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Define $h\colon \mathbb{Z}/\sim ~16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$

a. Prove $h$ is well defined.

b. Compute $h(a)$ where $a = \{[0]16, [3]16, [6]16\}$.

c. Compute $h^{-1}([10]24)$

Is the following correct:

a. $h$ is well defined since each for all $a$ $|h([a]16)|\equiv 1$.

b. $\{[0]24, [9]24, [18]24\}$

c. $\emptyset$

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    For a. how about for h of all equivalence classes of Z/~16 there is one equivalence class in Z/~24?2010-12-05

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Your answers to (b) and (c) are correct.

For (a), you need to prove that if a=b mod 16, i.e. $a-b=16n $ for some integer $n$, then $3a=3b$ mod 24, i.e. $3a-3b=24m$ for some integer $m$. Can you do that?