Is Associativity Assumed in the Proof of \(x^n x^m = x^{n+m}\)?

[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}

 

[Office_Shredder]

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]

What does xⁿ mean?

 

[HallsofIvy]

Yes, exactly what is your definition of xⁿ?

 

[evagelos]

Yes, exactly what is your definition of xⁿ?

What is your definition? to solve the problem

 

[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.

 

[HallsofIvy]

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 (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.</sup>

 

[pbandjay]

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]

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

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

<br /> x^n = x \cdot x^{n-1} \quad \text{ for } n \ge 1<br />

This gives

<br /> x^1 = x \cdot x^{1-1} = x \cdot 1 = x<br />

Rules follow by induction.

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

<br /> x^n = x \cdot x^{n-1} \quad \text{ for } n \ge 2<br />

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

 

[Integral]

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]

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 (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</sup>

<sup>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?</sup>

 

[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}

 

[evagelos]

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}

This is not a proof by induction .

 

[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

 

[Mentallic]

This is not a proof by induction .

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

prove that:

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

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

 

[hamster143]

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}

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.