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October 9, 2015
Posted by **sammy** on Tuesday, January 22, 2008 at 8:03pm.

- math (extra credit)! -
**Count Iblis**, Tuesday, January 22, 2008 at 8:27pmConsider first positive powers. Let's define:

f(a,n) = a^n

where a is a real number and n is a positive integer. This is well defined:

f(a,n) = a^n =

a*a*a...(n-factors in total).

Now, we haven't defined f(a,n) when n is not a positive integer, so, in theory, you are free to extend the function f(a,n) in any arbitrary way to other numbers. However, the function

f(a,n) has some nice properties and you want to preserve those when you extend the definition of f(a,n).

E.g.:

f(a,n+m) = f(a,n)*f(a,m)

If we want to extend the function to

n = 0 while not violating this equation, then we must choose f(a,0)= 1. To see this, take m = 0, in the above equation:

f(a,n) = f(a,n)*f(a,0) --->

f(a,0) = 1.

So, what mathematicians have done here (and in many other cases) is to take some of the properties of the function they want to extend to a larger set as the definition of the function (they uniquely define the function).

- math (extra credit)! -
**sammy**, Tuesday, January 22, 2008 at 8:46pmthx but this didnt help i need to know why n^0= 1

- math (extra credit)! -
- math (extra credit)! -
**Damon**, Tuesday, January 22, 2008 at 8:52pmHe told you, and I think it was the coolest answer yet!