A proof in powers

**evagelos** · Nov23-09, 09:24 PM

prove that:

for all ,n,m belonging to the natural Nos and x belonging to the real Nos;

$LaTeX Code: x^nx^m =x^{n+m}$

**Office_Shredder** · Nov23-09, 09:36 PM

Any ideas? Can you see why it's true for some small cases, e.g. n=2 or 3 and m=2 or 3?

**Integral** · Nov23-09, 09:47 PM

What does xⁿ mean?

**HallsofIvy** · Nov24-09, 05:42 AM

Yes, exactly what is your definition of xⁿ?

**evagelos** · Nov24-09, 07:54 AM

Originally Posted by HallsofIvy

Yes, exactly what is your definition of xⁿ?

What is your definition? to solve the problem

**g_edgar** · Nov24-09, 08:07 AM

To point of the question "What is your definition" is that to prove something about the operation x^n, you first need a definition for that operation. So start with that. If we do not know what definition is used in your book, it is hard for us to do any more that this.

**HallsofIvy** · Nov24-09, 08:45 AM

Originally Posted by evagelos

What is your definition? to solve the problem

That would be "intention". I am asking how you are defining "x to the n power". Definitions in mathematics are "working definitions"- you use the precise words of the definitions in proofs. Your first thought in proving anything about anything should be "what is its precise definition?"

The most common definition of xⁿ, for n a positive integer is a "recursive" one. x¹= x and xⁿ⁺¹= x(xⁿ).

With that you can prove x^mxⁿ= x^n+m by induction on m.

First prove: xⁿx¹= xⁿ⁺¹. That follows from the definition: since x¹= x, so xⁿx¹= xxⁿ= xⁿ⁺¹ from the second part of the definition.

Now suppose xⁿx^k= x[sup]n+ k[sup] (the "inductive hypothesis"). Then xⁿx^k+1= (xⁿx^k)x= (x^n+k)x= x^(n+k)+1= x^n+(k+1).

We do not "prove" that xⁿx^m= x^{n+ m} for m and n other than positive integers so much as we define the operation so that useful formula is true.

For example, n+0= n so in order to have [math]x[sup]n[sup]x⁰= xⁿ⁺⁰= xⁿ true, we must define x⁰= 1. (In order to go from xⁿx⁰= x⁰ to x⁰= 1, we must divide by xⁿ and so must require that x not be 0. x⁰ is defined to be 1 for x not equal to 0 and 0⁰ is not defined.)

n+ -n= 0 so in order to have [math]xⁿx^-n= x^n-n= x⁰= 1, we must [b]define[/quote] x^-n= 1/xⁿ and, again, must require that x not be 0.

**pbandjay** · Nov24-09, 09:55 AM

Originally Posted by HallsofIvy

we must define x⁰= 1.

Is it really necessary to define x⁰ = 1? If we already define that x^-n = 1/xⁿ then 1 = xⁿ/xⁿ = x^n-n = x⁰, and we get the x not equal to 0 thing for free.

**statdad** · Nov24-09, 12:56 PM

Originally Posted by evagelos

And if we define : $LaTeX Code: x^n = xx^{n-1}$

Then you have a choice. If you've already defined

, you can use

$LaTeX Code: x^n = x \\cdot x^{n-1} \\quad \\text{ for } n \\ge 1 $

This gives

$LaTeX Code: x^1 = x \\cdot x^{1-1} = x \\cdot 1 = x $

Rules follow by induction.

If you don't define the zero power to start then

$LaTeX Code: x^n = x \\cdot x^{n-1} \\quad \\text{ for } n \\ge 2 $

With

defined to be

. The other rules can be obtained in a manner similar to the first.

**Integral** · Nov24-09, 01:37 PM

Originally Posted by pbandjay

Is it really necessary to define x⁰ = 1? If we already define that x^-n = 1/xⁿ then 1 = xⁿ/xⁿ = x^n-n = x⁰, and we get the x not equal to 0 thing for free.

Unfortunatly you use the sum of exponents to arrive at this conclusion, this is the property to be proved.

