# Can you solve this !!

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• Feb 8th 2009, 03:33 PM
Banned for attempted hacking
Can you solve this !!
Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.
• Feb 8th 2009, 04:34 PM
Jester
Quote:

Originally Posted by ╔(σ_σ)╝
Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.

For the first one, let $u = \frac{x^{100}}{e^{x^{70}}}$, then

$\frac{x^{100}}{e^{x^{70}}} + \frac{e^{x^{70}} }{x^{100}} = u + \frac{1}{u}$

so you want to show $u + \frac{1}{u} > 2$ which is to show that

$u^2 - 2u + 1 > 0$ or $(u-1)^2 > 0$ which is true provide that $u \ne 1$. Now u is bounded above (using some calculus) so $u \le u_{max} < 1$ establishing the inequality.
• Feb 8th 2009, 04:43 PM
Banned for attempted hacking
Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.
• Feb 8th 2009, 04:47 PM
Jester
Quote:

Originally Posted by ╔(σ_σ)╝
Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.

I've been doing math for a while (Rofl)
• Feb 8th 2009, 04:56 PM
Banned for attempted hacking
Quote:

Originally Posted by danny arrigo
I've been doing math for a while (Rofl)

Cool

Now about the second one (Rofl).

Experience won't help.
• Feb 8th 2009, 06:06 PM
Banned for attempted hacking
Amazing no one can solve the second one ! (Thinking)
• Feb 8th 2009, 06:33 PM
ThePerfectHacker
Quote:

Originally Posted by ╔(σ_σ)╝
Prove that
cos 20 =/= a/b where a and b are integers .

This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$.
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.
• Feb 8th 2009, 06:37 PM
Banned for attempted hacking
Quote:

Originally Posted by ThePerfectHacker
This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$.
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.

$\cos 60 =/= \frac{\sqrt{3}}{2}$

$\cos 60 = \frac{\sqrt{1}}{2}$

Finish it.
• Feb 8th 2009, 06:47 PM
ThePerfectHacker
Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$.
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.
• Feb 8th 2009, 07:15 PM
Banned for attempted hacking
Quote:

Originally Posted by ThePerfectHacker
Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$.
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.

Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2
• Feb 8th 2009, 07:19 PM
ThePerfectHacker
Quote:

Originally Posted by ╔(σ_σ)╝
Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2

Sorry! That was a trivial error.
But the idea still works because: $4x^3 - 3x = \tfrac{1}{2} \implies 8x^3 - 6x -1 = 0$.
Apply rational roots theorem and conclude there are no rational solutions.