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I know it is somewhat trivial, but I am not sure what would be an acceptable way to submit the answer to this:

Let $a,b \in \mathbb{Z}$ .

Given the function $f: \mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}$, defined as follows: $f(a,b) := |a·b|$; prove that $f$ is associative.

($·$ is the usual product and || is the operation absolute value, both defined in $\mathbb{Z}$)

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    I don't know how formal your class is in terms of proof... Do you know or have you proven that $|a\cdot b|=|a|\cdot|b|$? – 2017-02-19
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    Maybe you can say that the product on $\mathbb{Z}$ is associative so that $f$ is also associative. Now, if you want to prove that the multiplication on $\mathbb{Z}$ is probably going to be tedious. – 2017-02-19
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    This is what I tried: $1,-2,3\in\mathbb{Z}$, so given that $|1·(-2)|·3|=|2·3|=6=|1·6|=|1·|(-2)·3|$, $f$ is asociative. I doubt this is good enough. – 2017-02-19
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    You can't use particular values to prove a general result. – 2017-02-19
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    Well, that's basically the problem. I can grasp it intuitively but I don't see how to express it appropriately. – 2017-02-19

2 Answers 2

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If $a\ge 0$ and $b\ge 0$ then $ab\ge 0$, $|a|=a$, $|b|=b$. Since $a\cdot b\ge 0$, $|a\cdot b|=a\cdot b=|a|\cdot|b|$

If $a\lt 0$ and $b\lt 0$ then $ab\gt 0$, $|a|=-a$, $|b|=-b$. Since $a\cdot b\gt 0$, $|a\cdot b|=-a\cdot -b=|a|\cdot|b|$

If $a\ge 0$ and $b\lt 0$ then $ab\lt 0$, $|a|=a$, $|b|=-b$. Since $a\cdot b\lt 0$, $|a\cdot b|=a\cdot -b=|a|\cdot|b|$

Similar for $a\lt0$, $b\ge 0$

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We need two properties of absolute value in order to prove $f$ is associative.

  • Property 1: $|a\cdot b|=|a|\cdot|b|$
  • Property 2: $\Big| |a| \Big|=|a|$

Here we go...

If $f$ is associative, then the following must be true. $$\Big||a\cdot b|\cdot c\Big|=\Big|a\cdot|b\cdot c|\Big|$$

By property 1, we have $$\Big| |a|\cdot|b|\cdot c\Big|=\Big|a\cdot |b|\cdot|c|\Big|$$

By property 1 again, we have $$||a||\cdot||b||\cdot |c|=|a|\cdot||b||\cdot||c||$$

By property 2, we get $$|a|\cdot|b|\cdot|c|=|a|\cdot|b|\cdot|c|$$

Done!

Update

Based on the comment of @pjs36, I better formalize the proof.

If $f$ is associative, then the following must be true. $$\Big||a\cdot b|\cdot c\Big|=\Big|a\cdot|b\cdot c|\Big|$$

We work with the left-hand side, $\Big||a\cdot b|\cdot c\Big|$, and work to make this the right-hand side.

\begin{align} \Big||a\cdot b|\cdot c\Big|&=\Big| |a|\cdot|b|\cdot c\Big|&\text{Prop 1 on $|a\cdot b|$}\\ &=\Big||a|\Big|\cdot\Big||b|\Big|\cdot |c|&\text{Prop 1}\\ &=|a|\cdot\Big||b|\Big|\cdot\Big||c|\Big|&\text{Prop 2}\\ &=\Big|a\cdot|b|\cdot|c|\Big|&\text{Prop 1}\\ &=\Big|a\cdot|b\cdot c|\Big|&\text{Prop 1} \end{align}

Therefore, $f$ is associative.

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    This is a good approach (provided the OP can use/prove the two properties) but I think it's not ideal to use equality the whole time -- it currently *looks* like you're showing $f(f(a, b),c) = f(a,f(b,c)) \implies [\text{true statement}]$. While I think each step is a biconditional, and thus a valid approach, it might give a false impression to someone who isn't aware of this (say, a beginner). It might be better to start with $f(f(a,b),c)$ and use the properties to end with $f(a, f(b,c))$. – 2017-02-19
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    Good point. I scratched out the associativity for myself, but didn't think about the teaching of proofs. Thanks! – 2017-02-19