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Here f is a function that returns the number of times a number is wholly divisible by 2, i.e. the number of times a number is divisible by 2 until it becomes an odd number.

Assumptions:

f accepts only zero or a positive integer
If x = 0 or x is an odd number then f(x) = 0

a, b and n are positive integers
a, b are odd
n is even
a > b

I would very much appreciate methods for this kind of problem. Thanks in advance!

2007-07-22 17:17:38 · 1 answers · asked by semyaza2007 3 in Science & Mathematics Mathematics

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2007-07-22 20:21:59 · update #1

I have posted my answer at http://www.geocities.com/sem26k/index.htm

2007-07-23 04:35:00 · update #2

1 answers

Problems like this are done by mathematical induction. The steps to solve them are:

1. Show that it's true for the first value--so in this case, n =2 would be the first value and f(2) = 1 so f(2) -1 =0.
2. Assume what you are trying to prove (that is the statement above with n.
3. Show that if it's true for n it's true for the next value also, which in this case would be n+2, n must be even.

The idea behind induction is that statements for all n fall down like a line of dominoes once these steps are followed. That is because the first one implies the second one, the second one implies the third, etc.

2007-07-22 18:06:22 · answer #1 · answered by pegminer 7 · 0 1

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