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I was wondering if my proof is right and if there are better methods of solving this proof?

My proof:

Assume $x \equiv_k y$ and $y \equiv_k z$ is true. This means that :

\begin{align}x &= y + kq,&&\text{where }\;q ∈\Bbb Z,\tag{1} \\ y &= z + ka,&& \text{where } a ∈ \Bbb Z. \tag{2} \end{align}

By substitution of $(2)$ into $(1)$ we get:

\begin{align} x &= z + ka + kq \\ &= z + k(a + q),&&\qquad \text{where }\: a ∈\Bbb Z \text{ and } q ∈ \Bbb Z \end{align}

This proves $x \equiv_k z$ is true. $\blacksquare$

Shaun
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jame_smith
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  • It's fine for me (except the formulæ code…). – Bernard Jan 04 '20 at 23:31
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    you have shown congruence to be a transitive relation – J. W. Tanner Jan 04 '20 at 23:36
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    Your argument is fine, but you might have mentioned that $(a+q)$ is an integer just before the "This proves..." punch line. Also, the symbol $\equiv_k$ is not standard, even though your use of it here makes it perfectly clear what you mean. – kimchi lover Jan 04 '20 at 23:40

1 Answers1

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Direct alternate proof.
k|x - y, k|y - z, k|x - y + y - z = x - z.

William Elliot
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