Discrete Math. Urgent help please.

**Juancd08** · April 29th 2009, 09:15 PM

Give a proof of the following implication: Let x be an arbritrary real number.

If 2x^2 + 2/x^2 = pi, then x^pi + x^(log10) = e^pi*pi^e

**Grandad** · April 29th 2009, 10:30 PM

Hello Juancd08

Originally Posted by Juancd08

Give a proof of the following implication: Let x be an arbritrary real number.

If 2x^2 + 2/x^2 = pi, then x^pi + x^(log10) = e^pi*pi^e

I think this is all to do with a false statement implying a statement which may be true or false. See the third and fourth lines of the Truth Table: Discrete mathematics/Logic/Page 2 - Wikibooks, collection of open-content textbooks.

$y=2x^2+\frac{2}{x^2}$

$\Rightarrow \frac{dy}{dx}=4x-\frac{4}{x^3}=0$ when $x = \pm 1, \pm i$

$\frac{d^2y}{dx^2}= 4+\frac{12}{x^4}>0$ when $x = \pm 1$

So if $x$ is real, the minimum value of $2x^2+\frac{2}{x^2}$ is $4$ , and therefore $2x^2+\frac{2}{x^2} = \pi$ has no real solutions.

So if $p(x)$ is the propositional function: $2x^2+\frac{2}{x^2}=\pi$ and $q(x)$ is the propositional function $x^{\pi} + x^{\log 10} = e^{\pi}\cdot\pi^e$ :

$\forall x \in \mathbb{R},\, p \Rightarrow q$ , since $p$ is always false

i.e. for all real values of $x, 2x^2+\frac{2}{x^2} = \pi \Rightarrow x^{\pi} + x^{\log 10} = e^{\pi}\cdot\pi^e$ .

Grandad

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