.3^{n\minus{}1 }}\minus{}1 \implies 2^{2.3^{n\minus{}1 }} \equiv 1\plus{}3^n.k \pmod {3^{n\plus{}1}}$
we need to prove $ k\equal{}1$.we use induction.
suppose that the congruency is true for all $ i \le n$.so we can suppose $ n \ge 2$.
we have
$ 3^{n} \mid \mid 2^{2.3^{n\minus{}1}}\minus{}1\equal{}(2^{2.3^{n\minus{}2 }}\minus{}1)((2^{2.3^{n\minus{}2 }})^2\plus{}(2^{2.3^{n\minus{}2 }})\plus{}1)$
and also we have $ 3^{n\minus{}1} \mid \mid 2^{2.3^{n\minus{}2}}\minus{}1,\frac{2^{2.3^{n\minus{}2}}\minus{}1}{3^{n\minus{}1}} \equiv 1 \pmod{3}$
so we just need to prove $ \frac{(2^{2.3^{n\minus{}2 }})^2\plus{}(2^{2.3^{n\minus{}2 }})\plus{}1}{3}\equiv 1 \pmod {3}$.or equivalently
$ (2^{2.3^{n\minus{}2 }})^2\plus{}(2^{2.3^{n\minus{}2 }})\plus{}1 \equiv 3 \pmod{9}$.which is easy to check!
DONE!


Do you mean trailing zeroes, or the number of zeroes in the whole product?

Please [b]do not[/b] use periods for multiplication signs. Learn $ \text{\LaTeX}$ instead.

Secondly, I don't think there is any way to find the number of zeros in that product, other than brute force (which is insane).

There are $ 500$ trailing zeroes (which we got by counting the factors of $ 5$) but there is no way to get the total number of zeroes without iterating algorithm and computer. Unless you want to multiply this and count them manually which is insane since $ 2009!$ has $ 5766$ digits.

there are 250 zeroes that are trailing

it is imposible to find the number of zeroes in the whole thing

No, there are $ 500$ of them:

$ \left \lfloor \frac{2009}{5}\right\rfloor \equal{}401$

$ \left \lfloor \frac{401}{5}\right\rfloor \equal{}80$

$ \left \lfloor \frac{80}{5}\right\rfloor \equal{}16$

$ \left \lfloor \frac{16}{5}\right\rfloor \equal{}3$

$ \left \lfloor \frac{3}{5}\right\rfloor \equal{}0$

$ 401\plus{}80\plus{}16\plus{}3\equal{}500$


$ 1$. Let ABC be an acute angled $ \bigtriangleup$; AD be the bisector of $ \angle BAC$ with $ D$ on $ BC$; and $ BE$ be the altitude from $ B$ on $ AC$. Show that $ \angle CED > 45^o$

$ 2$.Let $ a,b,c$be three natural numbers such that $ a < b < c$ and $ gcd(c \minus{} a,c \minus{} b) \equal{} 1$. Suppose there exists an integer $ d$ such that $ a \plus{} d,b \plus{} d,c \plus{} d$ form the sides of a right angled  $ \bigtriangleup$. Prove that there exist integers l,m such that $ c \plus{} d \equal{} l^2 \plus{} m^2$

$ 3$.Find all pairs $ (a,b)$ of real numbers such that whenever $ \alpha$ is a root of $ x^2 \plus{} ax \plus{} b \equal{} 0, \alpha^2 \minus{} 2$ is also a root of the equation.

 $ 4$.How many 6-digit numbers are there such that-:
a)The digits of each number are all from the set $ \{1,2,3,4,5\}$
b)any digit that appears in the number appears at least twice ?
(Example: $ 225252$ is valid while $ 222133$ is not)

$ 5$.A trapezium $ ABCD$ in which $ AB \parallel CD$, is inscribed in a circle with centre $ O$.Suppose the diagonals $ AC$ and $ BD$ of the trapezium intersect at $ M$,and $ OM \equal{} 2$.
a)IF $ \angle AMB \equal{} 60^o$ find with proof, the difference between the lengths of the parallel sides.
b)IF $ \angle AMD \equal{} 60^o$ find with proof, the difference between the lengths of the parallel sides.

