abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ 0 \to M \to N \to P \to 0$ an exact sequence of modules, with N isomorphic to the direct sum of M and P. Is it true that the sequence must split? I don't think so, but I can't find a counterexample.

You could take the natural homomorphism $ P\rightarrow M\oplus P\simeq N$ as section of $ N\rightarrow P$. But I am wondering which definition of [i]split exact[/i] you use, since your question is almost the defining property of split exact I know.

In my definition the exact sequence splits if N is the direct sum of M and P and the first map is the natural injection, or the sequence is naturally equivalent to this one. I'm asking whehter the fact that N is isomorphic to the direct sum of M and P is sufficient to say that the sequence is naturally equivalent to $ 0 \to M \to M \oplus P \to P \to 0$, where all maps are the natural ones.

[quote="Third Edition"]You could take the natural homomorphism $ P\rightarrow M\oplus P\simeq N$ as section of $ N\rightarrow P$.[/quote]
why should it be a section? I also think that there is a counterexample.

Yes, there is a counterexample. My fault, I was a bit hasty with my first reply.

There is the subtlety that the candidate of a section I have given does not induce the identity on P. This is exploited in detail in the following counterexample. It is held on an abstract level to see what is going on:

(A) Let $ G$ be an abelian group and $ n>1$ a natural number such that multiplication $ n\cdot$ is surjective, but not injective on $ G$.

The induced exact sequence \[ 0\rightarrow G_n\rightarrow G\stackrel{n\cdot}{\rightarrow} G\rightarrow 0\]gives the short exact sequence \[ 0\rightarrow G_n\rightarrow G_n\oplus(G\oplus G_n^{\mathbb N})\rightarrow G\oplus G_n^{\mathbb N}\rightarrow 0\,\,\, (S)\] with the maps $ a\mapsto (0,a,0)$ and $ (a,b,c)\mapsto (nb,a\oplus c)$, respectively. Then the alleged section from above turns out to be $ (b,c)\mapsto(0,b,c)\mapsto(nb,c)$ which is [b]not[/b] the identity on $ G\oplus G_n^{\mathbb N}$.

(S) does not split: Otherwise there is a projection $ \pi: G\rightarrow G_n$ inducing the identity on $ G_n$. By assumption, given $ a\in G_n$ there is some $ b\in G$ with $ a\equal{}nb$ which implies $ a\equal{}\pi(a)\equal{}n\pi(b)\in n\cdot G_n\equal{}\{0\}$. This contradicts the assumption that $ G_n$ is non-trivial.

... and there are a lot of examples of groups satisfying the requirements (A).

good example :). so we may take $ G \equal{} \mathbb{Q}/\mathbb{Z}$.


The problem was to solve:

$\cos^{-1}(x)+\tan^{-1}(x) = \frac{\pi}{2}$

[hide="Here is my method"]Taking tangents of both sides,

$\tan(\cos^{-1}(x)+\tan^{-1}(x)) = \tan(\frac{\pi}{2})$

$\frac{\tan(\cos^{-1}(x))+x}{1-\tan(\cos^{-1}(x))\cdot x}= \infty$

So to make the RHS $\infty$, the denominator of the LHS must be $0$ (I don't know if this is valid...):

$1-\tan(\cos^{-1}(x))\cdot x=0$

$1-\sqrt{1-x^{2}}= 0$

$1=\sqrt{1-x^{2}}$

$1=1-x^{2}$

$x^{2}= 0$

$x=0$

A quick check reveals that I am right. But I think I should have taken the cosine of both sides to avoid the infinity. Still, is my method right?[/hide]

Is this valid?

I don't see why it is invalid...It makes sense.

Just be careful when dealing with infinity.  It's not a quantity which can be used in the same way as a finite number.  $\tan^{-1}(\pi/2)$ isn't just $\infty$, it's in the very least $\pm \infty$.  It worked in this case, but you have to be very careful.  Also, remember that $\frac{0}{0}$ is NOT $\pm \infty$, but rather indeterminate, so if you see that any expression over something else tends towards infinity, then you cannot just set the bottom equal to zero.  If you did this for $\frac{x}{x}$, for example, you can't just say $x=0$.

The cosine would've been safer.  Alternately, what you could've done was taken the cotangent, which would've given a $0$ on the RHS, and then you would've set the numerator instead of the denominator to zero.  Same deal.


Find the smallest positive integer $ u$ such that there exists only one positive integer $ a$ and satisfies the inequality
\[ 20u < 19a < 21u \ \text{?}
\]

[quote="tdl"]Are $ a,u$ integer?[/quote]

Yes, it is. My bad... I forgot to put that when I translated the problem.

[hide="Solution"]
Dividing both 19, $ \frac{20u}{19}<a<\frac{21u}{19}$ Since there's only one integer $ a$, $ \frac{21u\minus{}20u}{19}\le 2$, so $ u\le 38$. From there, it's simple casework.[/hide]

None of you guys solved the question yet though (that is, produce ONE answer).

And 123456789's answer is wrong.

Are you sure it's wrong? I actually tried out every number from 0-9 to make sure, and unless you round, it's correct.

Oh, $ a$ is an integer! That's the mistake 1234 - whateveritis and I made.

We might as well rewrite the condition as \[ \frac{u}{19}< a\minus{}u< \frac{2u}{19}\] So the left should be just below an integer, and the other should be just above an integer. Now I feel like $ u\equal{}10$ and $ a\equal{}11$ should work...

[quote="Silverfalcon"]You weren't very wrong.

[hide="Answer"]
38

You could do it your way but simpler solution is to see that because 19 is a prime, 19|u. Now, if u = 19, then there is no integer between 20u/19 and 21u/19 (20 and 21). So, we go to next smallest one, which is 38.

20*38/19 = 40
21*38/19 = 42

41 is only possible value so $ a \equal{} 41$ and $ u \equal{} 38$. [/hide][/quote]Why is $ u \equal{} 10$ and $ a\equal{}11$ wrong? They certainly satisfy the inequality, and is less than $ u\equal{}38$...

19 doesn't have to divide $ u $.Let $ x=20u $ and $ y=21u $.Note that $ x \equiv u$ $ (mod  19) $ and $ y \equiv 2*u $ $ (mod  19) $.Now,if we want there to be exactly one $ a $ satisfying the conditions we need to fin such $ u $ that $ y<x $ $ (mod  19) $.First such $ u $ is obviously 10 so by simple trying if it works we obtain the solution.


function
Euler
number theory proposed
number theory

Made this up on my own, though it may be a problem somewhere else. Not sure if this is the proper forum, but here goes...

Let $\phi(n)$ be the Euler totient function. Show that $\sum_{k=1}^{\infty}\frac{1}{\phi(n_{k})}$ converges, where $n_{k}$ is any integer with at least $k$ distinct prime divisors.

Do I understand correctly: You first define $n_{k}$ as any number with $k$ distinct prime factors and then build the sum¿

That would make the problem rather easy: $\varphi(n_{k}) \geq 2^{k-1}$ by this (just use the product formula and $p-1 \geq 2$ for all primes $p \neq 2$).
Thus $\sum_{k=1}^\infty \frac{1}{\varphi(n_{k})}< \sum_{k=1}^\infty \frac{1}{2^{k-1}}=2$.

Yes, you understand correctly. Never claimed it was hard :D, though you made it look pretty simple. Figured I'd put it up as it had stumped a friend of mine for a little bit. Nice solution.


Hello everybody  :D

I just want to ask all of you: Which are your favorite math sites ?

I'm sorry if this is not the correct thread to post in this kind of questions.  :lol:

http://www.mathlinks.ro/  ¿does anyone know it? ajajja


Is this forum going to be locked during the USAMO?  I remember it was  locked during the AIME.

In general, when do they lock forums and why?

Yes, it will be. It'll probably lock sometime Mondy night or something. They lock it to prevent discussion before the test has actually ended. The locking for the USAMO has been discussed in another thread, but basically they want to keep it as safe and secure as possible from any possible cheating. I think it will open up Thursday.

[quote="bubala"]In general, when do they lock forums and why?[/quote]We usually lock the forums to make everyone understand that the problems are not to be discussed while the forum is locked (not even in other parts of the site!).

We will be locking this forum tomorrow morning, before the USAMO begins.  It will be unlocked again on Thursday.


[quote="cj"]For subject classes, there are midterm and final challenges.

For problem solving classes, I would like to have a mock test at the end of class.

For example, Mathcounts class will give a simulating mathcounts test,  and AMC 10/12 giving simulating AMC 10/12 tests.[/quote]

In the MATHCOUNTS class we do run a kind of mock countdown on the last day.

We have discussed the idea of creating mock tests for the problem series.  However, that takes a lot of time.  We are looking for ways to do that efficiently without having to significantly increase the fees for these classes which are already very inexpensive for the amount of time and effort involved.

For now, I'm not saying yes or no.

It does seem to me however that students can take past exams on their own.  Do you feel uncomfortable timing yourself on a contest exam?


It is problem 2.52, chapter II, "Sang tao bat dang thuc", published in 8-2006. Two solution will be shown in the next version. Here is one


algebra
polynomial
function
calculus
derivative
number theory
totient function

Let $\Phi_n$ be the $n$-th cyclotomic polynomial.

[b]a)[/b] prove that for all $n\geq 3$ one has $\Phi_n'(1) = \frac{\phi(n)}{2} \Phi_n(1)$ and $\Phi_n'(-1) = - \frac{\phi(n)}{2} \Phi_n(-1)$.

[b]b)[/b] show that $\Phi_n'(x) \neq 0 $ for all $x \in \mathbb{C} , |x|=1$.

[b]c)[/b] define additionally $\Phi_0(x) := x$ and find all $n$ for which $\Phi_n(x)-1$ is product of some (not necessary different) $\Phi_i$ ($i\in \mathbb{N}_0$).

(here $\phi$ means Euler's totient function and $f'(x)$ the (standard) derivation of the polynomial $f$)

[quote="ZetaX"]Let $\Phi_n$ be the $n$-th cyclotomic polynomial.

[b]a)[/b] prove that for all $n\geq 3$ one has $\Phi_n'(1) = \frac{\phi(n)}{2} \Phi_n(1)$ and $\Phi_n'(-1) = - \frac{\phi(n)}{2} \Phi_n(-1)$.

[b]b)[/b] show that $\Phi_n'(x) \neq 0$ for all $x \in \mathbb{C} , |x|=1$.

[b]c)[/b] define additionally $\Phi_0(x) : = x$ and find all $n$ for which $\Phi_n(x)-1$ is product of some (not necessary different) $\Phi_i$ ($i\in \mathbb{N}_0$).

(here $\phi$ means Euler's totient function and $f'(x)$ the (standard) derivation of the polynomial $f$)[/quote]

 :)

Some small heuristic for [b]c)[/b]:
There are more of them than one would expect:
$n=1$: doesn't work
$n=2$: works
$n=3$: works
$n=4$: works
$n=5$: works
$n=6$: works
$n=7$: works
$n=8$: works
$n=9$: works
$n=10$: works
$n=11$: works
$n=12$: works
$n=13$: works
$n=14$: works
Guess for the next ones ;)
[hide]
$n=15$: [b]doesn't work[/b]
$n=16$: works
$n=17$: works
$n=18$: works
$n=19$: works
$n=20$: works
$n=21$: [b]doesn't work[/b]
[...]
next ones that don't work (should be complete for $n\leq 121$ and correct):
[hide]
$n =$
$30 , 33 , 35 , 39 , 42 , 45 , 51 ,$
$55 , 57 , 60 , 63 , 65 , 66 , 69 ,$
$70 , 75 , 77 , 78 , 84 , 85 , 87 ,$
$90 , 91 , 93 , 95 , 99 , 102 , 105 ,$
$110 , 111 , 114 , 115 , 117 , 119 , 120$
Yes they get more ;)
The density of the working ones is indeed $0$.
[/hide]
[/hide]

Part (b) follows from the fact that the roots of the derivative of a polynomial lie in the convex hull of the roots of the polynomial.

Isn't $\Phi_n(-1)=1, \forall n \in \mathbb{N}$? Then the statement of the second question in part a) may be simpler.

No, but $\Phi_n(\pm 1)$ is $1$ in most cases.
See my post http://www.mathlinks.ro/Forum/viewtopic.php?t=25794 for the hopefully correct values of them at these points.

Oops, sorry, $\Phi_n(-1)=1$ holds for odd $n$ only.


geometry
geometric transformation
reflection
MATHCOUNTS

This year we only received the Team Report for our state and the list of top 57 individuals and 25 teams.

