question on 'divides'

Question

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 1

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.

Answer 2

For the ones which are not true you provide a simple counter example ,

For example for first one let $a=3$ and $b=5$ and $c=15$

As you see this is a counter example so the first statement is false.

For the true ones you have to prove them and it is sometimes lengthy.

Answer 3

Hint $ $ (i),(iii) are refutable with $c=a$, and (ii),(iv) are a special cases of $\,a\mid b, A\mid B\,\Rightarrow\, aA\mid bB\,$ (for (iv) use $\,a\mid b\,$ and $\,c\mid a\mid b)$