Prove the following assertion: ## ca\equiv cb \mod cn ##.

[Math100]

Homework Statement

Prove the following assertion:
If $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.

Relevant Equations

None.

Proof:

Suppose $a\equiv b \mod n$ and $c\mid n$.
Then $n\mid (a-b)\implies kn=a-b$ for some $k\in\mathbb{Z}$.
Since $c\mid n$, it follows that $ca-cb=ckn\implies ca-cb=k(cn)$.
Thus $ca\equiv cb \mod cn$.
Therefore, if $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.

 

[SammyS]

Homework Statement:: Prove the following assertion:
If $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.
Relevant Equations:: None.

Proof:

Suppose $a\equiv b \mod n$ and $c\mid n$.
Then $n\mid (a-b)\implies kn=a-b$ for some $k\in\mathbb{Z}$.
Since $c\mid n$, it follows that $ca-cb=ckn\implies ca-cb=k(cn)$.
Thus $ca\equiv cb \mod cn$.
Therefore, if $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.

The given information does not say that $c$ divides $n$ .

Furthermore, you don't need that in the place that you used it. All that's needed there is some basic algebra.

 

[Math100]

The given information does not say that $c$ divides $n$ .

Furthermore, you don't need that in the place that you used it. All that's needed there is some basic algebra.

Then should I write "Suppose $a\equiv b \mod n$ and $c>0$" in the first sentence? And how can I show/prove or where should I insert that $c\mid n$?

 

[SammyS]

Then should I write "Suppose $a\equiv b \mod n$ and $c>0$" in the first sentence? And how can I show/prove or where should I insert that $c\mid n$?

Not anywhere in this proof.

The problem does not state that $c\mid n$. So, as you might expect, it's not needed for the proof.

 

[Math100]

Suppose $a\equiv b \mod n$ and $c>0$.
Then $n\mid (a-b)\implies kn=a-b$ for some $k\in\mathbb{Z}$.
Note that $ca-cb=ckn\implies ca-cb=k(cn)$.
Thus $ca\equiv cb \mod cn$.
Therefore, if $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.

How about this revised proof above?

 

[SammyS]

Suppose $a\equiv b \mod n$ and $c>0$.
Then $n\mid (a-b)\implies kn=a-b$ for some $k\in\mathbb{Z}$.
Note that $ca-cb=ckn\implies ca-cb=k(cn)$.
Thus $ca\equiv cb \mod cn$.
Therefore, if $a\equiv b \mod n$ and $c>0$, then $ca\equiv cb \mod cn$.

How about this revised proof above?

Yes. That's good.