|Dec14-09, 04:21 PM||#1|
Prime number dividing fractions.
Let p be a prime number.
Let A be an integer divisible by p but B be an integer not be divisible by p.
Let A/B be an integer.
How do I prove that A/B is divisible by p?
This sounds like a simple question but I just can't get it. I'm doing it in relation to proving Fermat's little theorem. (a^p = a mod p for all integers a) I'm trying to understand why the binomial coefficients in the binomial expansion of (1+a)^n are all divisible by p (=0 mod p) for all the terms with powers [1, p-1].
|Dec14-09, 04:33 PM||#2|
Decompose A and B into prime factors. Since A/B is an integer all the B factors cancel A factors. However B did not contain p, so p remains a factor in A/B.
|Dec14-09, 04:54 PM||#3|
Thank you, mathman! That instantly resolved my question (and I was banging my head against it for like an hour)!
Damn that fundamental theorem of arithmetic!
|binomial, number theory|
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