Last year we also received:
- a copy of the tests and answers
- "Competition Statistics" page
-  "Score Distribution" - histograms for each test

Does this reflect what others have received?

Can anyone shed any light on this?

we got the tests and answers but i'm not sure about the others, at least our coach didn't let us see them...

If you post this concern on the mathcounts 'forrm', you are likely to get a reply from mathcoutns nationals since they do monitor that site.


suppose that
i) $\displaystyle \sum_{k=1}^n (a_k+b_k)$ converges, and
ii) $\displaystyle \lim_{n \rightarrow \infty} b_n =0 $.
Prove that the series $a_1+b_1+a_2+b_2+ ... $ converges.

Let $\sum_{k=1}^\infty (a_k+b_k) = L$. We are going to prove $a_1+b_1+a_2+b_2+\cdots=L$.

Given $\epsilon > 0$, there exists $N_1$ such that $\left|\sum_{k=1}^n (a_k+b_k)-L\right| < \frac\epsilon 2$ for $n>N_1$ and $N_2$ such that $|b_n|<\frac\epsilon 2$ for $n>N_2$. Let $N=\max\{N_1,N_2\}$.

Then for $n>2N$, the first $n$ terms of $a_1+b_1+a_2+b_2+\cdots$ is
(i) $a_1+b_1+a_2+b_2+\cdots+a_k+b_k (k>N)$
or
(ii) $a_1+b_1+a_2+b_2+\cdots+a_k+b_k+a_{k+1} (k>N)$

Case (i): since $k>N>N_1$, then $|(a_1+b_1+a_2+b_2+\cdots+a_k+b_k)-L| < \frac\epsilon 2 < \epsilon$ from the argument above.

Case (ii): since $k>N>N_1,N_2$, then $|(a_1+b_1+a_2+b_2+\cdots+a_k+b_k+a_{k+1})-L| \leq |(a_1+b_1+a_2+b_2+\cdots+a_k+b_k+a_{k+1}+b_{k+1})-L| + |b_{k+1}| < \frac\epsilon 2+\frac\epsilon 2 < \epsilon$.

liyi I think you mean $\left|\sum_{k=1}^n (a_k+b_k) - L \right| < \frac\epsilon 2$
my proof is esentially the same as liyi's, but I use some lemma  states that we can add infinitely many zeros to the converges series without lost its converges.. ;)
but actually what  I want to ask is this...
Consider  the harmonic series $\displaystyle \sum_{k=1}^n \frac{1}{k}$ it's clear that $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n}=0$ and the series
\[1-1 + (\frac{1}{2} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{3}) + ... \] converges to zero. But if we remove the parenthess
\[1-1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... \] has the sum $\frac{1}{n}$ if n is odd or $- \frac{1}{n}$ if n is even. So the series diveges.
It's "maybe" contradict the theorem that we just proved. I use $a_k = \frac{1}{k}$ and $b_k= \frac{1}{k}$
I must be wrong somewhere  :?  can you tell me where i'm wrong?

Well..It is not a counterexample.

$S_{2n} = 1-1+\frac{1}{2}-\frac{1}{2}+\cdots+\frac{1}{n}-\frac{1}{n} = 0$.
$S_{2n+1}=S_{2n}+\frac{1}{n+1} = \frac{1}{n+1}$.
So both $S_{2n}$ and $S_{2n+1}$ converge to 0.

What is wrong?

[quote="Diogene"]Let $L$ be the limit of  $\displaystyle S_n=\sum_{k=1}^n (a_k+b_k)$
Manifestly $a_1+b_1 + ... +a_{n-1}+b_{n-1}+a_n= S_{n}-b_n $ converges to $L$, and $a_1+b_1+ ... +a_{n-1}+b_{n-1}+a_n+b_n= S_n$ converges to $L$
 :cool:  Here you are !   :cool:[/quote]
There must be something wrong with me since I used $\varepsilon$-$N$.  :blush:  :blush:  :blush:

I can't understand what's the difference between $\sum_{k=1}^\infty (a_k+b_k)$ and $a_1+b_1+a_2+b_2+\cdots$. To my mind they are one and the same thing. Why am I wrong  :?

They're not one and the same thing, because they have different sequences of partial sums:

In one case, the partial sums are:

$a_1$
$a_1+b_1$
$a_1+b_1+a_2$
$a_1+b_1+a_2+b_2$
$a_1+b_1+a_2+b_2+a_3$ and so on.

In the other case, the partial sums are:

$a_1+b_1$
$a_1+b_1+a_2+b_2$
$a_1+b_1+a_2+b_2+a_3+b_3$ and so on.

The second sequence of partial sums is a subsequence of the first sequence. The convergence of a whole sequence implies the convergence of any subsequence, but it is possible to have a convergence subsequence of a divergent sequence.


geometry
3D geometry
prism
rectangle
Pythagorean Theorem

A rectangular floor is made up of identical square tiles. The floor is 40 tiles long and 30 tiles wide. If a straight line is drawn diagonally across the floor from one corner to the other, how many tiles will it cross?

I think it's sixty but I have no mathematical proof of it...

Let's say the tiles are one by one
So the length of the line form one corner to the other most be, by pythagorean theorem greater than 50.

[hide] I don't know if this proves anything. But looking at a 4 by 3 grid, we can see the diagonal intersects 6 of the tiles. $6*10=60$
So I think the answer is $\boxed{60}$
Ok, looking at a 8 by 6 grid, we can see that it is just 2 4 by 3 grids with 6 intersected tiles each and 2 4 by 3 grids which are not intersected. Going to a 12 by 9 grid, I believe there will be 3 4 by 3 grids with 6 intersected tiles each and 6 4 by 3 grids which are not intersected. Continuing up to 40 by 30, there will be 10 4 by 3 grids, 6 intersected tiles each. So $6*10=\boxed{60}$ tiles. :D [/hide]

Yeah that seems to make sense.  I couldn't find a general formula, but when I did specific cases I came up with $60$ as well.

[hide="explanation to the correct answer that everyone has"]
This is called inclusion-exclusion. What you want to do for a rectangle is take the number of vertical lines it will cross, add it to the number of horizontal lines it will cross, then subtract the number of vertices it will cross. The formula ends up as:

For an $a\times b$ rectangle, the diagonal will cross $a+b-\gcd{(a,b)}$ squares.

In this case, a=40 and b=30 so the squares it will cross is $40+30-\gcd{(40,30)}=70-10=60$.
[hide="expansion to higher dimensions"]
For higher dimensions, you have to include, exclude, include, exclude, include again, and so on. So for 3 dimensions you have:

For an $a\times b\times c$ rectangular prism, a space diagonal will cross $a+b+c-\gcd{(a,b)}-\gcd{(a,c)}-\gcd{(b,c)}+\gcd{(a,b,c)}$ $1\times1\times1$ cubes.
[/hide]
[/hide]

The problem with any formula is the case of an $n$ by $n$ grid... In that case it would only cross $n$ tiles, where the line would cross through only vertices.

Actually, in a $n$ by $kn$ grid, the same situation would occur, and the line would always cross $kn$ tiles.  In other situations it's difficult to come up with a formula...

[quote="Jhonny Ruiz"]I once saw a problem just like this one, the only difference was that it used a rectangular prism instead of this one just a two-dimensional rectangle.
Let's say a 3*4*5 prism,
Any idea of how to solve it...

[hide="hint"]I think that the formula to find the measure of the line that goes from one of the cornes to the other is...
$\sqrt{x^2+y^2+z^2}$[/hide][/quote]

So, using Hamster1800's technique, we have
[hide] $3+4+5-gcd(3,4)-gcd(4,5)-gcd(3,5)+gcd(3,4,5)=12-1-1-1+1=10$
So $\boxed{10}$ unit cubes is the answer. :D[/hide]

For generalization of PIE: http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html


f_n(x)=1+x+(x^2)/2!+...+(x^n)/n!

Prove that for any natural n,f_n doesnt have more than one real root

Source:Unknown (I mean,i know him,but you dont  :D )

i have seen a similar problem in my advanced algebra book.

At first, f_n'=f_{n-1},

so all roots of f_n are simple (if a is a multiple root, that a is a root of f_n'=f_{n-1} and so the root of of f_n-f_{n-1}=x^n/n!, so a=0, but f_n(0)=1).

Therefore, if f_n has at lest two real roots, it has two consecutive roots a and b such that f_n(x)<0 for a<x<b. Then, there exists c, a<c<b, such that f_n'(c)=0 (Rolle theorem). So, f_{n-1}(c)=0 and f_n(c)<0, then
c^n/n!=f_n(c)-f_{n-1}(c)<0. If n is even, it is already a contradiction. If n is odd, then n-1 is even, so f_{n-1} has not real roots, so f_{n-1}=f_n' is always positive. so f_n increases, so 0=f_n(b)>f_n(a)=0 - a contradiction.


Hint:

[hide]

What remainder does 60 divided by k^2 give if k^2 is bigger than 60?

[/hide]

Another hint: [hide]If 60 divided by k2 leaves a remainder of 6, then k2 must divide 60 - 6.[/hide]

i would've just forced it, checking all the squares. there's not a lot of them. but.......

[hide]you could do 60-6=b*x^2. just go on from there[/hide]


Let $A$ the set of numbers divisible by the number of their divisors. Then $n!$ belongs to $A$ for all $n>6$ and $A$ has density zero.

as for your first question. I think that the following statement is stronger:
let $a_{i}\le \frac{n}{i}$ then $a_{1}a_{2}\cdots a_{n}|n!$ :)


function
trigonometry
calculus
calculus computations

How to determine period of functions?:
$ y\equal{}e^{\minus{}x}sinx$
$ y\equal{}\frac{(sinx)^2}{2\plus{}cosx}$

Well, you know the periods of $ \sin x$ and $ \cos x$ are $ 2\pi$. so the second function has a period of $ 2\pi$. The first function: $ e^{x}$ is not periodic (unless you are using complex variables, in which case the period is $ 2\pi i$), so the function is not periodic.


hey guys, i think this question is stupid but i am still confused with this...

adsortion is a exothermic process, but rate of chemisorption increasses with temp???
WHYYY????? :huh:  :huh:  :huh:

Because activation energy is reqd to start the adsorption process. after this there is decrease in the amount adsorbed with increase in temp as it becomes exothermic, even though heat evolved is less compared to physical adsorption

but doesnt adsorption follow Le chattlier 's principle since it is also an equilibrium..

Physical adsorption is reversible....not chemical

is chemical adsortion a thermodynamic process or kinetic process?????

[quote="valeriummaximum"]Then how does chemisortion increase with temp....though adsorption is an exothermic process[/quote]

Chemisorption is not exactly an equilibrium reaction to follow Le Chat.. principle.

And you forgot Arrhenius equation???? Rate inc with inc in temp... Don't tell me you forgot that.

that's what i clearly asked in my earlier posts Shreyas, that whether chemisorptuion is KINETIC or THERMODYNAMIC......

All the procesess can be analysed both thermodynamically and kinetically there is nothing like a kinetic and thermodynamic process valerium. :roll:


Find all / prove that there are infinitely many primes $ p$ for which the following condition holds:

$ \forall _{a \in \{2,3,...,p \minus{} 2 \} } \exists _ { k(a) \in \mathbb {N} } : p | a^{2^{k(a)}} \plus{} 1$

Comment: My friend's told me that it's equivalent to certain known hypothesis. Which ?  :wink:

all these primes P are $ P\equal{}2^k\plus{}1$. If $ k\equal{}md$ had odd divisor d, then $ 2^{md}\plus{}1\equal{}(2^m\plus{}1)(2^{(d\minus{}1)m}\minus{}2^{(d\minus{}2)m}\plus{}...\plus{}1)$. Therefore work only Fermat's primes $ P\equal{}2^{2^k}\plus{}1.$ I think, that there are only finite Fermat's primes.

The question of the cardinality of the Fermat primes is open.  The only known Fermat primes occur when $ k \equal{} 0, 1, 2, 3, 4$.


probability
combinatorics unsolved
combinatorics

Hello everyone, if anyone could assist me with this question, it would be much appreciated!

I am a computer science student studying in part statistics for NL Holdem Poker games.

In NL holdem you start with 2 cards hidden to all other opponents.  Then 5 cards are eventually dealt as community cards.

For each 169 starting hands, I have written a program to iteratre through every combination of single opponents 2 cards against each combination of 5 flop cards.  This gives a total of:

50 choose 5 * 45 choose 2 iterations (2.1 billion-ish)

The program on each iteration counts (lets say for simplicity of the argument) the total wins and losses.  So say you held the cards AA it might come back with a number that you will win 80% of the time with 1 opponent (assuming the game is continued to the end regardless).

