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.