Solving Congruences: Proving a=b

ninjagod123 · Apr 21, 2010

Show that if a [tex]\equiv[/tex] b mod p for all primes p, then a = b.

JSuarez · Apr 21, 2010

Well, a - b must be divisible by all primes p. What is the only way for this to happen?

ninjagod123 · Apr 21, 2010

JSuarez said:

Well, a - b must be divisible by all primes p. What is the only way for this to happen?

Oh hmmm...

The only way is if (a - b) is zero. How would I formally write this up? I guess a - b can't be the product of all primes?

JSuarez · Apr 21, 2010

Every nonzero integer can only be divisible by a finite number of primes.

CRGreathouse · Apr 21, 2010

ninjagod123 said:

I guess a - b can't be the product of all primes?

In a sense, that's what 0 is. It's the "infinity" of the divisibility relation.

HallsofIvy · Apr 22, 2010

If [itex]a> b[/itex] then a- b is a positive number. Since there are an infinite number of primes, there exist a prime, p> a- b. Then p cannot divide a- b so [itex]a\ne b (mod p)[/itex].

If [itex]b> a[/itex] just use b- a instead of a- b.

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Solving Congruences: Proving a=b

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