Now the problem is, how to work out the numbers for n players.  Adding more players becomes to computationally expensive to perform.  I thought that probability^number of opponents would be correct, but I have been told that since the opponents cards are independant of each other this is only a good estimate.  So for example against 1 opponent, .80^1  = 80%, against 2 opponents .80^2 = 64% etc.

Is this true?  Will I have to iterate through all numbers of opponents as well if I want the exact figures?

Thanks for any help.

you need to write the program yourself, or you just want the results?  because there's programs out there that do this, pokerstove being one of them.

and it's true that what you were thinking would only give you an estimate, because like you said player 2's cards are not independent of player 1's.  consider these 2 situations.

1.  Player 1 has AA, player 2 has AA, player 3 has 66

2.  Player 1 has AA player 2 has KK, player 3 has 66

its clear player 3 has a higher chance of winning in case 1 than in case 2

I don't think I can help very much  :blush: , however I know a very good internet page where you will probably find the answer to your question and if some more come up http://www.pokerstrategy.org.uk/
it's not for playing, on the contrary, they show you how to, so I think you'l probably find what you're looking for.
good luck with your programme! :wink:

why are actually writing it? for fun, or is it kind of a homework??  :huh:

This seems junky, but how many cards of each kind hearts spade ectc. are in a card set? I missed one on a practice once, regarding probability of drawing heart and blah blah blah.  :(


Please visualize a figure as explaned.

$AB$ is a diameter of the given circle and $O$ is the centre. The diameter $AB$ is extended to a point $X$ which is far away from $B$.

$P$ is a point in the exterior of the circle.

The task is to construct a line from $P$ to $AX$ such that the constructed line from $P$ to $AX$ is perpendicular to it.

Conditions:
a) Only a long scale/ruler is available so you can draw straight lines as long as you wish
b) The ruler does not have measurement marks, so we cannot use it directly to make a square etc etc...
c) Of course, no compass etc also available.

I hope this is posted in right section...

[quote="shobber"]Are you asking to construct a perpendicular line through a given point outside a given circle to a given diameter?[/quote]

yes... but within the given constraints.

[quote="shobber"][quote="xxxyyyy"][quote="shobber"]$B$ is the othorcenter of $\triangle{APQ}$, thus $AB \perp PQ$.[/quote]
How do you prove that $B$ is the orthocenter of $\triangle{APQ}$ ?[/quote]
Because AB is the diameter.[/quote]

Just to confirm...
Let the point of intersection of circle and line $AP$ be $R$.
Let the point of intersection of circle and line $AQ$ be $S$.
Then, we have $\angle{ARB}$ $\angle{ASB}$ and as 90 degrees...

QED.


The curve $ y \equal{} ax^2 \plus{} bx \plus{} c$ goes through the points $ (1;1),(2;m),(m;2)$ , where $ m\in\mathbb{R} \minus{} \{2\}$ Find all real values for $ m$ so that the function admits minimum value.

Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, from the condition, we have
$ f(1) \equal{} 1\Longleftrightarrow a \plus{} b \plus{} c \equal{} 1\ \cdots [1]$
$ f(2) \equal{} m\Longleftrightarrow 4a \plus{} 2b \plus{} c \equal{} m\ \cdots [2]$
$ f(m) \equal{} 2\Longleftrightarrow am^2 \plus{} bm \plus{} c \equal{} 2\cdots [3]$

$ [2] \minus{} [1]: 3a \plus{} b \equal{} m \minus{} 1\ \cdots [4]$
$ [3] \minus{} [2]: (m^2 \minus{} 4)a \plus{} (m \minus{} 2)b \equal{} \minus{} (m \minus{} 2)$, since $ m\neq 2$, we have $ (m \plus{} 2)a \plus{} b \equal{} \minus{} 1\ \cdots [5]$

$ [5] \minus{} [4]: (m \minus{} 1)a \equal{} \minus{} m$, since $ m\neq 1$, we have $ a \equal{} \minus{} \frac {m}{m \minus{} 1}$, thus from $ [4]],\ [1]$, yielding $ b \equal{} \frac {m^2 \plus{} m \plus{} 1}{m \minus{} 1},\ c \equal{} \minus{} \frac {m^2}{m \minus{} 1}$.

$ f(x)$ has minimum value $ \Longleftrightarrow a > 0\ or\ a \equal{} b \equal{} 0$, for $ m\in\mathbb{R}$, $ m^2 \plus{} m \plus{} 1 > 0$, so the desired condition is

$ a > 0\Longleftrightarrow m(m \minus{} 1) < 0$ with $ m\neq 1$ and $ m\neq 2$, yielding the desired answer is $ \boxed{0 < m < 1}$.

Should $ a<0$ for the parabola to have a maximum value (other than infinity)?
(so that the parabola appears to have a upside-down U shape)

Ah, sorry, I read the question incorrectly (it asks for minimum instead of maximum)...

$ [3]\minus{}[2]$ killed the problem, I hadn't seen that.

Very clear solution, thank you kunny.


I was wondering weather anyone here will be watching the World Cup of Hockey, it starts today.

CANADA WON! I didn't watch it, but everyone in the school was wearing the Canadian flag the day after the final game and talking about it! :P  :)  :lol:

:D Canadian Pride
Watched all the games (even the placement round)
Too bad no more hockey :'(

[quote=nicarmt]you ruined it :rotfl:[/quote]

 :rotfl:


Let $c \geq 1$ be an integer, and define the sequence $a_1,\ a_2,\ a_3,\ \dots$ by \[ \begin{aligned} a_1 & = 2, \\ a_{n + 1} & = ca_n + \sqrt{\left(c^2 - 1\right)\left(a_n^2 - 4\right)}\textrm{ for }n = 1,2,3,\dots\ . \end{aligned} \] Prove that $a_n$ is an integer for all $n$.

Usually the trick is to introduce $a_{n-1}$

Here it works also  :D

$a_{n+2} = 2\ c\ a_{n+1} - a_n$

I think that tµtµ has said almost right.

My solution is like this:

[hide]\[\begin{array}{l}
 {a_{n + 1}} - c{a_n} = \sqrt {\left( {{c^2} - 1} \right)\left( {a_n^2 - 4} \right)}  \\
 a_{n + 1}^2 - 2c{a_n}{a_{n + 1}} + 4{c^2} + a_n^2 - 4 = 0 \\
 \end{array}\]
Also,
\[a_{n + 2}^2 - 2c{a_{n + 1}}{a_{n + 2}} + 4{c^2} + a_{n + 1}^2 - 4 = 0\]
Do substruction and organize,
\[\begin{array}{l}
 \left( {{a_{n + 2}} + {a_n}} \right)\left( {{a_{n + 2}} - {a_n}} \right) = 2c{a_{n + 1}}\left( {{a_{n + 2}} - {a_n}} \right) \\
 {a_{n + 2}} = 2c{a_{n + 1}} - {a_n} \\
 \end{array}\]

At last use the mathematical induction as
\[\begin{array}{l}
 {a_1} = 2 \\
 {a_2} = 2c \\
 \end{array}\]
When \[{a_n}\& {a_{n + 1}}\] are all integers
\[{a_{n + 2}}\] is to be an integer.

Proved.[/hide]

Both of their ideas are perfectly correct. I agree to both. Good job westwind and tµtµ.


How does the AP Calc test scoring work? thank you

45 multiple choice questions, 6 free-response questions.  For each multiple choice question you answer correctly, you get 1 point.  If you omit a question, you neither gain nor lose points.  If you answer wrong, you lose $ 1/4$ of a point.  So you are given a raw score defined by: # correct-$ 1/4$ # incorrect=raw score.  This raw score is then multiplied by 1.2 for a maximum possible score of 54.  Each free-response question is worth up to 9 points for 54 points total.  Your responses are graded by hand and compared to a rubric and points are awarded based on how complete/correct your solution is.

So the maximum total score is 108.  Typically to score a 5, you will need to score around 65-75 points or higher.  In other words, shoot for 37/45 multiple choice (assuming you get the other 8 wrong), and 35/54 on the free-response.  This will yield a score of 77, which is almost certainly a 5.

and last year accroidng to wiki over 40 percent got 5s and i think all you needed was 60ish/108 or something

good thing for me   :)

Keep in mind that the reason the number is so high is because most of the people who take AP Calc are rather good at math so it is likely that they will score quite well.


Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$, where $x$ is measured in degrees and $100< x< 200$.

[hide="hint2"]$\cos{(x+y)} + \cos{(x-y)} = 2\cos{x}\cos{y}$[/hide]
[hide="which means that"]$\cos{x}+\cos{y}=2\cos{\frac{x+y}{2}}\cos{\frac{x-y}{2}}$[/hide]
[hide="solution"] The left hand side is$\cos^3{3x}+\cos^3{5x}$ which becomes
$(\cos{3x}+\cos{5x})(\cos^2{3x}-\cos{3x}\cos{5x}+\cos^2{5x})$
by factoring.

Using the identity in the 'which means that' hint above on the first factor:
$(2\cos{4x}\cos{x})(\cos^2{3x}-\cos{3x}\cos{5x}+\cos^2{5x})=8 \cos^3{4x} \cos^3{x}$
Then, $\cos{4x}=0$ or $\cos{x}=0$, yielding $112.5, 157.5$ in the interval

Dividing by $2\cos{4x}\cos{x}$:
$\cos^2{3x}-\cos{3x}\cos{5x}+\cos^2{5x} = 4 \cos^2{4x}\cos^2{x}$

From the hint 'hint2', the equation becomes
$\cos^2{3x}-\cos{3x}\cos{5x}+\cos^2{5x} = (\cos{5x}+\cos{3x})^2$

Multiplying out the right hand side and then subtracting gives
$-3\cos{3x}\cos{5x} = 0$
so $\cos{3x}=0$ or $\cos{5x}=0$, yielding $150, 126, 162, 198$.

$112.5+157.5+150+126+162+198=906$.[/hide]

Lovely little problem - one of the few on the test I enjoyed.

[hide]

First, notice that . Expanding this last expression out and canceling out the two single cubed terms, we get

The constant is irrelevant, and we can see that any degree that makes a single one of the  on the RHS equal to  makes the whole thing . Since  has zeroes , then . Doing this all four times, we see that

The only values of  that fall in our desired range are . The sum of all of these is .[/hide]

So I posted this in another thread but thought it would be appropriate to put here -

[hide]
I really enjoyed 12, though I missed it to thinking $\cos 0^{\circ}=0$, etc. on the last step.

Here's my solution:

$\cos^{3} 3x + \cos^3 5x=8 \cos^{3}4x \cos^{3} x$
$\cos^{3}(4x-x)+ \cos^3 (4x+x)=(2 \cos 4x \cos x)^{3}$
$(\cos 4x \cos x+\sin 4x \sin x)^3+(\cos 4x \cos x- \sin 4x \sin x)^{3}=(2 \cos 4x \cos x)^{3}$
Let $\cos 4x \cos x=a$, $\sin 4x \sin x=b$.

Then $(a+b)^{3}+(a-b)^{3}=(2a)^{3}$, multiplying this out and simplifying gives

$a(a+b)(a-b)=0$

So you have three cases:
$\cos 4x \cos x=0$, so $\cos x=0$ or $\cos 4x=0$
$\cos 4x \cos x+ \sin 4x \sin x=0$, so $\cos 3x=0$
$\cos 4x \cos x- \sin 4x \sin x=0$, so $\cos 5x=0$.

Then find the values of $x$ in which $100<x<200$ that work for the three equations and add them up.[/hide]

Darn, forgot sum-to-product. Does anyone have a quick proof for it so that I can easily rederive it whenever I need it?

Use the product to sum identities to derive the sum to product identities ;)

Nice problem :)

Note that $\cos \theta + \cos \gamma = 2\cos \frac{\theta+\gamma}{2} \cos \frac{\theta-\gamma}{2}$. Now, let $\theta$ be $5x$ and $\gamma$ be $3x$. We get $\cos 3x + \cos 5x = 2\cos 4x \cos x$. This looks surprisingly similar to the right hand side, and we will use this to our advantage. Let $a=\cos 5x$ and $b= \cos 3x$. We know the left hand side is $a^3+b^3$, and by the expression we got previously, the right hand side is $(a+b)^3$. We now have $a^3+b^3=(a+b)^3$. Cancelling out terms, we get $3ab(a+b)=0$. From this, we get $3$ cases.

