Show that the follow statement is true:
If $ x \in \mathbb{R}$ such that $x^2+1=0$ then $x^4=\pi$
Constructive proof:
If $x,y$ $\in \mathbb{R}$ such that $x \lt y$, show that $\exists \ z\in\mathbb{R}$ such that $x\lt z\lt y$
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Hint: Show the first part is always wrong. Therefore the statement is true. – Loreno Heer Nov 07 '14 at 19:10
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I'm not sure what do do, I've only done proofs with even and odd numbers like x=2k and x=2k+1. I'm not sure how to get started for these two that involve all real numbers – janny Nov 07 '14 at 19:12
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@janny For the first one: Remember that if $p$ is false, then $p\Rightarrow q$ is always true. For the second one: take the average? – Akiva Weinberger Nov 07 '14 at 19:36
1 Answers
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-First one:
Hint: $x^2 + 1 = 0 \Rightarrow x = \pm i$ and $p \to q$ is true if $p$ is false and $q$ is false.
-Second one:
Hint: If $x<y \in \mathbb{R}$ take $z = \frac{x + y}{2}$.
Aaron Maroja
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For number 1, what is i? an integer?
For number 2, since it's a constructive proof, can I say that y=2, x=0, z=1, so 0<1<2 =x<z<y
– janny Nov 07 '14 at 19:36 -
$i^2 = -1$ is a complex number. In other words, in 1 there is no real number that its square is $-1$, the sentence is therefore false. – Aaron Maroja Nov 07 '14 at 19:38
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Thanks for the help, I just had a quick question. Why does z have to be (2+0)/2 instead of just "1"? – janny Nov 07 '14 at 19:47
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Because you want to show existence of $z$ for any $x<y \in \mathbb{R}$ you take, in particular, $x = 2$ and $y = 0$. – Aaron Maroja Nov 07 '14 at 19:48
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