Proof of x^{3/2} = \sqrt{x^{3}} is a typo?

[Hootenanny]

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

 

[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

 

[Hootenanny]

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

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. Thank's again.

~H

 

[dav2008]

I guess that proof only works for (positive) integer exponents.

I googled "properties of exponents proof" and found this webpage: http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html

That might explain it some more for non-integer exponents.

 

[HallsofIvy]

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

It is easy to show that, as long as n and m are positive integers, aⁿa^m= a^n+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]= a^nm. 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 a^xa^y= a^x+y and (a^x)^y= a^xy 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 guarantee that this formula is true for n negative, we must have aⁿa^-n= a^n-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= a^n-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 aⁿa^m= a^n+m for a= 0.

Similarly, if we want aⁿa^m= a^n+m for n or m negative, we must have aⁿa^-n= a^n-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 (a^m)ⁿ= a^mn true even when m and n are not integers, we must have (aⁿ)^1/n= a¹ so that a^1/n= \^n\sqrt{a} and then, that a<sup>m/n= ^n\sqrt{a^n.

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

 

[Hootenanny]

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

~H

 

[Gokul43201]

If we want (aⁿ)^m even when m is not a positive, we must have (aⁿ)^-n= a^n-n= a⁰= 1.

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

 

[Hootenanny]

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

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

~H