Case 1: $a=0$. In this case we need $\cos 5x =0$. By the specified interval, $500<5x<1000$, and the values which work are $90+180n$ for integers $n$. The ones in the range which work are $630, 810,$ and $990$. The values of $x$ for each of these are $126, 162,$ and $198$ respectively.

Case 2: $b=0$. If we have this, then we need $\cos 3x = 0$. Since $100<x<200$, we have $300<x<600$. The values which work are $90+180n$, where $n$ is an integer, and the ones in this interval are $450$, and dividing by $3$, we get $x=150$ for this case.

Case 3: $a+b=0$. Well we know that $a+b= 2\cos 4x \cos x$, so we need either $\cos 4x=0$ or $\cos x=0$. No values of $x$ in the specified range work, so we check $\cos 4x=0$. We know that $400 < 4x<800$, and the values which work are in the form $90+180n$ for integers $n$, so the ones in the interval are $450$ and $630$, which get us $x$ values $112.5$ and $157.5$ respectively.

Adding up all the values of $x$, we get $\boxed{906}$.

Solved with [b]sarun[/b], [b]js104[/b], [b]crazyeyemoody[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]

Let $a=\cos3x$ and $b=\cos5x$. Then, $a^3+b^3=(a+b)^3$ so $a=0$, $b=0$, or $a+b=0$. Therefore, $x=150$, $x=126$, $x=162$, $x=198$, $x=112.5$, or $x=157.5$, so the sum is $\boxed{906}$.


[hide]$(a^{-1})^{\frac{1}{b}}=x$ raise each side to -b^2 power it all cancel and you like get 1/x^(b^2)[/hide]

first post is right. kudos.

$\frac{1}{a^{b^{2}}}= a^{-b^{2}}$

 :P

[quote="skimnc"][quote="ckck"]If $(\frac{1}{a})^{\frac{1}{b}}=x$, find $a^{b}$ in terms of x and b. (ab not equal to 0, $a>0$ and $b>0$)[/quote]
Solution:
[hide]$(\frac{1}{a})^{\frac{1}{b}}=x$
$\ln (\frac{1}{a})^{\frac{1}{b}}=\ln x$
$-\frac{1}{b}\cdot \ln a=\ln x$
$b \ln a=-b^{2}\ln x$
$\ln a^{b}= \ln x^{(-b^{2})}$
$a^{b}= x^{(-b^{2})}$
$a^{b}= \boxed{(\frac{1}{x})^{b^{2}}}$[/hide][/quote]\

Very nice solution.  I was trying to do some exponent stuff but then I looked at your post and realized its possible to do it with logs and ln's. Nice solution man!


MATHCOUNTS
AoPS Books
geometry
angle bisector

Dear all,
Does anyone here have any tips to share on winning MathCounts & other State or National Maths COntest?

One of the best tips I could give would be to work for a lot of speed, since mathcounts is pretty much based on how fast you do the math.  On the sprint round, you should aim for less than 30 sec on the first ten, then 45 sec-1 min on the next ten, leaving you with ample time to solve the last ones, which are the hardest.  Also, don't check your work on a problem right after solving it.  If you feel good about it, most likely you got it right and you'd be wasting time.  If you don't feel good about it, then most likely you won't feel better if you waste your time checking it.  You'd be better off using the time remaining after you're done with all the problem to check your answers.  I'm quite sure most top national competitors are good at doing this.  I was, if I may say myself, extremely good at doing this, and got me pretty close to the top at nationals.

[quote="ZzZzZzZzZzZz"]I was, if I may say myself, extremely good at doing this, and got me pretty close to the top at nationals.[/quote]

Just out of curiosity, what's your name?  I might know you from last year's yearbook. :)

I don't think that's a necessary thing to ask.

Besides, winning MathCounts takes a lot of work and effort. You probably have to work problems from the AoPS books in the MATHCOUNTS bookstore. Those problems are pretty good problems to prepare for, especially the harder ones. Try to do them well with good accuracy and speed, because MathCounts is almost all about speed. In addition, you should also practice your mental math skills/skills w/o a calculator because that may help you in the Countdown Round. Finally, the main key is simply to always be practicing MATHCOUNTS leveled problems. And last of all, get a good night's sleep the day before the competition.

During the contest, you should also always check your answers if there's time. Don't lose very easy points just for making a careless error.

About the checking answers thing-If you're sure you have it right, don't spend the time right then checking it, and if you think it's wrong, don't spend minutes trying to do that problem. Instead, do them all, and then check. It's kind of a waste of time to check answers before you've finished all the questions.

Ugh practicing, I just basically look over problems and try to do the problems that I don't know how to do, I just skip the ones I know how to do...
I think it saves time and stuff.

I'm not sure if you should do that. First, it makes you better at it, and second, it'll help you do it faster, so incase a problem like that comes up (especially in countdown), you can get the answer fast.

you know why I don't do that, Peijin?  All I need to work on for most problems is to NOT MAKE STUPID MISTAKES.

Yeah alan makes a lot of "stupid mistakes".

no wonder Hao thinks he can be smarter than you.  :rotfl:

I wouldn't know about winning mathcounts because I haven't come close yet.  unless you count getting top 6th grader at states.  Just add on to what everyone else said this thing: Don't go in the contest thinking "I have to win."  You'll probably kill yourself over all the pressure.

I kill myself with pressure anyways.
-dies for the 781th time-

We're getting off topic-there's no need for that kind of insult/offensive talk there, ZhangPeijin and alanchoul.

I think it would be best to take a break and have some fun on hte day before a competition- Math is not like Vocabulary- You can memorize vocab within like 1-3 minutes. On the other hand, Math can take days to get good at a new concept thingy..............OR  you could just spend your last day recapping on formulas/ thereoms like angle bisector or bionomial, etc

As ppl said before me, try working through the problems, but not necessarily every one.  Then, try leaving 5 mins at the end of the sprint to check ur work.  It makes more sense on sprint to get all the questions u answer rite.  But, if u ever make the cd round, try going as fast as u can.  In cd, it's not worth the time to check ur answer when ur competing with some1 else.

for sum reason, that also works for me and I do the same thing.

[quote="abacadaea"][quote="gauss1181"]Right. That's also one of the tips for SAT test takers. People can treat MATHCOUNTS like the SAT, only, longer and harder.[/quote]

isnt mathcounts based on the SAT?[/quote]

Mathcounts isn't based on the SAT.  Mathcounts is definitely much harder, and if you gave a Mathcounts national competition to a bunch of people who scored a perfect 800 math on the SAT, only about .1% would get a perfect score, and a lot would get less than 50% correct.

[quote="andrewliu"][quote="penny_zf"]Hi Everyone,
Thank you for all your sharing and experiences in the MathCounts. The tips are very interesting and valuable. Will keep practicing to hone my speed.
I agree that it is quite difficult to cram at the last min for a major competition.
If you all can think of other tips or websites or competition strategies such as how to calm those nerves before a competition, do share them with me and the rest.
Thanks for now![/quote]
Hmmm.... personally, I like to chew gum before and during a test (don't ask me why- its probably "good luck" or it just tastes good- idk), circulate my blood (usually by slight jumps) and take deep breaths.....[/quote]

Strangely enough, I do that, too, but I chew gum because peppermint stimulates brain activity. :lol:

Just make sure u can do problems quickly. Try to do school tests in 15-20 minutes. Thats wat i do.

I quite frankly do not understand the point on this 'tips for winning' topic...seeing as you just have to do good...just know the stuff and there's nothing much else you can do. i've won a couple different competitions in mathcounts and your main focus should be *countdown* if you really care about *winning*. after all, it's something one must do...

hi - several on here said how chewing gum during a test relaxes them, most schools don't allow gum, but even if they did, do you think it relaxes the guy or girl next to you to hear chomp, chomp, chomp, and Worse, those cacklers, cracklers, bubble popping ones?? that would be like chalk on a blackboard sound to many -- really, quite inconsiderate - how'd you feel if someone sat and tapped their pencil loundly constantly on the table while you were taking the sprint round? Please do not chew gum in a test room full of anxious test takers just like yourself - silence should be the norm - please put others comfort above your personal needs - I realize you may never have thought about this before -that's why I'm posting

Um, guyinPA, are you like a test proctor or grader?  Because if you aren't then you probably don't understand that chewing gum doesn't make any sounds at all.  Unless, of course, as you said, they blew bubbles with it, but who in the world would actually have the time or concentration to blow bubbles while taking an intensive math test?  Just to let you know, unless if you know or are one of those people who chew their gum like buck-toothed horses (no offense), chewing gum regularly doesn't distract people nearly as much as pencil-tapping, which I agree is extremely annoying.  Just hope that clarifies a few points :D

mg - I was waiting on line at Disney last week and one of those "horse chewers" was next to me in line - I wasn't even taking a test and I wanted to scream at the lady "shut up"! -- she was [i]That[/i] loud (and Disney is not exactly quiet Christmas week) - so I was thinking about that when I read the gum chewing posts ... If she were next to me in a test,  arrrgghhhh!!!!!!!

Playing For The Win helps some, but don't memorize everything and then miss everything at State where you don't know how to do any problems.


A point $ P$ is randomly selected from the square region with vertices at $ (\pm 2, \pm 2)$. What is the probability that $ P$ is within one unit of the origin? Express your answer as a common fraction in terms of $ \pi$.

To find the probability, we need to find the ratio of the desired area to the whole. The square around it has area 4*4= 16.
The area we want is a circle of radius 1. This has area $ \pi$.
Our final answer is $ \pi / 16$

How is this level 19? This is literally just area of a circle and area of a square and done.

Idk
It has to do with your previous problems


If x+y and y+3x are two factors of
$ax^3-bx^2y+xy^2+y^3$
the third factor is.....

[hide="not at all sure"]This is a homo expression in x,y
expression=(y+x)(y+3x)(y+mx)
=$y^3+(m+3+1)y^2x+(3+m+3m)yx^2+3mx^3$
Simple solving gives $-3=m$
so third factor is (y-3x)?[/hide]

[quote="E^(pi*i)=-1"]I am not sure about this solution...

[hide]By using polynomial division on the two factors, we get a+b=0 and (1/3)(1+(1/3)(b+(1/3)a))=1, so b=9, a=-9.  We now divide in the two factors again and are left with -9x+3y.[/hide][/quote]
hmmmm..you are getting 3(y-3x) and I'm getting y-3x...... :huh:

Well since the $y^3$ term doesn't have a coefficient of $3$, I'd put my money on $y-3x$


function
integration
trigonometry
calculus
limit
real analysis
real analysis unsolved

Of a given real function defined over $ {\mathbb R}$ we know that $ f(c \plus{} h) \equal{} f(c) \plus{} f'(c)h \plus{} {1\over2}Ah^2 \plus{} o(h^2),$ $ A$ constant. Can we say that $ f^{\prime\prime}(c)$ exists? If not give a counterexample

This is false.

Let $ c \equal{} 0$ for convenience.

Let $ f(x) \equal{} \int_0^xt\sin\left(\frac1t\right)\,dt.$

Integration by parts gets us that

$ f(x) \equal{} \left. \minus{} t^3\cos\left(\frac1t\right)\right|_0^x \plus{} 3\int_0^xt^2\cos\left(\frac1t\right)\,dt$

Making the natural estimate of the right hand integral gets us that $ f(x) \equal{} O(x^3)$ as $ x\to0.$ So this satisfies the hypotheses with $ f'(0) \equal{} 0$ and $ A \equal{} 0.$

But $ f'(x) \equal{} x\sin\left(\frac1x\right)$ for $ x\ne 0$ and

$ f''(0) \equal{} \lim_{h\to0}\frac {f'(h)}{h} \equal{} \lim_{h\to 0}\sin\left(\frac1h\right)$ does not exist.

Yes, just an observation: in the last line you would say $ f^{\prime\prime}(0)\ldots$

The function $ f(x) \equal{} x^3\sin (1/x)$ has the same properties as Kent's example.


Let $A$ be a set of $n(k+1)$ points in $\mathbb R^{k}$, every $k+1$ of them being affinely independent.
Prove that we can partition these points into $n$ equal sets such that the simplexes or simplices determined by the sets do not intersect.

