Prove that if a|b and b|c then a|c using a column proof that has steps in the first column and the reason for the step in the second column.

Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then
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(i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$)
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(ii) if $a$ | $b$ and $a$|$bc$ for all integers $c$;
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(iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.

Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column
and the reason for the step in the second column.

My book is really vague. I’m not really sure what to do..

First I was thinking something like this: $a$|$b$ => $b=as$ and $b$|$c$ => $c=bt$ and $a$|$c$ => $c = au$
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$b + c + c = b + 2c$
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$=as + 2(bt+au)$

Yea then I just got lost.

Solutions Collecting From Web of "Prove that if a|b and b|c then a|c using a column proof that has steps in the first column and the reason for the step in the second column."

If you are allowed to use fractions then all inferences are immediate consequences of the fact that $\,\Bbb Z\,$ is closed under addition and multiplication – see below. If you are not allowed to use fractions then you can translate the below proofs to integers by clearing denominators, e.g. for $(i)$ we get $\ b=aj,\, c = ak\,\Rightarrow\, a(j + k) = b + c$

$( i)\quad \dfrac{b}a,\,\dfrac{c}a\in\Bbb Z\ \Rightarrow\ \dfrac{b}{a}+\dfrac{c}a = \dfrac{b+c}{a}\in\Bbb Z$

$(ii)\quad \dfrac{b}a,\ c\in\Bbb Z\ \Rightarrow\ c\, \dfrac{b}a = \dfrac{cb}a\in\Bbb Z$

$(iii)\ \ \ \dfrac{b}a,\,\dfrac{c}b\in\Bbb Z\ \Rightarrow\ \dfrac{b}a\dfrac{c}b = \dfrac{c}a\in\Bbb Z$

Note: $ $ if $\,b=0\,$ is allowed in $\,(iii)\,$ then you need to consider that case too.

See this
$$ a|b \implies b = ma \quad \rm and \quad a|c \implies c=na, \quad m,n \in \mathbb{Z} $$

which gives us

$$ b+c = (m+n)a = p a \implies a|(b+c),\quad p=n+m \in \mathbb{Z}. $$