Verify that the function is a injective

[Bashyboy]

Homework Statement

The function is $\phi " \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{24}$, where the rule is $\phi ([a]_{12}) = [2a]_{24}$. Verify this is a injection

Homework Equations

The Attempt at a Solution

Let $[x]_{12} ,[y]_{12} \in \mathbb{Z}_{12}$ be arbitrary. Suppose that $\phi([x]_{12}) = \phi( [y]_{12})$ is true, then

$[2x]_{24} = [2y]_{24}$ Two congruence classes are equal iff $2x \equiv 2y ~(\mod 24)$, which is by definition

$2x - 2y = 24k$, where there exists a $k \in \mathbb{Z}$

Here the difficulty I am facing. Can I just simply let $k=0$, which we give me $x=y$? If so, why?

 

[PeroK]

Homework Statement

The function is $\phi " \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{24}$, where the rule is $\phi ([a]_{12}) = [2a]_{24}$. Verify this is a injection

Homework Equations

The Attempt at a Solution

Let $[x]_{12} ,[y]_{12} \in \mathbb{Z}_{12}$ be arbitrary. Suppose that $\phi([x]_{12}) = \phi( [y]_{12})$ is true, then

$[2x]_{24} = [2y]_{24}$ Two congruence classes are equal iff $2x \equiv 2y ~(\mod 24)$, which is by definition

$2x - 2y = 24k$, where there exists a $k \in \mathbb{Z}$

Here the difficulty I am facing. Can I just simply let $k=0$, which we give me $x=y$? If so, why?

No. You would have to prove that k = 0.

Instead, why not simplify $2x - 2y = 24k$?

 

[exclamationmarkX10]

No, you can't just let $k = 0$ (try going through your arguments with $x=12$ and $y=0$). Remember what you are trying to show: You want to show that the congruence classes of $x$ and $y$ are equal, not $x=y$.

 

[Bashyboy]

Oh, heavens. This was much simpler than I thought it was. If I take $2x - 2y = 24k$ and divide by $2$, then I get $x-y = 12k$, which would eventually lead to $[x]_{12} = [y]_{12}$. Therefore, the function is an injection.