Prove that ## x\equiv 1\pmod {2n} ##

[Math100]

Homework Statement

If $x\equiv a\pmod {n}$, prove that either $x\equiv a\pmod {2n}$ or $x\equiv a+n\pmod {2n}$.

Relevant Equations

None.

Proof:

Suppose $x\equiv a\pmod {n}$.
Then $x=a+tn$ for some $t\in\mathbb{Z}$.
Now we consider two cases.
Case #1: Suppose $t$ is even.
Then $t=2m$ for some $m\in\mathbb{Z}$.
Observe that $x=a+tn=a+2nm$.
Thus $x\equiv a\pmod {2n}$.
Case #2: Suppose $t$ is odd.
Then $t=2m+1$ for some $t\in\mathbb{Z}$.
Observe that $x=a+tn=a+(2m+1)n=a+n+2mn$.
Thus $x\equiv a+n\pmod {2n}$.
Therefore, if $x\equiv a\pmod {n}$, then either $x\equiv a\pmod {2n}$ or $x\equiv a+n\pmod {2n}$.

 

[fresh_42]

Homework Statement:: If $x\equiv a\pmod {n}$, prove that either $x\equiv a\pmod {2n}$ or $x\equiv a+n\pmod {2n}$.
Relevant Equations:: None.

Proof:

Suppose $x\equiv a\pmod {n}$.
Then $x=a+tn$ for some $t\in\mathbb{Z}$.
Now we consider two cases.
Case #1: Suppose $t$ is even.
Then $t=2m$ for some $m\in\mathbb{Z}$.
Observe that $x=a+tn=a+2nm$.
Thus $x\equiv a\pmod {2n}$.
Case #2: Suppose $t$ is odd.
Then $t=2m+1$ for some $t\in\mathbb{Z}$.
Observe that $x=a+tn=a+(2m+1)n=a+n+2mn$.
Thus $x\equiv a+n\pmod {2n}$.
Therefore, if $x\equiv a\pmod {n}$, then either $x\equiv a\pmod {2n}$ or $x\equiv a+n\pmod {2n}$.

Yep. Nothing to add or complain about.

 

[Math100]

Yep. Nothing to add or complain about.

You've never complained about my proof.