# Indices proof

1. May 30, 2006

### Hootenanny

Staff Emeritus
Something I have been curious about, but never had the time to think about is, I know that;

$$x^{3/2} = \sqrt{x^{3}}$$

But I have never seen any proof of this. Does anyone have a good resource or can show me the proof here? It would be much appreciated.

~H

2. May 30, 2006

### dav2008

So are you asking why $$(a^m)^n=a^{mn}$$?

A simple explanation would just be to expand the left-hand side:

$$(a^m)^n$$
$$=a^m \cdot a^m \cdot...\cdot a^m$$ (n-times)
$$=a^{m+m+m+...+m}$$ (n times) since $a^m \cdot a^m=a^{m+m}$
$$=a^{mn}$$ since m+m+m+...+m n times is just m times n

Last edited: May 30, 2006
3. May 30, 2006

### Hootenanny

Staff Emeritus
Thank's yeah, I've just got it. Just as I was replying to this I found it in one of my old textbooks, guess I should look through my books more before asking stupid questions.:grumpy: Thank's again.

~H

4. May 30, 2006

### dav2008

Last edited by a moderator: Apr 22, 2017
5. May 30, 2006

### HallsofIvy

This is not so much a "proof" as an explanation of why we define
ax in certain ways.

It is easy to show that, as long as n and m are positive integers, anam= an+m. That's just a matter of counting the number of "a"s being multiplied.
Similarly, it is easy to show that, as long as n and m are positive integers, (an)[/sup]m[/sup]= anm. Again, that's just a matter of counting the number of "a"s being multiplied.

Those are very nice formulas! It would help a lot if axay= ax+y and (ax)y= axy for all x and y.

IF it were true that ana0= an+0, even when the exponent is 0, we must have an+0= an= ana0 and if a is not 0 we can divide by an to get a0= 1 as long as a is not 0.

Similarly, to guarentee that this formula is true for n negative, we must have ana-n= an-n= a0= 1: in other words that a-n= 1/an. Of course, we can only do that division if an is not 0: in other words if a is not 0.

If we want (an)m even when m is not a positive, we must have (an)-n= an-n= a0= 1. In other words, we must define a0= 1 as long as a is not 0. There is no way to define 00 that will make anam= an+m for a= 0.

Similarly, if we want anam= an+m for n or m negative, we must have ana-n= an-n= a0= 1 for all positive integers n: in other words, again dividing by an, a-n= 1/an as long as a is not 0.

As dav2008 pointed out, in order to have (am)n= amn true even when m and n are not integers, we must have (an)1/n= a1 so that a1/n= $\^n\sqrt{a}$ and then, that am/n= $^n\sqrt{a^n$.

In order to define ax for x irrational we require that f(x)= ax be continuous.

6. May 30, 2006

### Hootenanny

Staff Emeritus
Ahh, makes more sense now. Thanks both of you, it is much appreciated.

~H

7. May 31, 2006

### Gokul43201

Staff Emeritus
One of us is making a mistake with this bit. Is that a typo ?

8. Jun 1, 2006

### Hootenanny

Staff Emeritus
Didn't spot that, surely (an)-n = $a^{-n\cdot n}$?

~H