[i]or, if you prefer,[/i]

Let $A$ be a set of $3n$ points in $\mathbb R^{2}$ (the good ol' plane), every $3$ of them not being collinear.
Prove that we can partition these points into $n$ equal sets such that the triangles determined by the sets do not intersect.

proof:
the set of points has a convex hull
the convex hull can be triangulated


The numbers 1, 2, . . . , 12 have been written around a circle in an arbitrary order. In one step it is allowed to exchange places of two adjacent numbers if their absolute difference is greater than 1. Prove that it is possible in a finite series of moves to rearrange the numbers in the natural order.

move 11 around the circle in the direction that does not hit 10 so that it is adjacent to 12.
Then if 10 is not adjacent to 11, we move the 12-11 group around 11 leading so that hits 10.
If 9 is not adjacent to 10, we move the 12-11-10 group around, 10 leading, so that it hits 9.
Continue till the question is pwned.


complex analysis
complex analysis unsolved

$ z^{2}f'(z)\plus{}zf(z)\equal{}1$. Assuming that the equation has an analytic solution in the neighbourhood of 0, $ f(z)\equal{}\sum_{k} a_{k}z^{k}$, find the coefficients $ a_{k}$ recursively.

I know how to solve the equation directly, but when I substitute the series into the equation, I get:
$ \sum_{k} (k\plus{}1)a_{k}z^{k}\equal{}1$, and what I'm supposed to do now?

Well, depending on how much rigor you need in your solution...

The left- and right-hand sides of that equation are both series in $ z$.  If they're equal it's a pretty good guess that their [i]coefficients[/i] are also equal.  So $ a_0\equal{}1$, $ a_1\equal{}0$, etc.

Of course, this would only work if your equation is correct, which it is not.  If you could post the steps you took to get there, we could help you identify the error.  I think that probably you weren't so careful matching up like powers of $ z$.

The equation you wrote down does not have an analytic solution near zero- the left-hand side is guaranteed to have constant term zero.


In $ \triangle ABC$, $ H$ is the foot of the altitude from $ A$, $ M$ is the midpoint of $ AH$, $ D$ is the point of tangency of the incircle at $ BC$, and $ DM$ intersects the incircle again at $ N$. Prove that $ \angle BND \equal{} \angle CND$.

posted many times, for example here, :)

http://www.mathlinks.ro/viewtopic.php?t=301127&start=20
prob.9


modular arithmetic
Integer sequence
number theory
Divisibility
Sequence
IMO Shortlist

Prove that there exists two strictly increasing sequences $(a_{n})$ and $(b_{n})$ such that $a_{n}(a_{n}+1)$ divides $b^{2}_{n}+1$ for every natural n.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

It suffices to find $c_n,d_n$ s.t. $a_n|c_n^2+1,\ a_n+1|d_n^2+1$, because then, since $(a_n,a_n+1)=1$, we can find $b_n\equiv c_n\pmod {a_n},\ b_n\equiv d_n\pmod{a_n+1}$. Let's show that $a_n=5^{2n}$ works.

We have $a_n+1=5^{2n}+1=(5^n)^2+1$, so all we need to show is that there is some $c_n$ s.t. $5^{2n}|c_n^2+1$. We can show something more general: if $p$ is a prime of the form $4k+1$, then there is a $t_n$ s.t. $p^n|t_n^2+1$. We construct $t_n$ inductively. $t_1$ clearly exists, and assuming that $p^{n-1}|t_{n-1}^2+1$ we have to find $k\in\overline{0,p-1}$ s.t. $p^n|(t_{n-1}+kp^{n-1})^2+1$. This is reduced to $p|\frac{t_{n-1}^2+1}{p^{n-1}}+2k$, and it's clear that we can find such a $k$.

[b]OFFICIAL SOLUTIONS[/b]

[i]Solution 1:[/i]

[b]Lemma[/b]: If $a,c \in \mathbb{N}$ and $a^{2}|c^{2}+1$ then there exists $b \in \mathbb{N}$ such that $a^{2}(a^{2}+1)|b^{2}+1$.

[b]Proof[/b]: Indeed $a^{2}|(c+a^{2}c-a^{3})^{2}+1$ and $a^{2} + 1|(c+a^{2}c-a^{3})^{2}+1$, so we can take $b=c+a^{2}c-a^{3}$.

Using the lemma we see that it is enough to find strictly increasing sequences $(a_{n})$ and $(c_{n})$ such $a^{2}_{n}|c^{2}_{n}+1$ for every $n \in \mathbb{N}$. This can be realized by taking for instance $a_{n}=2^{2n}+1,c_{n}=2^{na_{n}}$. In this case

\[
c^{2}_{n}+1=(2^{2n})^{a_{n}}+1=(a_{n}-1)^{a_{n}}
\]
is divisible by $a^{2}_{n}$.

[i]Solution 2[/i]:

We will use factorizations over the ring $\mathbb{Z}[i]$. Take $(p,q,r)$ a Pythagorean triple $p^{2}=q^{2}+r^{2}$ and $a=p^{2}$. Then

\[
a(a+1)=(q^{2}+r^{2})(p^{2}+1)=(q+ri)(p+i)(q-ri)(p-i)
\]
must divide $(b+i)(b-i)$. Therefore we have to find a Gaussian integer x+iy such that

\[
(q+ri)(p+i)(x+yi)=b+i
\]
that is we have to find integers $x,y$ such that

\[
(pr+q)x+(pq-r)y=1
\]
It is easy to see that if $p,q,r$ are coprime then $pr+q$ and $pq-r$ are also coprime: if $d$ is a prime divisor of $pr+q$ and $pq-r$, then

\[
d|q(pr+q)-r(pq-r)=q^{2}+r^{2}=p^{2},
\]
hence $d|q$, which implies $d=1$. Thus the existence of $x,y$ is proven. The construction of the sequences $(a_{n})$ and $(b_{n})$ is now obvious.

Observe that since $b_n + (a_n)(a_n+1)$ works in lieu of $b_n$, we just need to find infinitely many $a_n$ such that there exists a $b_n$ where $a_n * (a_n + 1) | (b_n)^2 + 1$

Sorry to revive, but say we need $p^k | b_{n}^2 + 1$. Assume that there is a solution to $p | b_{n}^2 + 1$ for $b_{n}$. (quadratic reciprocity gives that these primes are of the form $1 \pmod 4$. Then since $b_n \not \equiv 0 \pmod p$, Hensel's lemma lets us lift these powers. Hence, it remains to show that we can find infinitely many $a_n$ such that $a_n$ and $a_n + 1$ have no prime factors which are $3 \pmod 4$ and are not multiples of $4$. But this is easy; indeed, observe that numbers of the form $4a^2 + 1$ have this property. Then just take $(4a^2+1)^2$ and $(4a^2+1)^2+1$ for $a_n$, and use Chinese Remainder theorem for the rest.

Let $a_n$ =$(q_n)^2$ where $(q_n)_{n\ge 1}$ is an increasing sequence of primes congruent to 1 mod 4.Let $a_n+1$ =$(p_1)^{\alpha_1}....(p_t)^{\alpha_t}$. Choose $b_n$ congruent to $r_i$ modulo $(p_i)^{\alpha_i}$ such that $(p_i)^{\alpha_i}$ divides $(r_i)^2 +1$ for $i$ equals $1,2,...,t$. also let $b_n$ congruent to $R_n$ modulo $(q_n)^2$ such that $(q_n)^2$ divides $(R_n)^2 +1$. Clearly by Hensel's Lemma such a construction is possible and such a  $b_n$ exists by Chinese Remainder Theorem.Hence we found such sequences.(We can make $b_n$ strictly increasing quite easily)

Consider Pell-type equation $x^2-5y^2=-9.$ Take $x_i=2b_i$ and $y_i=2a_i+1$ and the rest follows.

[hide="solution"]
If $p\mid b_n^2+1$, then $b_n^2\equiv -1\pmod{p}\implies p\equiv 1\pmod{4}$ or $p=2$. In addition, $v_2(b_n^2+1)\le 1$. Thus, $a_n(a_n+1)$ has only prime factors that are $1\pmod{4}$ and at most one factor of $2$.
In addition, given any $a_n$ such that $a_n(a_n+1)$ has only factors $1\pmod{4}$, then if $b_n<b_{n-1}$ then we can take instead $b_n' = b_n+ka_n(a_n+1)$ to ensure that $(b_n)$ is strictly increasing.

Thus, it remains to prove that there exists an infinite number of $a_n$ such that $a_n(a_n+1)$ has only factors $1\pmod{4}$. Take $a_n=(4n^2+1)^2$. By the lemma used in the first line, $a_n$ has only factors $1\pmod{4}$. In addition, $a_n+1=(4n^2+1)^2+1$ only has factors $1\pmod{4}$ for the same reason. Finally, $a_n$ is odd and $v_2(a_n+1)=v_2((\text{odd})^2+1=1$ so $v_2(a_n(a_n+1))=1$. Thus this works and we are done.[/hide]

My solution (along with some motivations): Note that it suffices to show that there exist two increasing sequences $x_n$ and $y_n$ such that $x_n^2 \equiv -1 \pmod{a_n}$ and $y_n^2 \equiv -1 \pmod{a_n+1}$, cause then, we can take the sequence $b_n$ with $b_n \equiv x_n \pmod{a_n}$ and $b_n \equiv y_n \pmod{a_n+1}$ (The existence of such a sequence $b_n$ is guaranteed by CRT). This works as $b_n^2 \equiv x_n^2 \equiv -1 \pmod{a_n}$ and $b_n^2 \equiv y_n^2 \equiv -1 \pmod{a_n+1}$, and so, using the fact that $\gcd(a_n,a_n+1)=1$, we get $b_n^2 \equiv -1 \pmod{a_n(a_n+1)}$.

The fact that $a_n+1 \mid y_n^2+1$ inspires us to take $a_n=t_n^2$, for some increasing sequence $t_n$. Then taking $y_n=t_n$ works. Notice that the condition that the sequences are increasing is superfluous, as after finding the sequences, we can simply arrange them in increasing order. Renaming $t_n$ as $a$, our problem is reduced to the following-
[quote=Remaining problem]Show that there exist infinitely many positive integers $a$, such that $-1$ is a quadratic residue modulo $a^2$.[/quote]
Suppose $p$ is a prime divisor of $a$. Then $\left(\frac{-1}{p} \right)=1$, where the fraction denotes Legendre's symbol. Using Euler's criterion, we get $$1 \equiv \left(\frac{-1}{p} \right) \equiv (-1)^{\frac{p-1}{2}} \pmod{p} \Rightarrow p \equiv 1 \pmod{4}$$
This motivates us to take $a=p$, such that $p \equiv 1 \pmod{4}$, which evidently gives that $-1$ is a quadratic residue modulo $a^2$. Hence, done. $\blacksquare$

One liner :) :
One can show that the pair of sequences $a_n=(4n^2+1)^2, b_n=(8n^3+6n)+(4n^2+1)^2*(8n^3-4n^2+6n-1)$ works by polynomial division, hence this pair of sequence is one of the desired pairs, hence proved.

[hide=Simple Solution] Let $a_i = p_i^2$, where $p_i$ is the $i$th prime $\equiv 1$ mod $4$. Then, for any prime divisor $q$ of $p_i^2(p_i^2 + 1)$ with $v_q(p_i^2(p_i^2 + 1)) = k$, $q \equiv 1$ mod $4$, and since $q^k$ has a primitive root $g$, by setting $b_i \equiv g^{\frac{(q - 1)q^{k - 1}}{4}}$ mod $q$, we have $b_i^2 \equiv -1$ mod $q^k$. Finally, by the Chinese Remainder Theorem, we can choose $b_i$ such that $b_i^2 \equiv -1$ mod $p_i^2(p_i^2 + 1)$ and can add $p_i^2(p_i^2 + 1)$ sufficiently many times to $b_i$ so that the sequence $b_n$ is strictly increasing.

Note that $a_n= {y_n}^2$ where $y_n$ is the n-th solution to $x^2-2y^2=-1$ works

Not that hard for an N3, considering that I already did two other problems (off AoPS) about divisors of $n^2+1$ beforehand.

From the problem, it is clear that for all $n$, we must have $$b_n^2=-1 \pmod {a_n(a_n+1)}.$$ It is well-known by quadratic reciprocity that only $-1$ is only a quadratic residue modulo $2$ or any power of a prime $1$ modulo $4$. Therefore $a_n(a_n+1)$ must overall only contain factors of primes $1$ modulo $4$ and at most one factor of $2$.

We claim that $a_n = p_n^2$, where $p_n$ is the $n$th prime equivalent to $1$ modulo $4$, follows the above condition. $a_n=p^2$ follows the conditions (and doesn't include a factor of $2$), while $a_n+1=p^2+1$ is itself one more than a square and therefore can only have factors of $1$ mod $4$ primes, with at most one factor of $2$, proving the claim.

