Solving Congruences: Proving a=b

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Show that if a \equiv b mod p for all primes p, then a = b.
 
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Well, a - b must be divisible by all primes p. What is the only way for this to happen?
 
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?
 
Every nonzero integer can only be divisible by a finite number of primes.
 
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.
 
If a> b 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 a\ne b (mod p).

If b> a just use b- a instead of a- b.
 

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