Can anyone show me how to prove this statement? I'm very confused. I have tried substitute numbers into each x, k and m, it seems this statement is true but it needs to be proved by mathematical methods.  Thank you!
123953
Can someone please show me how to prove this statement? $$\forall x\in\mathbb N (x &gt; 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$$ I can only assume $m = 3$ and when $x &gt; 1$.
57494
Let $x\in\mathbb{Q}$ with $x&gt;0$.  Prove that we can find $n\in\mathbb{N}^*$ and distinct $a_1,...,a_n \in \mathbb{N}^*$ such that $$x=\sum_{k=1}^n{\frac{1}{a_k}}$$
79185
Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
123714
Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
46073
Show by combinatorial arguments that:   $$\sum_{k = 0}^m \binom{m}{k}\binom{n}{r + k} = \binom{m + n}{m + r}$$  Hello ,  I want to prove this argument by double counting method, i kinda have an idea of proving by comb args. but not by double counting... Can someone give any clues or ideas? thanks :)
83386
Can someone explain to me the proof of $${n+1\choose k} = {n\choose k} + {n\choose k-1}$$?
332
I got quite some trouble trying to prove this.  $$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n&gt;m\geq 0$$  I tried using $$\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$$ and then changing the summation boundaries as $k=0, n=n-m$, but that took me nowhere... I would really appreciate help. Thank you.
87995
If $\gcd(m,n) =1, $ then $m\mathbb{Z}/ mn\mathbb{Z} \cong \mathbb{Z}/(n)$.   Here is my proof to the a/m problem. But I did not make use of $\gcd(m,n) =1.$ Could anyone hence point out and explain the mistakes in my proof? Thank you.   $\text{Define}\ \phi: m\mathbb{Z}/ mn\mathbb{Z} \to \mathbb{Z}/(n)$ by $\phi(ms + mn\mathbb{Z})= [s']_n, \text{where}  \ s \in [s']_n \ \text{and} \ s' \in \{0,...,n-1\}$  $\phi$ is well-defined:  $mx + mn\mathbb{Z} = my + mn\mathbb{Z}$  $\implies mx = my + mr,$ for some $r\in \mathbb{Z}$  $\implies n | x-y$  $\implies [x]_n = [x']_n= \phi(mx + mn\mathbb{Z}) = [y]_n = [y']_n= \phi(my+ mn\mathbb{Z})$  $\phi$ is injective:  It suffices to prove that $\ker(\phi) =\{mn\mathbb{Z}\}.$  Since $\phi(0+mn\mathbb{Z}) = [0]_n, \ker(\phi) \neq \emptyset.$ Given $x = ms+mn\mathbb{Z} \in \ker(\phi),\ $ $ \phi(x)= [s]_n = [0]_n$ $\text{hence}\  s = nk,  \text{for some}\  k \in \mathbb{Z}.$  Clearly, $mnk + mn\mathbb{Z}\subseteq mn\mathbb{Z}.$ Given $y \in mn\mathbb{Z},$ let $y = mnk + (mnc -mnk),$ for some $c \in \mathbb{Z}.$ Then $y \in mnk + mn\mathbb{Z},$ i.e. $mnk + mn\mathbb{Z} = mn\mathbb{Z} $ and hence $\ker(\phi) =\{mn\mathbb{Z}\}.$  $\phi$ is homomorphism:  Let $a = ms + mn\mathbb{Z}, a' = ms'+ mn\mathbb{Z} \in m\mathbb{Z}/mn\mathbb{Z}. $ Then, $\phi(a+a')= \phi(m(s+s') + mn\mathbb{Z}) $  $\exists r \in \{1,...,n-1\}$ such that $[r]_n = [s+s']_n$ and so $\phi(m(s+s') + mn\mathbb{Z}) = \phi(mr + mn\mathbb{Z}) = [r]_n = \phi(a) +_n \phi(a').$   $\phi$ is surjective:  $\text{Given} \ [r]_n \in \mathbb{Z}/(n),\  \text{let}  \ x= mr + mn\mathbb{Z}. \ \text{Then}   \ \phi(x) =[r]_n$
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
72327
How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...  Thanks in advance...
27132
I want to prove that if $d \mid n \in \mathbb{N}$, then $\varphi(d) \mid \varphi(n)$.  It's given that $d \mid n$, so we know that $n = dm$, for some $m \in \mathbb{Z}$. Now, I want to show that $\varphi(d) \mid \varphi(n)$, i.e. that $\varphi(n) = k\varphi(d)$, for some $k \in \mathbb{Z}$. Drawing a blank of where to go from here.
25053
$\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that  $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?  Thanks in advance.
82769
I get some troubles with this problem, could you help me? Given prime numbers $p$ and $q$ prove $\exists k\in\mathbb{Z}$ such that $pn^q+qn^p+kn$ is a multiple of $pq,~\forall n\in\mathbb{Z}$.
22623
I'd like to have my proof verified and if possible, to see other solutions that are interesting.  Proof: Suppose $n(n+1)$ is a square. Then we write $$n(n+1) = \prod_{p} p^{c(p)}$$ where $c(p) = a(p) + b(p)$ are such that \begin{align*} n &amp;= \prod_p p^{a(p)} \\ n+1 &amp;= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, $c(p)$ is even for all primes $p$. As $(n,n+1)=1$ for all $n$, it must be that $a(p)$ and $b(p)$ are even for all primes $p$ and moreover, $a(p) = 0$ whenever $b(p)&gt;0$ and reversely.  This indicates that both $n$ and $n+1$ are squares. This is impossible as there are no consecutive squares in the natural numbers.
47211
I wanted to prove the following:  if $x_n \to x$ then $y_n \to x$ where $$ y_n = {x_1 + \dots + x_n \over n}$$  Please can you tell me if my proof is correct? My proof is this:  Let $\varepsilon &gt; 0$. Fix $N$ such that $n &gt; N$ implies $|x_n - x| &lt; {\varepsilon \over 2}$.   Then $\left |{1 \over n} \sum_{k=N+1}^{N + n} x_k - x \right | &lt; {\varepsilon \over 2}$. Now let $M$ be such that ${|x_1 + \dots + x_N | \over M} &lt; {\varepsilon \over 2}$ and $M &gt; N$. Then  $$ \left | \sum_{k=1}^M {x_k \over M} - x   \right | \le  \left | \sum_{k=1}^N {x_k \over M}   \right | +  \left | \sum_{k=N+1}^M {x_k \over |M-N|}  - x   \right | &lt; \varepsilon $$ Here the proof is finished. But one can observe:  It is possible that $y_n$ converges even if $x_n$ doesn't: If $x_{2n} = 0$ and $x_{2n + 1} = 1$ then $x_n$ does not converge but $y_n \to {1 \over 2}$.
23402
(a) Prove $(k,n+k)=1$ iff $(k,n)=1$. (b) Is it true that $(k,n+k)=d$ iff $(k,n)=d$? (all variables are integers, (a,b) means gcd(a,b))  For (a), I can only get so far.  $(k,n+k)=1\ \Rightarrow\ kq_1+(n+k)q_2=k(q_1+q_2)+nq_2=1$.  Note that we don't yet have that theorem (from a different class) that says something like if $d=(a,b)$, then $d$ is the smallest (or is it biggest) number which can be made by $ak_1+bk_2$, (which I'm not quite sure how it would help, anyway). So I'm stuck.
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.