Considering every maximal prime power factor of $p_n^2(p_n^2+1)$, if $b_n$ is equal to a certain residue modulo that prime power, then that prime power will divide $b_n^2+1$. Therefore, by CRT, there will exist infinitely many $x$ so that $a_n(a_n+1)$ divides $x^2+1$. Therefore, we can make $b_n$ increasing, finishing the proof.

ISL marabot solve.

Let $p_n$ denote the $n$th prime that is $1\pmod 4$ (there are infinitely many prime $1\pmod 4$ by Dirichlet's).

Consider $a_n=p_n^2$.

Then $a_n(a_n+1)=p_n^2(p_n^2+1)$.

So $b_n^2\equiv -1\pmod {p_n^2(p_n^2+1)}$, so we need $b_n^2\equiv -1\pmod {p_n^2}$ and $b_n^2\equiv -1\pmod {p_n^2+1}$.

Lemma: For any prime $q\equiv 1\pmod 4$, there exists a positive integer $s$ so that $q^2\mid s^2+1$ or in other words $s^2\equiv -1\pmod {q^2}$.
Proof: Let $f(x)=x^2+1$. Then by the Law of Quadratic Reciprocity, there exists an $r$ so that $q|f(r)$. Now note that $f'(r)=2r\not\equiv 0\pmod q$, so by Hensel's Lemma, there is a positive integer $s$ so that $q^2|f(s)=s^2+1$.
$\blacksquare$

Let $s$ be a residue $\pmod{p_n^2}$ so that $s^2\equiv -1\pmod{p_n^2}$.

Consider $b_n\equiv s\pmod{p_n^2}$ and $b_n\equiv p_n\pmod{p_n^2+1}$. It's clear that this satisfies $b_n^2\equiv -1\pmod{a_n(a_n+1)}$.

Note for any $n$, we can take $b_n$ to be arbitrarily large, larger than $b_{n-1}$, so we can construct the sequence $(b_n)$ that's strictly increasing.

Doesn't $a_n=n$ and $b_n=\sqrt{n^2+n-1}$ work? This solution feels to obvious to be right.

Or do $a_n$ and $b_n$ have to be integers?

bump

can anyone confirm or deny that my sequences work?

Note that we can always increase $b_n$ by $a_n(a_n+1)$ and $a_n(a_n+1)$ will still divide $b_n^2+1$. So, the condition that $(b_n)$ is increasing doesn't matter since it is always possible to make it increasing if it isn't. So, this problem is equivalent to showing that there's a strictly increasing sequence $(a_n)$ such that $-1$ is a quadratic residue modulo $a_n(a_n+1)$. This is equivalent to showing that for arbitrarily high integers $a$, $-1$ is a quadratic residue modulo $a$ and $a+1$.

Assume towards a contradiction that there are a finite number of primes $p$ such that $-1$ is a quadratic residue modulo $p$. Let $p_1, p_2,\ldots, p_n$ be these primes. Then, let $q$ be a prime such that $q|(p_1p_2\cdots p_n)^2+1$. Then, $-1$ is a quadratic residue mod $q$, but $q$ is relatively prime to $p_1, p_2, \ldots, p_n$, a contradiction. So, there are infinitely many primes $p$ such that $-1$ is a quadratic residue mod $p$.

Let $p$ be an odd prime such that $-1$ is a quadratic residue mod $p$. Then, there exists $a$ such that $a^2\equiv-1\pmod{p}$. Then,
$$(a-2^{-1}a^{-1}(a^2+1))^2\equiv a^2+4^{-1}a^{-2}(a^2+1)^2-2\cdot a\cdot 2^{-1}a^{-1}(a^2+1)\equiv a^2-2\cdot a\cdot 2^{-1}a^{-1}(a^2+1)\equiv a^2-(a^2+1)\equiv -1\pmod p^2$$
so $-1$ is a quadratic residue modulo $p^2$. Clearly, $-1$ is a quadratic residue modulo $p^2+1$. Since $p^2$ can be arbitrarily large, this completes the proof.

Edit: @above I interpreted it as meaning integers but it never specifies that so technically your solution works

It comes down to finding infinitely many $a_n$ for which $-1$ is a quadratic residue mod both $a_n$ and $a_{n}+1.$ Call a number $n$ good if $-1$ is a quadratic residue mod $n$. If coprime $m,n$ are good then $mn$ is good.
\\$~$\\
$1,2$ are good but no other powers of $2$ are. However, for primes $p>2,$ if $-1$ is a quadratic residue mod $p$, no new restrictions are added for higher powers of $p$ so $-1$ is a quadratic residue mod $p^2.$
\\$~$\\
Prime $p$ is good iff $\left(-1\right)^{\left(\frac{p-1}{2}\right)}\equiv 1\pmod n$ iff $p\equiv 1\pmod 4.$ We claim that $p^2$ and $p^2+1$ are both good. Clearly $p^2+1$ only contains one power of $2.$
\\$~$\\
$p^2$ being good is trivial, so let us show that if odd $q\mid p^2+1$ then $q\equiv 1\pmod 4.$ If $d$ is the order of $p$ mod $q$ then $d\mid 4$ but $d\neq 2$ which implies $d=4.$ However, $d\mid q-1$ so $q\equiv 1\pmod 4$ as desired.

Note that all we have to do is find some not necessarily increasing sequence $(b_n)$, as we can add $a_n(a_n+1)$ to $b_n$ allowing us to make it as big as possible. Thus the problem is equivalent to finding infinitely many $a$ for which $-1$ is a QR $\pmod{a}$ and $\pmod{a+1}$. We claim that any number $a$ with no $3 \pmod 4$ prime factor works. Since $a^2+1$ only has $1 \pmod 4$ prime factors, the following lemma will finish by CRT.
[color=#f00][b]Lemma:[/b][/color] $-1$ is a QR $\pmod{p^n}$ for any prime $p \equiv 1 \pmod{4}$.
[color=#f00][b]Proof:[/b][/color] We use induction on $n$. The base case is well known. For the inductive step, assume there exists $r^2 \equiv -1+mp^n \pmod{p^{n+1}}$. We then have that $$(p^n \frac{m}{2r} - r)^2 \equiv p^{2n} \frac{m^2}{4r^2} - m p^n + (-1+mp^n) \equiv -1 \pmod{p^{n+1}},$$ so the lemma is proven and we are done.

@above i don't believe
[quote]any number $a$ with no $3 \pmod 4$ prime factor[/quote]
works? take $a = 65$, then $a$ has only $1 \pmod{4}$ factors but we can clearly see that $-1$ is a NQR $\pmod{66}$...

Consider the sequence of numbers $p_n$ that is defined to be the $n$th prime that is $1 \pmod{4}$ (by Dirichlet, infinitely many exist.) I claim $a_n = p_n^2$ works. For this it suffices to show that $-1$ is a quadratic residue $\pmod{p_n^2}$ and $\pmod{(p_n^2+1)}$, as then we define $b_n$ to be sufficiently large and satisfy $b^2_n \equiv -1 \pmod{p_n^2(p_n^2+1)}$.

[color=#f00][b]Claim: [/b][/color] $-1$ is a quadratic residue $\mod p_n^2$.

[i]Proof.[/i] It is known that $-1$ is a quadratic residue $\mod p_n$ - now use Hensel's lemma to lift to $\mod p_n^2$. $\blacksquare$

[color=#f00][b]Claim:[/b][/color] $-1$ is a quadratic residue $\mod p_n^2 + 1$.

[i]Proof.[/i] It is well-known that $p_n^2 + 1$ has no factors that are $3 \pmod{4}$. For each odd prime factor of $p_n^2 + 1$, we may (using Hensel's lemma and the fact that $-1$ is a QR for all primes that are $1 \pmod{4}$) lift to a sufficient exponent of the prime factor. To finish, use CRT to stitch these prime factors together (and $2$, using that $-1 \equiv 1$ is a quadratic residue $\mod 2$) to obtain that $-1$ is a QR $\mod p_n^2+1$, as desired. $\blacksquare$

For the reasons mentioned above, we are done. $\square$

Edit: the solution is wrong since 4 isn’t square free. Will fix later.

Apparently this stronger version that there are infinite $(a,b)$ such that $a^2+a = b^2+1$ is also true.

Note that it suffices to prove there are infinite integer solutions to $a^2+a \mid b^2+1$ as we can just start with any $(a_1,b_1)$ and repeatedly choose $a_k>b_{k-1}$.

It suffices to find infinitely many $(a,b)$ such that
\[a^2+a = b^2+1 \iff a = \frac{-1 \pm \sqrt{4b^2+5}}{2}\]
However, since there are infinite solutions to the Pell Equation $c^2-4b^2 = 5$ (since $(c,b) = (3,1)$ works), we can just take that value of $b$ to get
\[a = \frac{c-1}{2}\]
Since $c$ must be odd by the Pell Equation, $a$ is a positive integer, so there are infinitely many $(a,b)$ as desired.

The fact that $b$ is increasing is not really relevant, since we can just increase it by some multiple of $a_n(a_n+1)$ if it is not. Thus, the problem essentially asks us to prove that there are infinitely many $n$ for which $-1$ is a quadratic residue mod $n(n+1)$.\\

We claim that $n=p^2$ for some prime $p\equiv1\pmod{4}$ works. We claim that $p^2(p^2+1)$ only has 1 mod 4 prime factors and 2, which would solve the problem. Suppose $q$ is an odd prime such that $q\mid p^2+1$. This means that $-1$ is a quadratic residue mod $q$, which means that $q$ is 1 mod 4. Thus, $p^2(p^2+1)$ only has 1 mod 4 prime factors, so $-1$ is a quadratic residue, done.

The problem basically forces the solution; it's just a matter of getting details to work.

We pick $a_n = 5^{2n}$ for each $n$, such that all divisors of $a_n(a_n+ 1)$ are $1$ mod $4$. Furthermore, there exists a $b_n$ such that $b_n^2 \equiv -1 \pmod {p^k}$ modulo every $1$ mod $4$ $p \mid a_n(a_n + 1)$, [hide=which is well known]This follows say from Fermat Christmas and multiplicativity of sum of squares. There exists a nontrivial $a^2+b^2 = p$ and hence a nontrivial solution $a_k^2+b_k^2 = p^k$ by the recursion $a_{k+1} = a_ka-b_kb$, $b_{k+1} = ab_k + a_kb$. Here $\gcd(a_kb_k, p) = 1$. Then we may take $b_n \equiv \frac{a_k}{b_k} \pmod {p^k}$.[/hide] by Chinese Remainder theorem. It follows that this sequence works.

The condition that $(b_n)$ is increasing doesn´t matter as we can always add a suficientely large factor of $a_n(a_n+1)$ to $b_n$. We claim that $-1$ is a quadratic residue $\pmod {p^2(p^2+1)}$, which is enough to finish;  for a $p\equiv 1\pmod 4$ prime, it suffices to show that a prime divisor of $p^2(p^2+1)$ is $1\pmod 4$ which it´s true as $p^2$ has only $1\pmod 4$  prime factors as well as $p^2+1$


Hi.
I need some help about a problem -- irrationality of e^n.
[hide]proof that for all n(n is natural number) e^n is irrational[/hide]
We know Fourier's proof about e. (ofcourse we can do this with transcendent number concept)
And I applicate it to  (to solve problem for) e^n. But I can't get anything.
Can someone help me about it?

Thanks..

[quote="#5-1Mole"]As e is transcedent, it isn't a integer root of any number, so any of its natural degrees can't be integer.[/quote]

Ofcourse guy, I know it.. But I'm asking do we have any way to proof it without analysis?
could anyone suggest a sollution?
thanks

(Sorry, I know you already know that but  FWIW ....)

 .There is  a (very) simple proof when n=1, :)  and also when  n=2 (There is a post in AoPS  with that proof - you might have to do search)
For general n I  have not seen any simple proof, but the it is  fairly well known (very similar to pi is irrational proof)  for example,  one you can see here:
http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality.html

Here is a cut and paste from my earlier post wrt to proof that $ e^2$ is irrational-  which does not require calculus.

[hide]
Proof by contradiction:

Assume $ e^{2}$ and be written and $ p/q$ where $ p$ and $ q$ are positive integers.

