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I’m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don’t understand. In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.) In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.). When I do it I get $2^{(ab)(b) – (b)(b)}$ and assume this is as far as it can be taken. Can someone please help me understand what steps he is taking to to get his answer?

Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then $2^n$ − 1 is not prime. Proof. Since n is not prime, there are positive integers a and b such that a < n, b < n, and n = ab. Let x = $2^b$ − 1 and y = 1 + $2^b$ + $2^{2b}$ +· · ·+ $2^{(a−1)b}$. Then

xy = ($2^b$ − 1) · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)

Step 1.) = $2^b$ · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) − (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)

Step 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +···+$2^{ab}$) − (1 + $2^b$ + $2^{2b}$ + ···+$2^{(a-1)b}$)

Step 3.) = $2^{ab}$ − 1

Step 4.) = $2^n$ − 1.

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    $\begingroup$ Are you familiar with how $x^c\times x^d = x^{c+d}$? Are you familiar with how $(a-1)b + b = ab$? $\endgroup$ – JMoravitz Jun 10 '17 at 16:00
  • $\begingroup$ My title should have read multiplying $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$. Sorry about that. $\endgroup$ – maybedave Jun 10 '17 at 16:00
  • $\begingroup$ @JMoravits. Not really and now I'm feeling a little dumb because from what you are saying, this is actually correct, no? $\endgroup$ – maybedave Jun 10 '17 at 16:01
  • $\begingroup$ For integer values of $c$ and $d$, notice that $x^c\times x^d = \overbrace{\underbrace{x\times x\times x\times \cdots \times x}_{c~\text{times}}\times \underbrace{x\times x\times \cdots \times x}_{d~\text{times}}}^{c+d~~\text{times}}$. (The property can be extended to real values of $c$ and $d$ as well, look up a more in depth proof for that). As for why $(a-1)b+b=ab$, this is algebraic manipulation., $(a-1)b+b=ab-1b+b=ab-b+b=ab+(-b+b)=ab+0=ab$ $\endgroup$ – JMoravitz Jun 10 '17 at 16:07
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$$2^{(a-1)b}2^b=2^{(a-1)b+b}=2^{ab-b+b}=2^{ab}.$$

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  • $\begingroup$ Ok, so what what you guys are saying is that I really should be adding the b to (a-1)b rather than multiplying it in. This is A BIG HELP! Thank you! $\endgroup$ – maybedave Jun 10 '17 at 16:04
  • $\begingroup$ @maybedave It's $x^rx^s=x^{r+s}$... $\endgroup$ – Lord Shark the Unknown Jun 10 '17 at 16:05
  • $\begingroup$ "rather than multiplying it in" You absolutely should not multiply them! Notice $64 =4*16=2^2*2^4=2^{2*4}=2^8=256$. Doesn't work. But $64= 2^2*2^4 = (2*2)*(2*2*2*2) = (2*2*2*2*2*2) = 2^6 = 64$. Addition has to work. Notice $2^4*2^4 = (2*2*2*2)*(2*2*2*2) = [(2*2*2*2)(2*2*2*2)(2*2*2*2)(2*2*2*2)] = 2^{4*4}$ just doesn't make any sense at all. $\endgroup$ – fleablood Jun 10 '17 at 16:18
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Using the laws of exponents, $2^{(a-1)b}2^b = 2^{(a-1)b+b} = 2^{ab}$

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Hint:

The correct rule is:

$$ 2^b 2^{(a-1)b}=2^{(a-1)b+b} $$

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Basic rules $a^m a^n = (a*a*a*a*a........*a[n \text{ times}])*(a*a*...... *a[m \text{ times}]) = a*a*a*.......*a [n+m\text { times}] = a^{n+mm}$.

So $2^{(a-1)b}*2^b = (2*2*2*2*2........*2[(a-1)b \text{ times}])*(2*2*...... *2[b \text{ times}]) = 2*2*2*.......*2 [(a-1)b+b\text { times}] = 2^{(a-1)b + b}$

Meanwhile $(a-1)*b + b = a*b - 1*b + b = (ab) + (-b + b) = ab + 0 = ab$.

So $2^{(a-1)b}2^b = 2^{(a-1)b + b} = 2^{ab}$

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