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I was asked to proof if the above statement is True or give a counter example if it is False.

However, I say it is True.

Reason: since if m divides a-b then we say a is congruent to b or a is congruent to b modulo iff m|(a-b).

is my reason correct or their is a Valid reasoning and proof to the above statement?

Bill Dubuque
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Surdz
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    What is your definition of $a \equiv b \bmod m$? Some would say you're being asked to prove a definition, which doesn't make sense. – KCd Apr 09 '16 at 23:09
  • Yes, it follows immediately from the definition of the $!\bmod$ congruence relation - see the linked dupe. – Bill Dubuque May 03 '23 at 18:08

1 Answers1

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Let $a = k_1m + r_1$ and $b = k_2m+ r_2$ by the Division Algorithm where the $k$'s and $r$'s are unique integers such that $m>r_1\ge 0\land m>r_2 \ge 0$.

$$a-b = (k_1m + r_1) - (k_2m+r_2)$$ $$a-b = m(k_1 - k_2) + r_1 - r_2$$

Now, if the LHS is a multiple of $m$ by assumption, so is the LHS. So, $m|r_1 - r_2$. But if this must happen $r_1 = r_2$ which implies that $a \equiv b \pmod m$.

TheRandomGuy
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