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.

Question

Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (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: $a\mid b\Rightarrow b=as$ and $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$ $$ $$ $=as + 2(bt+au)$

Yea then I just got lost.

Answer 1

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.

Answer 2

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

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

Answer 3

According to the definition of division, $b=as$ and $c=bt$

$$bc=asbt$$

$$c=\frac{asbt}b$$

$c=a(st) \implies s$ and $t$ are integers which makes $st$ also an integer this means $a|c$

Answer 4

Let $a,b,c\in\mathbb{Z}$, then

1) if $a\mid b$ we can write $b=a\cdot k$ for some $k\in\mathbb{Z}$. Similarly for $c$ we can say that $c=a\cdot l$ for some $l\in\mathbb{Z}$. Now multiplication distributes over addition we have $b+c=ak+al=a(k+l)$ which is obviously divisible by $a$ since it has $a$ as a factor.

2) if $a\mid b$ then $b=a\cdot k$ for some $k\in\mathbb{Z}$. So $bc=akc=a(kc)$ since multplication is associative and the last expression is clearly divisible by $a$.

3) if $a\mid b$ and $b\mid c$ so we can write $b=ka, \ k\in\mathbb{Z}$ and $c=bl,\ l \in\mathbb{Z}$. Combining these two gives $c=bl=(ka)l=(ak)l=a(kl)$ since multiplication is both commutative and associative. The last expression is clearly divisible by $a$.