-1
$\begingroup$

I know at the back of my head by exhaustion that this proof is true. But I just don't understand how do I proof this on paper. Here's my work so far:

$$ \begin{align*} n &= 2a \\(-1)^{2a} &= 1 \end{align*} $$

I don't know how to approach this problem after that.

  • 0
    but what if $n=2a+1$? so this is clearly not true. – 2017-01-12
  • 0
    Hint: a property of exponents is $x^{ab}= (x^a)^b$ – 2017-01-12

3 Answers 3

5

If you're told that $n$ is even, then the final steps are $$ (-1)^{2a} = ((-1)^2)^a = 1^a = 1 $$ If you're not told that $n$ is even, then you can't prove that $(-1)^n = 1$, because it isn't true in general. Specifically, you're not allowed to say that $n = 2a$ aith $a$ being an integer. If you try anyways, $a$ could be a fraction, and you must be very careful with negative numbers raised to fractional exponents. I strongly advise you to avoid the concept completely.

2

it is clear that we have $$(-1)^{2m}=1$$ and $$(-1)^{2n+1}=-1$$ since $$((-1)^2)^m=1$$ and $$(-1)^{2n+1}=(-1)^{2n}\cdot (-1)^1=-1$$

  • 0
    I can solve up to that, if you read my description, but I needed assistance after that. – 2017-01-12
  • 0
    i have edited my post – 2017-01-12
1

We can write as $$(-1)^n=(-1)^{2a}=\left((-1)^2\right)^a=(1)^a=1$$