Prime number dividing fractions.

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SUMMARY

The discussion centers on proving that the fraction A/B, where A is an integer divisible by a prime number p and B is not, is also divisible by p. The user connects this proof to Fermat's Little Theorem, specifically the relationship a^p ≡ a (mod p) for all integers a. The resolution involves recognizing that the prime factorization of A contains p, while B's factorization does not, leading to the conclusion that p remains a factor in the integer result of A/B. The discussion highlights the importance of the fundamental theorem of arithmetic in understanding these divisibility properties.

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  • Understanding of prime numbers and their properties
  • Familiarity with integer divisibility concepts
  • Knowledge of Fermat's Little Theorem
  • Basic principles of binomial expansion and coefficients
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  • Explore the fundamental theorem of arithmetic and its implications
  • Learn about binomial coefficients and their properties in modular arithmetic
  • Investigate integer factorization techniques and their applications
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Mathematicians, students studying number theory, and anyone interested in the properties of prime numbers and their applications in proofs and theorems.

Jolb
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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].
 
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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.
 
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!
 

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