# Verify that the function is a injective

## 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

## 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?

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## 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

## 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##?

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 $x=y$.

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.