
#1
Feb1004, 05:29 PM

P: 297

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
Feb1004, 05:41 PM

Sci Advisor
HW Helper
P: 2,538

"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
Feb1104, 10:57 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

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 "k^{n}" that you are using. The definitions I know give k^{0}= 1 as part of the definition. Often a definition starts by defining k^{1}= k, the recursively defining k^{n}= k*k^{n1} but that only works for the natural numbers. Since, for m, n natural numbers, we have k^{n}*k^{m}= k^{n+m}, in order to extend k^{n} to include 0 and still have that true, we must define k^{0}= 1 in order that k^{n+0}= k^{n}k^{0}= k^{n}. 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. 



#4
Feb1104, 11:08 AM

P: 297

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



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