# A proof in powers

1. Nov 23, 2009

### evagelos

prove that:

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

$$x^nx^m =x^{n+m}$$

2. Nov 23, 2009

### Office_Shredder

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

3. Nov 23, 2009

### Integral

Staff Emeritus
What does xn mean?

4. Nov 24, 2009

### HallsofIvy

Yes, exactly what is your definition of xn?

5. Nov 24, 2009

### evagelos

What is your definition? to solve the problem

6. Nov 24, 2009

### g_edgar

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.

7. Nov 24, 2009

### 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 xmxn= xn+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 xnxk= xn+ k (the "inductive hypothesis"). Then xnxk+1= (xnxk)x= (xn+k)x= x(n+k)+1= xn+(k+1).

We do not "prove" that xnxm= xn+ 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]xnx0= 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= xn-n= x0= 1, we must define[/quote] x-n= 1/xn and, again, must require that x not be 0.

Last edited by a moderator: Nov 24, 2009
8. Nov 24, 2009

### pbandjay

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

Last edited: Nov 24, 2009
9. Nov 24, 2009

### statdad

Then you have a choice. If you've already defined $$x^0 = 1$$, you can use

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

This gives

$$x^1 = x \cdot x^{1-1} = x \cdot 1 = x$$

Rules follow by induction.

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

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

With $$x^1$$ defined to be $$x$$. The other rules can be obtained in a manner similar to the first.

10. Nov 24, 2009

### Integral

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

11. Nov 24, 2009

### evagelos

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

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

12. Nov 25, 2009

### Mentallic

Wouldn't this be sufficient?:

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

13. Nov 26, 2009

### evagelos

This is not a proof by induction .

14. Nov 26, 2009

### epkid08

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

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

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

15. Nov 26, 2009

### Mentallic

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

16. Nov 26, 2009

### hamster143

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.

Last edited: Nov 26, 2009
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