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If I have the map $f: \mathbb{Z}_{20} \rightarrow \mathbb{Z}_{15}$ where $x + 20\mathbb{Z} \mapsto x + 15\mathbb{Z}$, I have to determine whether this is well defined or not. After staring at it for only a couple minutes, I am pretty sure it is not well defined. Now, I must find a contradiction to back up my answer. Not sure how to do that though. Maybe the wording of the problem is just confusing me.

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$f(1 + 20\mathbb{Z}) = 1 + 15\mathbb{Z}$. But $1 + 20\mathbb{Z} = 21 + 20\mathbb{Z}$ and $f(21 + 20\mathbb{Z}) = 21 + 15\mathbb{Z} = 6 + 15\mathbb{Z} \neq 1 + 15\mathbb{Z} $. –  Jonas Aug 23 '13 at 0:45

2 Answers 2

Recall what it means for $f$ to be well-defined:

$$x + 20 \mathbb{Z} = y + 20 \mathbb{Z} \implies x + 15 \mathbb{Z} = y + 15 \mathbb{Z}$$

Now if $x + 20\mathbb{Z} = y + 20\mathbb{Z}$, then $x - y \in 20 \mathbb{Z}$, so $20 | x - y$.

On the other hand, $x + 15 \mathbb{Z} = y + 15 \mathbb{Z}$ is the same as saying that $15 | x - y$.

This should guide you in finding a counterexample.

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Okay so what about this contradiction. Suppose I start with something like $18$.Then, $18 + 20\mathbb{Z} = 38 + 20\mathbb{Z}$, but this implies that $18 + 15\mathbb{Z}$ is $3$, and $38 + 15\mathbb{Z}$ is $8$. –  user91154 Aug 23 '13 at 0:57
    
@user91154 Yes, that is a valid counterexample. –  T. Bongers Aug 23 '13 at 0:58

It is not well defined. $15=35$ (mod $20$), but then we get $0=f(15)=f(35)=5$ (mod $15$), which isn't true.

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