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Let $a,b \in \mathbb{Z}$ and suppose that $a | b$ and $b |a$. Then $a=b$ or $a=-b$

My attempt

$a|b \Rightarrow b=k_1a$ for some $k_1$

$b|a \Rightarrow a=k_2b$ for some $k_2$

By substitution $a=k_1k_2a \Rightarrow a(1-k_1k_2)=0$

So, we have two cases:-

Case $(1)$

$k_1k_2=1 \Rightarrow k_1=k_2=\pm 1$

Thus, $a=\pm b$

Case $(2)$

I am stuck here, as $a=0 \Rightarrow b=0$.

But then division by $0$ doesn't make sense. What to do?

Note

This question is not a duplicate as it explicitly asks the question for $\mathbb{Z}$ and not $\mathbb{N}$ or nonzero integers.

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    In number theory we define *divisibility* by $\, a\mid b \iff an = b\ $ for some $n\in\Bbb Z.\,$ In particular $\, 0\mid b\iff b = 0.\ $ Now you can continue your proof. Note the statement concerns the *relation* "divides", not the *operation* "division" (though they are related, they are not identical). – 2017-02-06

2 Answers 2

1

If $a=0$ then, since $a\mid b$ by hypothesis , we must have that $0=ak=b$ for some $k\in\mathbb{Z}$.

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If you read $a | b$, then you can suppose that $a \ne 0$

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    Read the question carefully. I specifically ask what happens when $a=0$. – 2017-02-06
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    @user330477 Just for clarification, when we supposed that $a|b$ and $b|a$ then its understood that $a\neq 0$ and $b\neq 0$. You don't need to ask question like what will happen if $a=0$ because that is not possible to happen. – 2017-02-06
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    The condition $a | b$ is given, which means that $a$ cannot be zero. Otherwise the assumptions will be false. – 2017-02-06
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    @user330477 Zero is divided by everything... – 2017-02-06
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    @ΘΣΦGenSan So this means that case 2 doesn't arise as $a \neq 0$ when $a|b$. So, my proof is ok without that – 2017-02-06
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    The definition of a|b that I use allows a=0. Since I would say that a|b means "There exists k with ak=b" which is true with a=0, b=0 and k=1 or anything else, i.e. 0|0 is true. Of course, that fits the statement OP is trying to prove (see mathbeing's answer). – 2017-02-06
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    @Chessanator Now I am completely confused – 2017-02-06
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    @user330477 Precisely, you don't need case 2. Look, you already have $$a(1-k_1k_2)=0.$$ So, since $a\neq 0$, you are forced to conclude that $1-k_1k_2=0$ ,that is, $1=k_1k_2$ and hence $k_2|1$. Hence, $k_2=1$ or $k_2=-1$. – 2017-02-06
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    @user330477 There are two sensible definitions of a|b on the integers. The one which I now think you are using (given your confusion at my comment) and which the other commentators are using is "Consider the fraction b/a. If it is a whole number then a|b". This definition doesn't allow a=0 because then you would have division by zero. If this is the defintion you are using then GenSan's comment finishes your proof. – 2017-02-06
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    @user330477 The other definition is the one I gave in my previous comment: it agrees with the other definition everywhere but also allows 0|0. Experience has taught me that definitions like that are better because they treat zero more naturally. Once again, mathbeing was also using this second definition and gave the conclusion of the proof using it in their answer. – 2017-02-06