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Old 2009-07-19, 23:25   #12
xkey
 
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Quote:
Originally Posted by Mini-Geek View Post
Is this assuming k and n are integers? If so, I don't see how this could be.
think n = 0
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Old 2009-07-19, 23:30   #13
Primeinator
 
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Quote:
Originally Posted by Mini-Geek View Post
Is this assuming k and n are integers? If so, I don't see how this could be.
It makes sense- if you acknowledge K to be an integer, and n to be an integer, then 2^n is always an integer, multiplied by another integer (k) will always be an integer.

Last fiddled with by Primeinator on 2009-07-19 at 23:30
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