# Defining things by induction

1. Feb 10, 2004

### Ed Quanta

For k is an element of the set of natural numbers, and m,n are elements of the set of natural numbers or the set of zero, I have to prove that

1)k^n is an element of the set of natural numbers
2)k^mk^n=k^(m+n)
and
3) (k^m)^n= k^(mn)

I'm pretty confident that these can be shown by induction using the set of whole numbers. But it was never shown in class that k^0=1. Is there any way I can go about showing this? If anyone can just give me some feedback that would be much appreciated as it has been this entire semester. Thanks.

2. Feb 10, 2004

### NateTG

"and m,n are elements of the set of natural numbers or the set of zero"

Isn't particularly clear.

Regardless, if you're defining exponentiation inductively, then you still need a k^0 or k^1 case that is handled in some different way.

Effectively, k^0=1 by definition.

3. Feb 11, 2004

### HallsofIvy

Staff Emeritus
By " and m,n are elements of the set of natural numbers or the set of zero", I think you mean m, n are natural numbers or zero (the phrase "the set of zero" is unclear). Normally that's referred to as the "set of whole numbers".

The key thing you should be thinking about is the precise definition of "kn" that you are using. The definitions I know give k0= 1 as part of the definition.

Often a definition starts by defining k1= k, the recursively defining kn= k*kn-1 but that only works for the natural numbers. Since, for m, n natural numbers, we have kn*km= kn+m, in order to extend kn to include 0 and still have that true, we must define k0= 1 in order that kn+0= knk0= kn. Of course, to do that, you must also limit k to being a positive number. I assume you are limiting k to be a positive integer.

Last edited: Feb 11, 2004
4. Feb 11, 2004

### Ed Quanta

Yeah, I'm sorry. I forgot to say that k is defined as a natural number. Thanks bro, I think I got the proof.