Search found 268 matches

by nayel
Sat Mar 05, 2016 11:38 pm
Forum: Algebra
Topic: Inequality with xyz=1
Replies: 1
Views: 711

Inequality with xyz=1

Let $x,y,z>\frac 12$ satisfy $xyz=1$. Prove that
\[\frac{1}{2x-1}+\frac{1}{2y-1}+\frac{1}{2z-1}\ge 3.\]
by nayel
Sun Nov 22, 2015 8:22 pm
Forum: Number Theory
Topic: Factorial divisible by Mersenn Numbers
Replies: 1
Views: 650

Re: Factorial divisible by Mersenn Numbers

Bang's theorem tells us that if $n\neq 1,6$ then $2^n-1$ has a primitive prime divisor $p$. If $p\le n$ then $p$ divides $2^{p-1}-1$ which is less than $2^n-1$, a contradiction. So one only needs to check $n=1,6$ of which only $n=1$ works.
by nayel
Sun Nov 22, 2015 7:29 am
Forum: Number Theory
Topic: #Number Theory
Replies: 2
Views: 847

Re: #Number Theory

Your claim does not hold if $b=0$ so I'll assume that $ab\neq 0$. Then $p$ must be odd. Suppose that $p^2=a^2+2b^2$. Then $(p+a)(p-a)=2b^2$, so $p\pm a$ are both even. Set $p+a=2x$ and $p-a=2y$. Then $2xy=b^2$, implying that $b$ is even. Set $b=2z$. Then $xy=2z^2$. Note that $(x,y)$ divides $2p$ and...
by nayel
Sun Nov 22, 2015 7:14 am
Forum: Number Theory
Topic: prove it!!!
Replies: 2
Views: 638

Re: prove it!!!

Let $S=\{a_1x_1+\cdots+a_nx_n:x_1,\dots,x_n\in\mathbb Z\}$. Let $d'$ be the smallest positive element of $S$. Prove the following:

(i) $d$ divides $d'$.

(ii) $d'$ divides $d$.

So $d=d'\in S$ and your conclusion will follow.
by nayel
Sun Nov 22, 2015 7:12 am
Forum: Number Theory
Topic: Sequence
Replies: 3
Views: 956

Re: Sequence

It's a divergent series. Neither the infinite sum nor the infinite product has a finite value. So it does not even make sense to say that they are equal. More details: https://en.wikipedia.org/wiki/Harmonic_ ... thematics)
by nayel
Wed Apr 15, 2015 12:43 am
Forum: Higher Secondary Level
Topic: Functional equation
Replies: 1
Views: 977

Re: Functional equation

joy_li wrote:$\Rightarrow -c(x-y) \neq -cx-cy$
Why? What if $c=0$?
by nayel
Sat Mar 21, 2015 7:23 pm
Forum: Geometry
Topic: Congruence
Replies: 1
Views: 712

Re: Congruence

Hint:
There is too much symmetry here.
by nayel
Mon Feb 23, 2015 3:55 am
Forum: Higher Secondary Level
Topic: Vectors around Regular Polygon
Replies: 4
Views: 1211

Re: Vectors around Regular Polygon

Two words:
Complex numbers
by nayel
Sat Feb 21, 2015 1:13 am
Forum: Algebra
Topic: Generating Z^2
Replies: 0
Views: 477

Generating Z^2

Let $a=(a_1,a_2)\in\mathbb Z^2$ where $a_1$ and $a_2$ are coprime. Show that there exists $b=(b_1,b_2)\in\mathbb Z^2$ such that any element of $\mathbb Z^2$ can be written as a $\mathbb Z$-linear combination of $a$ and $b$. Terminology: 1. A $\mathbb Z$-linear combination of $a$ and $b$ is anything ...
by nayel
Fri Feb 20, 2015 3:11 am
Forum: Algebra
Topic: Binomial and power of 4(or 2?)
Replies: 3
Views: 915

Re: Binomial and power of 4(or 2?)

Same idea from the other thread gives a stronger bound:
\[
\begin{align*}
&2^n=(1+1)^n=\binom n0+\cdots+\binom nn\le (n+1)\binom{n}{\lfloor n/2\rfloor}\\
\Rightarrow &4^n\le (n+1)^2\binom{n}{\lfloor n/2\rfloor}^2<(n+1)(2n+1)\binom{n}{\lfloor n/2\rfloor}^2
\end{align*}
\]