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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 $(-1)^{2k}(-1)$.

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    $\begingroup$ This questions needs more upvotes. The user has showed his work which made it easy to find his mistake and help him. $\endgroup$ – Git Gud May 11 '13 at 11:45
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$( -1)^n = ( -1 )^{2m+1} \color{red}= ( ( -1)^2 )^{m+1}$

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 $q>0$?

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  • $\begingroup$ Ah, thank you. The proof then just died. $\endgroup$ – usernvk May 11 '13 at 11:45
  • $\begingroup$ Whoa, never encountered those are signs in a math book before. $\endgroup$ – usernvk May 11 '13 at 11:47
  • $\begingroup$ @usernvk Which sings (I guess you mean symbols) are you talking about? $\endgroup$ – Git Gud May 11 '13 at 11:47
  • $\begingroup$ $\in$, $\Bbb N$ and $\Bbb Z$. What do they mean? $\endgroup$ – usernvk May 11 '13 at 11:49
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    $\begingroup$ @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 $\Bbb Z^+$. $\endgroup$ – user31280 May 11 '13 at 11:57
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You made an error in the 2nd equality. The relevant property of exponentiation here is: $$ a^{m+n}=a^m a^n $$

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

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