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Suppose that $n$ is an even integer.

Then $n = 2r$ for some $r \in\mathbb{Z}$.

Hence $(−1)^n =(−1)^{2r} = 1$ because $2r$ is even.

Therefore $(−1)^n = 1$ for any even integer $n$

Verification?

InsideOut
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SunshineTS
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    Basically, you need to write $(-1)^n=(-1)^{2r}=((-1)^2)^r=(+1)^r=1$. – barak manos Aug 23 '16 at 12:11
  • @barakmanos $\sqrt{a}=a^{\frac{1}{2}}$, while $a^{-2}=\frac{1}{a^2}$ – edm Aug 23 '16 at 12:24
  • @edm: Ooops, long hours :) – barak manos Aug 23 '16 at 12:28
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    This is completely circular. Why does $(-1)^{2r} = 1$? You say it is because $2r$ is even. That's what you are trying to prove! You can't use what you are trying to prove to prove it! – fleablood Aug 23 '16 at 20:58
  • Prove all numbers are less than 523. Let $n$ be a number. Let $n =r+1$. $r+1 < 253$ because it is a number. Therefore all $n < 253$. What is wrong with my proof? My proof is exactly the same reasoning as your proof. – fleablood Aug 23 '16 at 21:03

3 Answers3

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You didn't prove anything yet. I am sorry, but your current "solution" is worth $0$ points.

Why?

You say

$(-1)^{2r} = 1$ because $2r$ is even.

But how do you know that if $2r$ is even, $(-1)^{2r} = 1$?

You cannot just assume the conclusion!

That's not how mathematical proofs work. Remember that what you want to prove is the statement:

If $n$ is even, then $(-1)^n=1$.

In your current "proof", you basically say,

I already know that if $2r$ is even, then $(-1)^n=1$.

But you don't know that because that is exactly what you want to prove.

Hint:

For the actual solution, remember that ${\left(a^b\right)}^c = a^{bc}$.

5xum
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You know $$(-1)^2=1$$, therefore $$(-1)^{2r} = ((-1)^{2})^r = 1^{r} = 1 $$ for any r.

user361972
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$$ (-1)^2 = (-1) \times (-1) = 1 $$ hence

$$\forall r \in \mathbb{R}, \qquad (-1)^{2r} = \left((-1)^2\right)^r = 1^r = 1.$$

Matt
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