elementary number theory - question on 'divides'

Let $a,b,c>0$ be natural numbers. Consider the following statments:

i) if $a\nmid b$ and $b |c$ then $a\nmid c$

ii) if $a |b$ and $b |c$ then $ab |bc$

iii) if $a |c$ and $b |c$ then $ab |c$

iv) If $a |b$ and $b |c$ and $c |a$ then $ac |b^2$

Question: Determine whether each statement is true or false.

$q_1,q_2,q_3$ are natural numbers

So for i) a is not a factor of b, and b divides c, say $c=q_1b$ so in the case when a is a factor of $q_1$ this is false.

for ii) a divide b implies $a |b=aq_1$ and b divides c so $aq_1 |c=aq_1q_2$ and as $aaq_1 |aq_1aq_1q_2$ which is true..

for iii) a divides c implies $a |c=aq_1$, b divides c implies $b |c=bq_2$ so ab does not divide c when a is not a factor of $q_2$ or b is not a factor of $q_1$ so false

for iv) a divides b so $a |b=aq_1$ b divides c $b=aq_1 |c=aq_1q_2$ and c divides a $aq_1q_2 |a$ which implies $q_1,q_2$ are 1 so this means that a,b and c must beequal so this is always true.

This seems like a really long way to do this is it right and is there a nicer way to get this done?

Thanks

Answer

You have the right idea for all of them. However, to show that (i) and (iii) are not true, you must give specific examples of $a,b,c.$

Your proof of (iv) looks optimal, though your proof of (ii) could be improved a bit. Once you get to $c=aq_1q_2,$ it follows directly that $$bc=baq_1q_2=abq_1q_2,$$ so that $ab\mid bc,$ as desired.