# Thread: Power law proof

1. ## Power law proof

We all know a^m x a^n = a^(m+n) and it is easy to show why for integer m and n.

How would you prove the law for non-integer m and n?

For rational m and n this would be equivalent to showing

a^(p/q) x a^(r/s) = a^(p/q+r/s) for integer p,q,r and s.

I can do this in the reverse direction, i.e. show that

a^(p/q+r/s) = a^(p/q) + a^(r/s) by combining p/q+r/s as (ps+rq)/qs

and using the fact that ps and rq are integers and using the law for integers.

But how do you show the law is true working from a proof in the forward direction?

What about irrationals?

How would you show that a^pi x a^e = a^(pi+e) say?

Chris Robinson

2. Originally Posted by crobbo52
We all know a^m x a^n = a^(m+n) and it is easy to show why for integer m and n.

How would you prove the law for non-integer m and n?

For rational m and n this would be equivalent to showing

a^(p/q) x a^(r/s) = a^(p/q+r/s) for integer p,q,r and s.

I can do this in the reverse direction, i.e. show that

a^(p/q+r/s) = a^(p/q) + a^(r/s) by combining p/q+r/s as (ps+rq)/qs
Well, you mean "a^(p/q+ r/s)= a^(p/q) x a^{r/s}", not "+".

and using the fact that ps and rq are integers and using the law for integers.

But how do you show the law is true working from a proof in the forward direction?
What do you mean by "in the forward direction". You have shown that
a^(p/q+ r/s)= a^(p/q)+ a^{r}{s}. There is no "direction" in "="- it doesn't matter on which side you did the calculation, they are equal.

I would point out that this (or any proof that a^(x+y)= a^x a^y is "circular". We really define a^(1/n) in such a way as to make this true.

What about irrationals?

How would you show that a^pi x a^e = a^(pi+e) say?

Chris Robinson
How do you define a^e? Usually we define exponential powers "by continuity". That is if $r_1, r_2, ...,r_n, ...$ is a sequence of rational numbers converging to x, we define $e^x$ to be limit of the sequence $e^{r_1}, e^{r_2}, ..., e^{r_n}, ...$. That would, of course, require showing that sequence does converge and, for any sequence of rational numbers converging to x, that sequence will converge to the same thing, but that's not too difficult.

From that, let $b_1, b_2, b_3, ..., b_n, ...$ be a sequence of rational numbers converging to the irrational number x and let $c_1, c_2, c_3, ..., c_n, ...$ be a sequence of rational numbers converging to the irrational number y. Then $a^{b_1+ c_1}, a^{b_2+ c_2}, a^{b_3+ c_3}, ..., a^{b_n+ c_n}$ converges to $a^{x+ y}$. But each term in that sequence is the same as $a^{b_i}a^{c_i}$ because all $a_i$ and $c_i$ and that sequence converges to $a^xa^y$.

3. ## Power law

Thanks for your reply, HallsofIvy.

Perhaps I should give the question some more context. I am thinking of an A level pupil who might say:

'I can see a^5 x a^3 = a^(5+3), but how do you know it works for
a^1.2 x a^1.8 = a^(1.2+1.8)?

I should not have talked about proof, it would be better to ask 'how do you show that a^m x a^n can be written as a^(m+n) for non-integer m and n as well as integers.'

I take your point about direction being meaningless with an '='sign, but I am talking about convincing a school kid. Starting with the desired answer and working back to the original will not satisfy.

I had another think, and realised how to do this in the 'forward direction' ( pardon me!).

I don't know how to write proper algebra in the forum so this will look clumsy I'm afraid.

For rational m and n write m as p/q and n as r/s.

Write a^(p/q) as qth root (a^p) and a^(r/s) as sth root (a^r).

a^(p/q) x a^(r/s) = qth root (a^p) x sth root (a^r).

Power up each side q times.

(a^(p/q) x a^(r/s))^q = a^p x (sth root (a^r))^q
= a^p x sth root (a^rq)

Similarly power up s times.

(a^(p/q) x a^(r/s)^qs = a^ps x a^rq = a^(ps + rq)

Take the qs th root

a^(p/q) x a^ (r/s) = a^ ((ps+rq)/qs) = a^(p/q + r/s). Done!

This only uses power law stuff seen at GCSE level.

Your treatment of irrationals is beyond me at the moment, but I will persevere!

Thank you again,

Chris Rob