# Thread: Modular solutions question

1. Originally Posted by flyingsquirrel
For example : $\displaystyle x \equiv 0\,[2] \Rightarrow 2x \equiv 0\,[2]$ but the other way is wrong : $\displaystyle 2x \equiv 0\,[2] \not \Rightarrow x \equiv 0\,[2]$

If we multiply the modulo and the congruence by 2, both ways are correct : $\displaystyle x \equiv 0\,[2] \Leftrightarrow 2x \equiv 0\,[4]$
Because 2 has no inverse in Z/2Z.

Here, we multiplied by the inverse of a number, so it'll always work.

2. Originally Posted by flyingsquirrel
For example : $\displaystyle x \equiv 0\,[2] \Rightarrow 2x \equiv 0\,[2]$ but the other way is wrong : $\displaystyle 2x \equiv 0\,[2] \not \Rightarrow x \equiv 0\,[2]$

If we multiply the modulo and the congruence by 2, both ways are correct : $\displaystyle x \equiv 0\,[2] \Leftrightarrow 2x \equiv 0\,[4]$
Originally Posted by Moo
Because 2 has no inverse in Z/2Z.

Here, we multiplied by the inverse of a number, so it'll always work.
It has nothing to do with inverses in the forward direction. If $\displaystyle a\equiv b(\bmod n)$ then it means $\displaystyle n|(a-b)$ then certainly $\displaystyle n|c(a-b)$ so $\displaystyle n|(ac-bc)$ thus $\displaystyle ac\equiv bc(\bmod n)$.

3. Originally Posted by ThePerfectHacker
It has nothing to do with inverses in the forward direction. If $\displaystyle a\equiv b(\bmod n)$ then it means $\displaystyle n|(a-b)$ then certainly $\displaystyle n|c(a-b)$ so $\displaystyle n|(ac-bc)$ thus $\displaystyle ac\equiv bc(\bmod n)$.
I think we were more talking about the reversal of the implication lol, a little off-topic

Page 2 of 2 First 12