**evagelos** · Nov24-09, 08:40 PM

Originally Posted by HallsofIvy

That would be "intention". I am asking how you are defining "x to the n power". Definitions in mathematics are "working definitions"- you use the precise words of the definitions in proofs. Your first thought in proving anything about anything should be "what is its precise definition?"

The most common definition of xⁿ, for n a positive integer is a "recursive" one. x¹= x and xⁿ⁺¹= x(xⁿ).

With that you can prove x^mxⁿ= x^n+m by induction on m.

First prove: xⁿx¹= xⁿ⁺¹. That follows from the definition: since x¹= x, so xⁿx¹= xxⁿ= xⁿ⁺¹ from the second part of the definition.

Now suppose xⁿx^k= x[sup]n+ k[sup] (the "inductive hypothesis"). Then xⁿx^k+1= (xⁿx^k)x= (x^n+k)x= x^(n+k)+1= x^n+(k+1).

We do not "prove" that xⁿx^m= x^{n+ m} for m and n other than positive integers so much as we define the operation so that useful formula is true.

For example, n+0= n so in order to have [math]x[sup]n[sup]x⁰= xⁿ⁺⁰= xⁿ true, we must define x⁰= 1. (In order to go from xⁿx⁰= x⁰ to x⁰= 1, we must divide by xⁿ and so must require that x not be 0. x⁰ is defined to be 1 for x not equal to 0 and 0⁰ is not defined.)

n+ -n= 0 so in order to have [math]xⁿx^-n= x^n-n= x⁰= 1, we must [b]define

x^-n= 1/xⁿ and, again, must require that x not be 0.[/quote]

And if we define : $LaTeX Code: x^n = xx^{n-1}$ how would you do the proof then?

**Mentallic** · Nov25-09, 09:48 AM

Wouldn't this be sufficient?:

$LaTeX Code: x^mx^n=(\\underbrace{xxx...x}_{m.times})(\\underbrac e{xx...x}_{n.times})=\\underbrace{xxxx...x}_{(m+n). times}=x^{m+n}$

**evagelos** · Nov26-09, 05:16 PM

Originally Posted by Mentallic

Wouldn't this be sufficient?:

$LaTeX Code: x^mx^n=(\\underbrace{xxx...x}_{m.times})(\\underbrac e{xx...x}_{n.times})=\\underbrace{xxxx...x}_{(m+n). times}=x^{m+n}$

This is not a proof by induction .

**epkid08** · Nov26-09, 06:37 PM

$LaTeX Code: x^m = \\prod^m_{k=1}x$

$LaTeX Code: x^n = \\prod^n_{k=1} x$

$LaTeX Code: x^mx^n = [\\prod^m_{k=1}x][\\prod^n_{k=1} x] = \\prod^{m+n}_{k=1} x$

**Mentallic** · Nov26-09, 10:24 PM

Originally Posted by evagelos

This is not a proof by induction .

I wasn't informed that a specific tool to prove this was necessary:

Originally Posted by evagelos

prove that:

for all ,n,m belonging to the natural Nos and x belonging to the real Nos;

$LaTeX Code: x^nx^m =x^{n+m}$

**hamster143** · Nov26-09, 10:55 PM

Originally Posted by Mentallic

Wouldn't this be sufficient?:

$LaTeX Code: x^mx^n=(\\underbrace{xxx...x}_{m.times})(\\underbrac e{xx...x}_{n.times})=\\underbrace{xxxx...x}_{(m+n). times}=x^{m+n}$

Depends on the definition of x^n. There are multiple possible definitions, which happen to be equivalent for real x and natural n, and that fact by itself is a theorem. In a general groupoid, there are distinct concepts of "left power" and "right power" depending on which way you compute, and the notation "xxx..x" is meaningless because the order of operations is unspecified.

Your solution, your notation and your definition implicitly assume that real multiplication is associative. The implicitness of the assumption makes it a poor proof (a good proof would express everything explicitly).

HallsofIvy's proof is a little better, because, even though it uses associativity (also implicitly), the definition of x^n is written down explicitly without relying on the property.