So we have $ p e^{ \minus{} 1} \equal{} q e$

Or $ 0 \equal{} q e \minus{} p e^{ \minus{} 1}$
Now let us take $ n > p \plus{} q$ and multiply both sides with $ n!$

$ 0 \equal{} n! \left( \sum_{k \equal{} 0}^{\infty}\frac {q \minus{} p ( \minus{} 1)^{k}}{k!}\right)$
= $ n! \left( \sum_{k \equal{} 0}^{n }\frac {q \minus{} p ( \minus{} 1)^{k}}{k!}\right) \plus{} n! \left( \sum_{k \equal{} n \plus{} 1}^{\infty }\frac {q \minus{} p ( \minus{} 1)^{k}}{k!}\right)$

First part is obviously an integer, the absolute value of the second part:

 $ \left\lvert n! \left( \sum_{k \equal{} n \plus{} 1}^{\infty }\frac {q \minus{} p ( \minus{} 1)^{k}}{k!}\right) \right \rvert$
$ \le (q \plus{} p) \left( \sum_{k \equal{} n \plus{} 1}^{\infty }\frac {n!}{k!}\right)$
	$ \equal{} (p \plus{} q)\sum_{k \equal{} n \plus{} 1}^\infty \frac {1}{(n \plus{} 1)(n \plus{} 2)\dotsb k}$
  	$ \le (p \plus{} q)\sum_{k \equal{} n \plus{} 1}^\infty (n \plus{} 1)^{n \minus{} k}$
  	$ \equal{} (p \plus{} q)\sum_{t \equal{} 1}^\infty (n \plus{} 1)^{ \minus{} t}$
  	$ \equal{} (p \plus{} q)\frac {1}{n}$

Which is less than 1, the only integer less than 1 is zero, so second part has to be zero for [b]every[/b] large arbitrary $ n$ (But one can easily show that this is not the case   -- (sum from  n to infinity is not equal to sum from n+1 to infinity as nth term is not zero etc ..)

Contradiction!

So $ e^{2}$ can not be written as $ p/q$

QED[/hide]

[hide]I have seen "elementary" prrof  for e^4  also... [/hide]


geometry
3D geometry
tetrahedron
symmetry
pyramid
combinatorial geometry

[img]http://www.artofproblemsolving.com/Community/Images/Problems/P17.gif[/img]

If you've seen this one before, please let someone else tackle it.

[hide]For any collection of 6 points as described, we can choose 4 of them (A, B, C, and D) such that the other two lie outside tetrahedron ABCD (choose the one with smallest volume for example). I will refer to plane ABC as plane d, and similarly with the other three planes. I will also say that a point P is "inside" plane d if it is on the same side of plane d as point D, and the point is "outside" if it is on the opposite side of the plane (excuse the lousy notation). Since a point can be inside or outside of each of 4 planes, except a point can't be outside all four planes, there are 2^4-1 = 15 regions. Also, since E and F (the other two points) are outside tetrahedron ABCD, there are only 14 regions to consider.

Let E and F be the fifth and sixth of these points. Notice that if E is outside plane a but inside the other three planes, then the blue segment AE passes through the interior of red triangle BCD. Similar arguments work with the other 3 similarly defined regions. If E is inside planes a and b but outside c and d, then blue segment AB passes through the interior of red triangle CDE. Again, this also works for the other 5 similar regions. Thus, we have dealt with 10 of the 14 regions.

The last set of 4 regions ar the ones where E and F are outside three of the planes and inside the fourth. There are two cases:

Case I: both E and F are inside a and outside b, c, and d.

For this case, consider the tetrahedron EBCD. If F is inside this tetrahedron, rename E and F, so that F is outside of tetrahedron EBCD. Since A is inside this tetrahedron and F is outside the tetrahedron, segment FA must intersect one of the four faces of the tetrahedron.

Case II: E is inside a and outside the other three, while F is inside b while outside the other three.

For this case, F must lie outside tetrahedron EBCD, while A must be inside it. So, as before, FA must intersect one of the 4 faces of the tetrahedron.

Whew![/hide]

Cool approach - can anyone find a similar solution by considering the convex hull of the 6 points?

[quote="rrusczyk"]can anyone find a similar solution by considering the convex hull of the 6 points?[/quote]

Richard's hint leads to a short, sweet, marvelous proof, which the margins of this posting will not contain.

Meanwhile, I had been contemplating this totally different approach based on a 2D convex hull and coloring edges.  It's a bit unwieldy, and is incomplete.  Does anybody care to complete my proof?

Consider a projection of the 6 points onto any plane in which the points form a 2-dimensional, 6-point convex hull (i.e., a hexagon), with the three main diagonals of the hexagon intersecting at a single point.

With six points in space, there are a total of 6C3/2=10 pairs of triangles to be considered.  In this proof, 4 pairs of triangles are trivial and need not be considered.  Thus, I concern myself primarily with the 6 pairs shown in the attached illustration.  In the projection onto the chosen plane, each of the 6 pairs of triangles intersect to form a quadrilateral (color-coded in the diagram).  Each quadrilateral is disjoint from the others, and the union of the six quadrilaterals forms a smaller hexagon inside the convex hull.

Considering the triangles in 3-space, modes of overlap of one pair of triangles are shown in the illustration.  These triangles may overlap, without piercing each other in two ways.  Or one triangle of a pair may pierce the other in four ways.  These triangles may also mutually pierce each other (i.e., interlock with each other) in additional ways, not shown.  Though not shown, the same logic holds for the other five pairs of triangles I want to consider.

My proof is by contradiction.  Let's try to find a case where none of the triangles pierce each other.

As is evident by the diagram of piercing modes, a pair of triangles will *not* pierce each other if opposite sides of the quadrilateral are both above the other pair of opposite sides or if they are both below the other pair of opposite sides.  I.e., if the quadrilateral's sides were labelled a, b, c, and d in order, then if sides b and d are above sides a and c there is no piercing.  Likewise if sides b and d are below sides a and c there is no piercing.  In all other cases, one triangle will pierce the other.

I attempt to "color" each side of each of my six quadrilaterals to avoid piercing.  Start with the quadrilateral with sides abcd.  We arbitrarily choose side a to be "low".  Then side c must also be low.  And sides b and d must be "high".  We proceed to the quadrilateral with sides cefg.  We have already assigned side c to be low, so side f must also be low.  Sides e and g must be "high".  Continue in this manner for quadrilaterals with sides ghjk, klmn, npqr, and rstd.

The final "coloring" of the sides of the quadrilaterals is shown in the illustration.

We have found a case where we can color the quadrilaterals to avoid piercing.  It would appear we have disproven the hypothesis.  However, the coloring I just performed was for the special case where the three main diagonals of the hexagon intersect at a single point.  Note that the coloring requires the diagonals to change from being "high" to "low" where they intersect in the center of the hexagon.  If we return to the 3-dimensional space, this will occur only if the lines which project to form the main diagonals of the hexagon meet at a single point in 3-space.  If they do, then the coloring is valid.  But then we also have two lines meeting at a single point, which defines a plane.  There must be at least 4 points in that plane, which violates a condition of the problem.

This proof is incomplete, as I have not considered the case where the 2-D convex hull cannot be enclosed by a hexagon.  We must consider a non-convex hull or we must consider enclosure by a smaller convex polygon.  I believe the extension would be straightforward, but without it, this isn't really a proof.

Choose any 2 points, A and B.  Hereafter, we view the space containing the points along axis AB, with point A nearer to the viewer, and point B farther from the viewer (and directly behind point B).

Choose any of the remaining points.  Label it C.  Considering point A to be the center, we proceed counterclockwise from point C, labeling the remaining points D, E, and F.

Consider triangles ACE and BDF.  There are two topologically distinct situations.  In one situation, none of the points C, D, E, or F are within the other triangle.  The intersection of the projection of these triangles is a quadrilateral.  In the other situation, one of the points, C, D, E, or F is within the other triangle.  The intersection of the projection of these triangles is a smaller triangle.  Note that we illustrate the case where the projection of point D is inside the projection of triangle ACE, but similar arguments would apply for the other cases.

We will prove the theorem by contradiction.  Assume no leg of the "Blue" triangle passes through the inside of the "Red" triangle.  By symmetry, the "Red" and "Blue" triangles are interchangable.  Therefore, we look at mutual pairs of triangles using all 6 of our distinct points.

Let's look at (the mutual pair) of triangles ACE and BDF from our viewpoint.  Clearly, in the immediate vicinity of point A, triangle ACE is in front of triangle BDF.  If no part of triangles ACE and BDF are to penetrate each other, the entire region of overlap (a quadrilateral or a triangle) of triangle ACE must be in front of triangle BDF.  [The left side of the attached diagram shows this situation.]

Let's turn to the mutual pair of triangles ADF and BCE from our viewpoint.  Clearly, in the immediate vicinity of point A, triangle ADF is in front of triangle BCE.  If no part of triangles ADF and BCE are to penetrate each other, the entire region of overlap of triangle ADF must be in front of triangle ACE.  [The right side of the attached diagram shows this situtation.]

But wait!  In the case of a quadrilateral overlap region, the upper-left illustration shows that line segment DF must pass behind line segment CE.  The upper-right illustration shows line segment DF must pass in front of line segment CE.  This contradiction cannot be resolved.  [Note that line segments DF and CE cannot intersect, or we would have four points coplanar.]

In the case of a triangular overlap region, the lower-left illustration shows that point D must be behind triangle ACE.  The lower-right illustration shows that point D must be in front of triangle BCE.  Points ABCE form a tetrahedron.  To avoid a contradiction, point D must be inside the tetrahedron.  Point F is outside the tetrahedron.  Line DF must penetrate a triangular face of the tetrahedron.

i'm not sure if this is right, but can't you just say that if for every one of the twenty ways to choose the triangles, the conditions fails, then every one of those triangles must be a face of the convex hull.  But can't the convex hull only have around eight or so faces?

Have to be careful with that - you could have two triangles such that they aren't linked and neither is a face of the convex hull.  See if you can find such an example.  You may be able to get to a proof with this approach, but your counting will have to take these cases in to account.

[quote="rrusczyk"]can anyone find a similar solution by considering the convex hull of the 6 points?[/quote]

Since this thread has been made nonsticky,  here is a proof based on your hint...

Lemma 1:  All faces of every convex hull wrapping any set of 4 or more of the 6 points will be a triangle.

Proof of Lemma 1:  All points on a face lie in the same plane.  If any face were a polygon of more than 3 sides, there would be at least 4 vertexes coplanar, which violates a condition of the problem.  [End of Lemma 1]

Lemma 2:  No pair of lines joining any subset of 4 distinct points will intersect.

Proof of Lemma 2:  The intersection of the 2 lines defines a plane.  So there would exist 4 points in the same plane, which violates a condition of the problem.  [End of Lemma 2]

Wrap the 6 points by a convex hull.  Choose any point, P, on the hull.  Discard the existing hull, and wrap the remaining 5 points with a new convex hull, H.  Point P will lie outside the convex hull, H.

We'll either have 4 or 5 points on the hull.  We consider those cases separately.

Case 1:  There are 4 points on the convex hull.  Since H wraps five points, there must be a point, Q, inside the tetrahedron.  By Lemmas 1 and 2, line segment PQ (blue) passes through a triangular face (red) of the tetrahedron.

Case 2:  There are 5 points on the convex hull.  There are two classes of convex polyhedra containing five vertices.  We consider those separately.

Case 2a:  The 5 points may form a pyramid with a square base.  By Lemma 1, this case cannot occur.

Case 2b:  The 5 points may form a dipyramid, with a common triangular base.  The line segment (blue) between the two apexes passes through the interior of the common triangular base (red).

End of proof.


inequalities
search
inequalities proposed

Cezar, try to be a little more original next time. Just taking $x=1/a$ and so on in Vasc's inequality does not make a new inequality.

Harazi, did I say something about "original" problem? :?

I wasn't speaking about original problems, but rather about  original choosing of problems to be posted. Useless problems are not very interesting.

harazi, do you mind showing me the link to vasc's inequality? I'm interested in the solution.

Search for the topic "nice ineq like Cauchy" by manlio.


A container of eye drops contains 30 ml. The drops are used three times daily, and each time 0.05 ml is used in each eye. How many days will the container last?

each time we use it, we use a total of 0.1ml of the eye drop
each day, we use it three times, so eac day, we use 0.3 ml

to get our answer, we divide 30 by 0.3 and get our answer of $ \boxed{100}$ days


geometry
rectangle
analytic geometry
graphing lines
slope
Pythagorean Theorem

[hide]Draw line AC.  It has slope $-1/3$.  So,the fold line, line EF, has slope $3$.  Now, Use the Pythagorean theorem to get $\sqrt{10}$ . [/hide]

[quote="#H34N1"]In rectangle $ABCD$, $AB=3$ and $BC=9$. The rectangle is folded so that points $A$ and $C$ coincide, forming the pentagon $ABEFD$. What is the length of segment $EF$? Express your answer in simplest radical form.

