Mathematical Induction: Proof for all Positive integers

LiquiGels

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Jan 25, 2010
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Hey,

I have this problem that I need to proof using mathematical induction.

The problem:
For all the positive integers k, k could be written as the product of some odd number and some power of two.
k = (2i + 1) x (2^e), where i and e are both natural numbers and e is the exponent of 2.

My work up till now:
I know that k could turn out to be even or odd and I've proved the base case to hold for k = 1, while i and e equal 0. But I cannot form the inductive step for each of the separate cases (even and odd)..Any ideas? :?:
 
k+1 = 2^n(2m+1) + 1

Either n=0 or n>0.

If n=0, k+1 = 2(m+1).
Note m+1 < k+1, hence we write m+1=2^b(2a+1)
k+1 = 2^(b+1)(2a+1) and we're done

If n > 0, k+1=2^n(2m+1)+1 = 2^(m+n)+2^n+1
if not already proved in some manner, prove this: even # + even # + odd # = odd #

hence k+1 is odd => k+1 = 2^0(2p+1)
 
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