Is [a]^-1 = [0] if and only if p|a?

[zoner7]

Homework Statement

in Zp, where a does not equal 0, that [a][x] = [a][y] is the same as [x] = [y].

This was a question on a test I just finished. Just curious how wrong my answer is. So I sort of ignored the congruence class part. I just said:

* [a]^-1 represents the inverse of [a] *

[a][x] = [a][y]
[a]^-1[a][x] = [a][y][a]^-1 (by associative law)
[1][x] = [1][y] (by the law of inverses)
[x] = [y] (since 1 is the identity element under multiplication in congruence classes).

so yea... did i do anything right?

thanks for the help in advance

 

[Office_Shredder]

At the very least mention why you think [a]^-1 should exist. (it's non-zero). In fact, this is technically correct but I suspect more detail was expected (since if you know Z_p is a field, this is a useless question). Here's how you probably were supposed to do it (I left two details to fill in)

[a][x] = [a][y] if and only if p|(ax-ay) if and only if p|a(x-y)

p|a(x-y) implies p|a or p|(x-y) (why?)

p does not divide a (why?)

Hence p|(x-y) and this gives [x]=[y]

and you can run the same thing backwards

EDIT TO ADD: I forgot to mention I'm assuming p is prime here, as otherwise the question's wrong

 

[zoner7]

p|a(x-y) implies p|a or p|(x-y) (why?)
This is true, because since p is prime, we are able to apply Euclid's lemma.

p does not divide a (why?)
I honestly couldn't tell you why p does not divide a.

so... did I do anything correct at all...?

 

[Office_Shredder]

p does not divide a (why?)
I honestly couldn't tell you why p does not divide a.

Go back and read the question carefully... what do you know about a?

so... did I do anything correct at all...?

Nothing you did was wrong per se, it's just that you didn't support why there should exist a congruence class of the form [a]^-1, which was probably what was expected of you (alternatively, and easier is the argument I posted).

 

[zoner7]

Go back and read the question carefully... what do you know about a?
/QUOTE]

ahh, got you.

A does not equal 0. but why does that mean that p cannot divide a. any prime p divides into 0.

 

[Office_Shredder]

By definition

[a] = [0] if and only if p|a