[img]7355[/img][/quote]
You can label side DF as $x$ and then FC (or FA) will be $9-x$. Then, solve for $x$ with pythag: $3^{2}+x^{2}=(9-x)^{2}\rightarrow 9+x^{2}=81-18x+x^{2}\rightarrow 18x=72\rightarrow \boxed{x=4}$.

Then, draw a perpendicular line from E to FA/FC and then call the point of intersection Z. EZ is now 3, and then FZ is $9-2x$ or $9-8=\boxed{1}$.

Finally, use pythag again: $\sqrt{3^{2}+1^{2}}=\boxed{\boxed{\sqrt{10}}}$


If the roots of the quadratic equation x^2 + px + q = 0 are rationals, where p , q are integers, show that roots are integers.

Since $p^2-4q$ is an integer, its square root can be either integer or irrational (imaginary numbers are out of question in this problem). By the condition of the problem, it must be integer, hence $p^2-4q=m^2$ where $m\in\mathbb{Z}$. This means that $p$ and $m$ are of the same parity as the difference of their squares is even. But if they are of the same parity, then $\frac{-p\pm m}{2}$ are integer numbers.

Thanks Farenhajt, I was thinking in a different way. There cannot be two rationals whose both sum and product are integers.


If $A\in\mathcal M_{2}(R)$ and $A^2=Tr(A)\cdot A\Rightarrow A^3=Tr^2(A)\cdot A$ ?  :?

\[ A^3=A^2\cdot A=Tr(A) \cdot A\cdot A=Tr^2(A)\cdot A  \]
 :lol:


What is the number of positive factors of 648?

The prime factorization of 648 is $ 2^3 \cdot 3^4$. Thus the number of factors is $ (3\plus{}1)(4\plus{}1) \equal{} \boxed{20}$, because any factor of a number must have each of its prime factors, so it could have 0 through 3 as a power for the 2 and 0 through 4 as a power for the 3.


abstract algebra
real analysis
calculus
integration

Hi I was wondering if anyone could recomend me good book about Fourier series, some that explains things like Dirichlet kernel, strong convergence, Lebesgue integrals and such (Riemman-Lebesgue theorem and such). I would need a book that is concise, and that can help me with understanding that theory.

Thanx!

What about Schaum's outlines on Fourier Series  :roll:

I don't know, Schaum has good examples and all, but the thing my professor said at the lectures are faar more complex than what you can find in Schaum (which I have :D)

Well, there's also the 3. volume of Stein and Sharkarchi that I hear is good


limit
real analysis
real analysis unsolved

Fix a real number x_o > 0, and define a sequence {x_n}_n >= 0 recursively by setting x_n+1 = x_n + sqrt(x_n), for all n >= 0.

1. Prove that lim x_n as n -> oo = + oo

2. Prove that lim x_n / n^2 as n -> oo = 1 / 4 .

I would do this:

1) $x_{n+1}-x_{n}=\sqrt{x_{n}}>0$, hence the sequence increases.

By way of contradiction we assume that it is bounded , i.e. the sequence converges, then passing to the limit in the recurrence relation we get that $l=0$, where $l=\lim x_{n}$ which is definitely a contradiction with the fact that the sequence increases.

2) First let us observe  that $\lim \frac{x_{n+1}}{x_{n}}=1+\lim \frac{1}{\sqrt{x_{n}}}=1$.

We calculate your limit by Cesaro Stoltz lemma:

$L=\lim \frac{x_{n}}{n^{2}}=\lim \frac{x_{n+1}-x_{n}}{2n+1}=\lim \frac{\sqrt{x_{n}}}{2n+1}$.
Another application of the same lemma yields

$L=\lim \frac{\sqrt{x_{n+1}}-\sqrt{x_{n}}}{2}=\lim \frac{x_{n+1}-x_{n}}{2(\sqrt{x_{n+1}}+\sqrt{x_{n}})}=\lim \frac{\sqrt{x_{n}}}{2(\sqrt{x_{n+1}}+\sqrt{x_{n}})}$
$L=\frac{1}{4}$,

 based on the preceding observation.


Find an $n$ positive integer in which the sum of the digits is $10$,

and the sum of the digits of the square of $n$ is $100$.

Well...

The number has to be in the form $10+9n$, where $n$ is an odd integer.

Hmm...

[hide]
Well, we know that $n\geq \sqrt{199999999999}$ (that's 11 9's!), so the only reasonable approach would be a clever construction, rather than brute force. This should work.
\[n=\sum_{k=1}^{10}10^{2^{k}}\]
The sum of the digits of $n$ is clearly 10, since we have ten 1's distributed over different digits. To see that the condition on $n^{2}$ is met, observe that
\[n^{2}=\sum_{a,b=1}^{10}10^{2^{a}+2^{b}}=\sum_{k=2}^{11}10^{2^{k}}+2\sum_{a<b}10^{2^{a}+2^{b}}\]
So $n^{2}$ has 45 2's and 10 1's, making the sum of the digits equal to 100.
[/hide]

Very nice, (the above [hide="soution"]
Well, we know that $n\geq \sqrt{199999999999}$ (that's 11 9's!), so the only reasonable approach would be a clever construction, rather than brute force. This should work.
\[n=\sum_{k=1}^{10}10^{2^{k}}\]
The sum of the digits of $n$ is clearly 10, since we have ten 1's distributed over different digits. To see that the condition on $n^{2}$ is met, observe that
\[n^{2}=\sum_{a,b=1}^{10}10^{2^{a}+2^{b}}=\sum_{k=2}^{11}10^{2^{k}}+2\sum_{a<b}10^{2^{a}+2^{b}}\]
So $n^{2}$ has 45 2's and 10 1's, making the sum of the digits equal to 100.
[/hide] ...
The number above, is rather large though (1024 digits), similar logic, but one can get a much  [hide="smaller"]18 digit [/hide]number which has the same property.

For those,  who  want to   peek,  to see a smaller number:

[hide]One smaller number is: $10111111111$ (Of course, one can insert a few zeros here or there and that will work too) [/hide]


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let R be a commutative ring with identity such that all submodules of a free module are free. Prove that R is PID.

it immediately follows that every ideal (= a submodule of $R$ over $R$) is a principal ideal, so it's enough to show that there are no zero divisors. assume a is a zero divisor. the submodule $Ra$ is free and has a generator $b$, say $b = ra$. since $a$ is a zero divisor, this is also true for $b$. but this implies $b=0$ since $\{b\}$ is linearly independant over $R$, and hence $a=0$.

besides, the converse is also true: every submodule of a free module over a PID is free.


function
calculus
derivative
integration
trigonometry
real analysis
real analysis unsolved

Let $f,g$ be two real functions such that $f$ si differentiable and $g$ is derivative and $f'(x)=g(f(x))$. Show that f is monotone.

What do you think about the case $f=0$ ?

I not meant strictly monotone. Monotone = $f(x)\ge f(y), \forall x \ge y$. $f=0$ is monotone. Now prove that every function which respects that condition is monotone.

What is the hypothesis on $g$ ? I think I could prove the result assuming $f$ differentiable and $g$ only continuous.

$g$ is derivetive i.e. $\exists G'(x)=g(x),\forall x \in \mathbb{R}$. So, $g$ may not be continuous. You can post your soution however.

In the following I assume $g$ to be continuous (as a result, $f'$ is $C^{1}$ and $g$ has a primitive). I also assume that the set of the zeroes of $g$ is discrete.

If $g \circ f=0$, then $f'=0$ and $f$ is constant, and in particular monotonic. Otherwise, $\exists \alpha \in \mathbb{R}, g \circ f (\alpha) \neq 0$. Without loss of generality, we can assume $g \circ f (\alpha)>0$. By continuity of $g \circ f$, there is $\epsilon>0, \forall x \in ]\alpha-\epsilon,\alpha+\epsilon[, f'(x) = g \circ f (x)>0$.

Now let $G$ be a primitive of $g$. We have $G(f(t))'=g(f(t))f'(t)=f'(t)^{2} \geq 0$. Thus $\Delta: u \rightarrow \int_{f(\alpha)}^{f(u)}{g}$ is increasing, and $f$ is strictly increasing on $[\alpha,\alpha+\epsilon[$. If $f$ is increasing on $[\alpha,+\infty[$, then we are done. Otherwise let $S>\alpha$ be the (finite) least upper bound of the reals $x$ such that $f$ is increasing on $[\alpha,x]$. By continuity of $f$, $f$ will be increasing on $[\alpha,S]$ and $f'(t)=g(f(t)) \geq 0$ there. Also it is clear that $f'(S)=g(f(S))=0$ (by continuity of $f'$ and definition of $S$).

But by definition of $S$ and continuity of $f$, there is $\eta>0$ such that $\eta+S >\alpha$ and $f(\alpha)<f(\eta+S)<f(S)$. Since $\Delta'=f'^{2}$ is not identically $0$ on $[S,S+\eta]$, we must have $\Delta(\eta+S)>\Delta(S)$. But this is absurd since $g$ is nonnegative on $[\alpha,S]$.

The same reasoning can be reproduced on $]-\infty,\alpha]$, and yields to the result.

There is actually a problem with this proof. This is the assertion: $f(\eta+S)<f(S)$, which is not necessarily true since $g$ can fluctuate very much up and down on $[S,S+\eta]$. That problem vanishes if we suppose that the set of the zeroes of $g$ is discrete.

[quote="xirti"]I don't understan your notations. The intervals $[S+\niu,S]$ and $[S,S+\niu]$ connot exist at the same time. Since $\niu>0$ you must have only the interval $[S,S+\niu]$.[/quote]

Yeah sorry. Edited.

In fact the problem is simpler.. denote by $G$ the primitive of $g$. Then $\left( Gof\right)'=f'^2$. Assume that $f(a)=f(b) \Rightarrow Gof(a)=Gof(b)$ and since $(f')^2 \ge 0$ we have that $Gof$ is constant on $[a,b]$. We can easily prove that f is monotone.

[quote="xirti"]In fact the problem is simpler.. denote by $G$ the primitive of $g$. Then $\left( Gof\right)'=f'^2$. Assume that $f(a)=f(b) \Rightarrow Gof(a)=Gof(b)$[/quote]

No need to assume that, it's always true.

[quote="xirti"] and since $(f')^2 \ge 0$ we have that $Gof$ is constant on $[a,b]$. We can easily prove that f is monotone.[/quote]

$Gof$ is increasing, but I don't think you can assert it is constant.

sorry,I haven't read julien_santini yet.
Anyway the same problem have been posted by me [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%5C%27%28x%29%3Dg%28f%28x%29%29&t=67252]here[/url]
My approach is two consider:
If $f$ is not monotone.
Then to get $x_1,x_2$,such that
$f(x_1)=f(x_2)$ and $f'(x_1)\neq f'(x_2)$
then contradiction with $g$ is a function.
Anybody can complete it?

[quote="julien_santini"][quote="xirti"]In fact the problem is simpler.. denote by $G$ the primitive of $g$. Then $\left( Gof\right)'=f'^2$. Assume that $f(a)=f(b) \Rightarrow Gof(a)=Gof(b)$[/quote]

No need to assume that, it's always true

[quote="xirti"] and since $(f')^2 \ge 0$ we have that $Gof$ is constant on $[a,b]$. We can easily prove that f is monotone.[/quote]

$Gof$ is increasing, but I don't think you can assert it is constant.[/quote]

First one: I assumed that there exists a,b such that $f(a)=f(b)$. :P
Second one:  You are wright $Gof$ is increasing. But it increses from $G(f(a))$ to .... $G(f(a))=G(f(b))$ over $[a,b]$. If a function increses over an interval from the value $a$ to the value $a$. I say that it is constant :P  :P  :P  :P  :P
And my proof works.

[quote="xirti"]First one: I assumed that there exists a,b such that $f(a)=f(b)$.[/quote]

Now I get what you mean. Cool proof !

hi,I am sorry.
But How to do the problem if there no condition that g is continuous?

[quote="xirti"]In post #9 I did not used the continuity of $g$. Only the differentiability of $G$. And the fact that if a function has nonnegative derivative it is non-decreasing.[/quote]
but why $g$ has its  primitive ?
Sorry,can you explain?

[quote="xirti"]Since $g$ is a derivative function, there must exist $G$ such that $G'=g$. Continuous functions have primitives, but there exist discontinuous functons which have primitives. Example:
$f=\begin{cases}\sin\left(\frac 1 x\right)\text { for } x\neq 0 \\ 0 \text{ for }x=0\end{cases}$.[/quote]
Thanks very much.
But how to do if without the condition:" $g$ is a derivative function" because in my book doesn't have this condition. :(