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How can I prove this theorem? I thought of using the definition of divisibility that if $a | b$ if and only if there is an integer $c$ such that $b = ac$, but I have no idea how to do this.

Prove: For all a, b, and c integers with $a ≠ 0,$ if $a | b$ and $b | c$, then $a^2 | bc.$

lone student
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dalerottt
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    If a|b and b|c, then what can you say about a|c? Can you combine that with what you know to get the answer? – Saegusa Jun 29 '21 at 21:36
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    $b=ak_1$ and $c=bk_2$ implies $bc=\ldots$ – Prasun Biswas Jun 29 '21 at 21:39
  • Hints: 1) Divisibility is transitive. 2) If $w|x$ and $y|z$, then $wy|xz$. – Geoffrey Trang Jun 29 '21 at 21:49
  • $$a,b≠0$$

    $$\begin{cases} \frac ba \in\mathbb Z \ \frac cb \in\mathbb Z\end{cases} \implies \frac ba \times \frac cb=\frac ca \in\mathbb Z $$

    $$\frac{bc}{a^2}=\frac ba× \frac ca \in\mathbb Z.$$

     – lone student Jun 29 '21 at 22:18
  • Just do it. What does $a|b$ mean? It means that there is an integer, $m$ so that $b=ma$. And what does $b|c$ mean? It means there is an integer $n$ so that $c = bn$. So if $c = bn$ and $b=am$ then what does $bc$ equal? Is there an integer $K$ so that $bc=a^2K$? – fleablood Jun 29 '21 at 22:51
  • By transitivity of divisibility $,\color{#c00}a\mid b\mid \color{#c00}c,\Rightarrow, \color{#c00}{a\mid c},,$ so $,\color{#0a0}{a\mid b} ,\Rightarrow, \color{#c00}a\cdot \color{#0a0}{a\mid b}\cdot \color{#c00}c,$ by the linked Divisibility Product Rule $\ \ $ – Bill Dubuque Jun 30 '21 at 00:13

1 Answers1

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Firstly, welcome to Math StackExchange!

Let $a,b,c \in \mathbb{Z}$ with $a \neq 0$.

Suppose $a \mid b$ and $a \mid c$. We need to show that $a^2 \mid b \cdot c$.

Since $a \mid b$, we have $ b = a \cdot k_1$ with $k_1 \in \mathbb{Z}$, by the definition of divisibility.

For $a \mid b$, we also have $ c = a \cdot k_2$ with $k_2 \in \mathbb{Z}$.

Multiplying $b$ by $c$:

$ \begin{equation} \label{eq1} \begin{split} b \cdot c & = (a \cdot k_1) \cdot (a \cdot k_2) \\ & = a^2 \cdot k_1 \cdot k_2 \\ & = a^2 \cdot k_3 \text{, with } k_3 = k_1 \cdot k_2 \in \mathbb{Z} \\ \end{split} \end{equation} $

Therefore, we have $a^2 \mid b \cdot c$ by the definition of divisibility.

I hope this helps you!

Daniel Sehn Colao
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  • Hi, thanks so much for the help. Should I assume that (a | b) and (a | c) even though the question says it is (a | b) and (b | c)? – dalerottt Jun 29 '21 at 23:40
  • Yes, @dalerottt. The only thing that the question gives you is that "For all $a, b,$ and $c$ integers with $a \neq 0$". The other part is an implication (if-then statement), you need to prove it. Please, consider accepting my answer if I helped you. – Daniel Sehn Colao Jun 30 '21 at 00:12
  • Yes, it helped, thank you very much. I understand better. – dalerottt Jun 30 '21 at 00:24
  • @dalerottt: You wouldn't be assuming $a\mid c$. It follows from the given premises. $a\mid b$ and $b\mid c$ together imply $a\mid c$. – Prasun Biswas Jun 30 '21 at 03:56