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Let $m, n \in N$ and $a, b \in Z$ so that $ m\mid n$ and $a \equiv b \mod n.$ Then $a \equiv b \mod m$

MeredithGrey
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    $a\equiv b \mod n \implies a = b + n\lambda ,, $. Since $m \mid n $ , we have $n = m\alpha.$ So we get $$ a = b + m\alpha \implies \color{#0b2}{a\equiv b \mod m}$$ – The Demonix _ Hermit Nov 26 '19 at 16:35
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    Consider an example $26 \equiv 5 \pmod {21}$ because $26 = 5 + 21$. And $21= 37$ so $26 = 5 + 37$. So $26\equiv 5\pmod 3$ and $26 \equiv 5 \pmod 7$. – fleablood Nov 26 '19 at 16:47
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    $m|n$ means $m$ divides into $n$ or that there is an integer $k$ so that $n = km$. $a \equiv b\pmod n$ means $n|(a-b)$ or $n$ divides evenly into $a-b$. So you are asked to prove if $m$ divides evenly into $n$ and $n$ divides evenly into $a-b$ then ...... – fleablood Nov 26 '19 at 16:51

2 Answers2

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just do it.

$a \equiv b \pmod n \implies$

$n|a-b$.

And $m|n$ and $n|a-b$ so $m|a-b$.

So $a\equiv b \pmod m$

.....

Alternatively.

$a \equiv b \pmod n\implies$ there is a $k $ so that $a = b + kn$.

And $m|n \implies$ there is a $j$ so that $n= jm$.

So $a=b + k(jm) = b+(kj) m$.

So $a \equiv b\pmod m$.

.........

Consider an example:

$26 \equiv 5 \pmod {21}$.

we know that is true because $26 = 5 + 21$.

But $26 \equiv 5 \pmod 7$ because $26 = 5 + 3*7$.

And $26 \equiv 5 \pmod 3$ because $26 = 5 + 7*3$.

.....

Does that not make sense?

fleablood
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To say $m|n$ means there exists some natural number $x$ such that $mx=n$.

To say $a \equiv b$ $mod n$ means that $(a-b)=yn$ for some integer $y$.

We want to show there exists some value $z$ such that $(a-b)=zm$.

Since we know $(a-b)=yn$ and $mx=n$, we can substitute like so: $(a-b)=y(mx)$. Observe that $z=yx$

EDS
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