9
$\begingroup$

I ask only because my textbook infers this in an example. Where should I go to learn more about this?

I'm trying to learn mathematics by Induction but my knowledge of simplifying algebraic equations is crippling me.

Thanks.

  • 4
    Someone should probably mention that the formula you are asking about is usually taken to be the formal definition of integer exponentiation (it is a definition by recursion.) So if you are having trouble finding a formal proof of it, that's why.2012-10-30

2 Answers 2

15

By the rules of exponentiation,

$x^{k} \times x = x^{k+1}$.

If $k$ is an integer, $x^k = \underbrace{x \times x \times \cdots \times x}_{k \textrm{ times}}.$

So $x^k \times x = \underbrace{x \times x \times \cdots \times x}_{k \textrm{ times}} \times x = \underbrace{x \times x \times \cdots \times x}_{k+1 \textrm{ times}}.$

  • 0
    unless $x$ is a Grassmann number ;)2012-10-30
8

$2^{k+1}$ is $2$ multiplied with itself k+1 times. $2\cdot2^k$ is $2$ multiplied $k$ times with itself and an additional $2$ makes it multiplied $k+1$ times with itself.

Also a look at http://en.wikipedia.org/wiki/Exponentiation may help.

  • 0
    Oh, it makes sense. I was confused, but it is quite simple. Since multiplying 2 by $2^k$ is just like increasing k by 1. Very easy stuff. Thanks.2012-10-30