Prove ## a/d\equiv b/d \mod n/d ##.

[Math100]

Homework Statement

Prove the following assertion:
If $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$, then $a/d\equiv b/d \mod n/d$.

Relevant Equations

None.

Proof:

Suppose $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$.
Then $n\mid (a-b)\implies mn=a-b$ for some $m\in\mathbb{Z}$.
Note that $a=id\implies \frac{a}{d}=i$, $b=jd\implies \frac{b}{d}=j$, and $n=kd\implies \frac{n}{d}=k$ for some $i, j, k\in\mathbb{Z}$.
Thus $id-jd=m(kd)\implies i-j=mk\implies \frac{a}{d}-\frac{b}{d}=m\frac{n}{d}\implies \frac{a}{d}\equiv \frac{b}{d} \mod \frac{n}{d}$.
Therefore, if $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$, then $a/d\equiv b/d \mod n/d$.

 

[fresh_42]

Homework Statement:: Prove the following assertion:
If $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$, then $a/d\equiv b/d \mod n/d$.
Relevant Equations:: None.

Proof:

Suppose $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$.
Then $n\mid (a-b)\implies mn=a-b$ for some $m\in\mathbb{Z}$.
Note that $a=id\implies \frac{a}{d}=i$, $b=jd\implies \frac{b}{d}=j$, and $n=kd\implies \frac{n}{d}=k$ for some $i, j, k\in\mathbb{Z}$.
Thus $id-jd=m(kd)\implies i-j=mk\implies \frac{a}{d}-\frac{b}{d}=m\frac{n}{d}\implies \frac{a}{d}\equiv \frac{b}{d} \mod \frac{n}{d}$.
Therefore, if $a\equiv b \mod n$ and the integers $a, b, n$ are all divisible by $d>0$, then $a/d\equiv b/d \mod n/d$.

That's fine.

Edit: Only remark: the product "id" is a bit unlucky because $id$ often abbreviates the identity function.

 

[Math100]

That's fine.

Edit: Only remark: the product "id" is a bit unlucky because $id$ often abbreviates the identity function.

I agree. I haven't thought about that before.