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
Hello Juancd08I 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.
$\displaystyle y=2x^2+\frac{2}{x^2}$
$\displaystyle \Rightarrow \frac{dy}{dx}=4x-\frac{4}{x^3}=0$ when $\displaystyle x = \pm 1, \pm i$
$\displaystyle \frac{d^2y}{dx^2}= 4+\frac{12}{x^4}>0$ when $\displaystyle x = \pm 1$
So if $\displaystyle x$ is real, the minimum value of $\displaystyle 2x^2+\frac{2}{x^2} $ is $\displaystyle 4$, and therefore $\displaystyle 2x^2+\frac{2}{x^2} = \pi$ has no real solutions.
So if $\displaystyle p(x)$ is the propositional function: $\displaystyle 2x^2+\frac{2}{x^2}=\pi$ and $\displaystyle q(x)$ is the propositional function $\displaystyle x^{\pi} + x^{\log 10} = e^{\pi}\cdot\pi^e$:
$\displaystyle \forall x \in \mathbb{R},\, p \Rightarrow q$, since $\displaystyle p$ is always false
i.e. for all real values of $\displaystyle x, 2x^2+\frac{2}{x^2} = \pi \Rightarrow x^{\pi} + x^{\log 10} = e^{\pi}\cdot\pi^e$ .
Grandad