The only sense I can make of this question is that $m = dq$ and $a \cong b (mod\,m)$ is the same as $as+mt=b$, or $as+(dq)t=b$. I feel as though proving that this holds since $m$ is just a multiple of $d$ is not enough.
Any help is appreciated.
Asked
Active
Viewed 33 times
0
-
Why is it not enough? – José Carlos Santos Nov 07 '19 at 15:44
-
$s$ doesn't happen to be 1 ? – Nov 07 '19 at 15:45
-
My understanding of this is shaky at best, so maybe it is enough to prove this, but I don't see how $a \cong b$ for the modulo of some number still holds for a divisor of that number. – Maksymilian5275 Nov 07 '19 at 15:47
2 Answers
0
As $a \equiv b (\mod m)$, so $m$ divides $a-b$, by definition.
Now as $d$ divides $m$ and as $m$ divides $a-b$, so $d$ too divides $a-b$, and hence $a \equiv b (\mod d)$, as required.
Saaqib Mahmood
- 26,762
0
$$a\equiv b\bmod m\implies a=mx+b\implies a-b=mx$$ and $$a-b=mx\land m=de\implies a-b=dex$$