# Basic Mathematics. Trouble with proof, powers and odd numbers.

Greets,

In the exercises, at the end of chapter 1.4, Basic Mathematics, Serge Lang

6) Prove: If $n$ is odd, then $\quad (-1)^n = -1$

How? The working I did

\begin{align}( -1)^n &= ( -1 )^{2m+1}\\ &= ( ( -1)^2 )^{m+1}\\ &= ( ( -1 )( -1 ) )^{m+1}\\ &= ( 1 )^{m+1} \end{align}

Can it get from here to $-1$?

The book shows

Let $n = 2k + 1$. Then

\begin{align}(-1)^n &= (-1)^{2k+1}\\ &= (-1)^{2k}(-1)\\ &= 1 \cdot (-1)\\ &= -1 \end{align}

I'm having trouble with how $(-1)^{2k+1}$ became $(-1)^{2k}(-1)$.

• This questions needs more upvotes. The user has showed his work which made it easy to find his mistake and help him. – Git Gud May 11 '13 at 11:45

## 2 Answers

$( -1)^n = ( -1 )^{2m+1} \color{red}= ( ( -1)^2 )^{m+1}$

Wrong. Note that $2(m+1)=2m+2$.

$(-1)^n = (-1)^{2k+1} \color{blue}= (-1)^{2k}(-1)$, for the blue equality, answer this question: how do you define $p^q$, where $p,q\in \Bbb Z$ and $q>0$?

• Ah, thank you. The proof then just died. – usernvk May 11 '13 at 11:45
• Whoa, never encountered those are signs in a math book before. – usernvk May 11 '13 at 11:47
• @usernvk Which sings (I guess you mean symbols) are you talking about? – Git Gud May 11 '13 at 11:47
• $\in$, $\Bbb N$ and $\Bbb Z$. What do they mean? – usernvk May 11 '13 at 11:49
• @GitGud $\Bbb N$ is not the set of positive integers but the set of natural numbers (some people include $0$ in $\Bbb N$). The set of positive integers is better denoted with $\Bbb Z^+$. – user31280 May 11 '13 at 11:57

You made an error in the 2nd equality. The relevant property of exponentiation here is: $$a^{m+n}=a^m a^n$$

• Solved! I completely forgot this. Wonder what I was doing during this part of the book. – usernvk May 11 '13 at 12:18