$ 6.$ Prove that:
a)$ 5 < 5^{1/2} \plus{} 5^{1/3} \plus{} 5^{1/4}$
b)$ 8 > 8^{1/2} \plus{} 8^{1/3} \plus{} 8^{1/4}$
c)$ n > n^{1/2} \plus{} n^{1/3} \plus{} n^{1/4} \ \forall$ integers $ n\geq9$

[quote="abhishekj"]hey, when are the results for rmo-07 expected to be out?[/quote]
the chosen few have been informed

well that surely depends on the state which u are in  :wink:

hey anyone from delhi region having any info abt results??? please do reply

Where can I find solutions?
Thanks  :lol:
P.S:Is the answer for #4, 1055?

No the answer is 1405.
The soln to that problem is found in the foll. link :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169944 .


function
trigonometry
real analysis
real analysis unsolved

Check if the sequence of functions $ \sum f_n$ converges uniformly, if $ {f_n}$ is:

1) $ {nx\over\(n^{2}x^{2}+1} - {(n+1)x\over\(1+(n+1)^{2}x^{2}}$ x in R

2) $ {n^{2}(1-cos{x^{2}\over\(n^{2}})} , 2pi\geq|x|$

1) $ \forall x: \ S_n(x) \equal{} \sum_{k \equal{} 0}^{n} f_k(x) \equal{} \minus{} \frac {(n \plus{} 1)x}{1 \plus{} (n \plus{} 1)^2x^2} \to 0, n\to \infty$,
but
$ \|S_n\| \equal{} \sup_{x} \left| \frac {(n \plus{} 1)x}{1 \plus{} (n \plus{} 1)^2x^2} \right| \ge \left. \frac {(n \plus{} 1)x}{1 \plus{} (n \plus{} 1)^2x^2} \right|_{x \equal{} \frac {1}{n \plus{} 1}} \equal{} \frac12.$ So $ S_n$ converges to 0, but not uniformly.

2)

$ \|n^2(1 \minus{} \cos\frac {x^2}{n^2})\| \equal{} \max_{x\in[ \minus{} 2\pi,2\pi]} |n^2(1 \minus{} \cos\frac {x^2}{n^2})| \equal{} \left. n^2(1 \minus{} \cos\frac {x^2}{n^2}) \right|_{x \equal{} 2\pi} \equal{} n^2(1 \minus{} \cos\frac {4\pi^2}{n^2}) \equal{} 2n^2\sin^2\frac {2\pi^2}{n^2} \le \frac {8\pi^4}{n^2}$. And $ \sum_{k \equal{} 1}^{\infty} \frac {1}{k^2} < \infty$.
So, by Dirichlet test, we get:  $ \sum f_n$ converges uniformly on $ [ \minus{} 2\pi,2\pi]$


algebra
polynomial
quadratics
real analysis
real analysis unsolved

hello everyone,

I need to find how many solutions the following equation has:

${x^2+\sqrt{bx^4+cx^2+d}+e=0}$

I dont care what they are, just how many.

Thanks

hello, write your equation in the form
$ \sqrt{bx^4\plus{}cx^2\plus{}d}\equal{}\minus{}x^2\minus{}e$. By squaring both sides we get
$ bx^4\plus{}cx^2\plus{}d\equal{}x^4\plus{}e^2\plus{}2x^2e$. This is equivalent with
$ x^4(b\minus{}1)\plus{}x^2(c\minus{}2e)\plus{}d\minus{}e^2\equal{}0$. By setting $ x^2\equal{}t$ you get a quadratic equation $ t^2(b\minus{}1)\plus{}t(c\minus{}2e)\plus{}d\minus{}e^2\equal{}0$, which is easily to solve.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, write your equation in the form
$ \sqrt {bx^4 \plus{} cx^2 \plus{} d} \equal{} \minus{} x^2 \minus{} e$. By squaring both sides we get
$ bx^4 \plus{} cx^2 \plus{} d \equal{} x^4 \plus{} e^2 \plus{} 2x^2e$. This is equivalent with
$ x^4(b \minus{} 1) \plus{} x^2(c \minus{} 2e) \plus{} d \minus{} e^2 \equal{} 0$.
[/quote]

As you can see mickel your equation is a 4th degree polynomial which means that has 4 solutions(some may be real numbers and some complex)
Generally a polynomial with degree $ n$  has $ n$ solutions (real or complex or both )

There are [i]at most[/i] $ 4$ solutions (even counting multiplicity and even if $ b \neq 1$).  The use of the square root sign constrains $ bx^4 \plus{} cx^2 \plus{} d$ to be a non-negative real, so some of those solutions may not solve the original equation.

With $ b\equal{}1$, there is a case with infinitely many solutions: $ e<0$, $ bx^4\plus{}cx^2\plus{}d\equal{}(x^2\plus{}e)^2$.