negative exponents

**manyarrows** · April 1st 2009, 07:55 PM

Is there a mathmatical rational for the choosing of negative exponents to represent fraction reversal or was is it just a matter of convention.

**Grandad** · April 1st 2009, 10:44 PM

Hello manyarrows

Originally Posted by manyarrows

Is there a mathmatical rational for the choosing of negative exponents to represent fraction reversal or was is it just a matter of convention.

Yes, there's every mathematical reason in the world! Like being able to continue to use the laws that apply to positive exponents. For instance, the simplest and most basic rule is:

$x^a \times x^b = x^{a+b}$

This is obviously the case for positive values of $a$ and $b$ , because we can simply write the meanings in full:

$x^a \times x^b = \underbrace{x \times x \times \dots \times x}_{a\,x's} \times \underbrace{x \times x \times \dots \times x}_{b\,x's} = \underbrace{x \times x \times \dots \times x}_{(a+b)\,x's}$

If we want to continue to use this law for negative powers (whatever they may mean), then we shall be forced to conclude that $x^{-b}$ will have to be defined as $\frac{1}{x^{b}}$ . Why? Well, if the law above continues to remain in force, then (and we'll assume for now that $a>b$ ):

$x^a \times \color{red}x^{-b} \color{black}= x^{a+(-b)} = x^{a-b}$

But we already know that $x^a \times \color{red}\frac{1}{x^b}\color{black} =\frac{x^a}{x^b} = x^{a-b}$ , by 'cancelling' all the $x$ 's in the denominator with $b$ of the $x$ 's in the numerator.

And therefore we shall insist that $x^{-b}$ and $\frac{1}{x^b}$ mean the same thing.

If that all seems too complicated, and you want a simpler reason, just look at the pattern of powers of 2 below, and ask yourself what happens if you continue the pattern, dividing by 2 each time:

$2^3 = 8,\, 2^2 = 4,\, 2^1 = 2,\, 2^0 = ?,\, 2^{-1} = ?,\, 2^{-2}= ?,\, ...$

Grandad

**HallsofIvy** · April 2nd 2009, 11:33 AM

I just want to point out that if we want that very nice property, $x^ax^b= x^{a+b}$ , then we want $x^0x^a= x^{0+a}= x^a$ so that we must have $x^0= 1$ .

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