# 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

• You say 'consider' these statements, ... but do you have to prove these statements? – Bram28 Nov 11 '18 at 16:40
• no just whther they are true or false – Carlos Bacca Nov 11 '18 at 16:41
• Ah! OK, can you make that more clear in your post? – Bram28 Nov 11 '18 at 16:43
• For statements that are false, it suffices to give a counterexample. E.g., for iii), let $a=b=c=2$, so that $2\mid 2$ and $2\mid 2$, but $4\not\mid2$. – Barry Cipra Nov 11 '18 at 16:48
• An alternative (possibly nicer) way to show that iv) is true is to note that for positive integers, if $m\mid n$, then $m\le n$. Hence if $a\mid b$ and $b\mid c$ and $c\mid a$, then $a\le b\le c\le a$, which implies $a=b=c$, so that $ac=b^2$ and thus $ac\mid b^2$. – Barry Cipra Nov 11 '18 at 18:02

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.

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.

• Thanks, is what i ahve done a proof in the cases when they are true? – Carlos Bacca Nov 11 '18 at 16:51
• Sure, that is what you do. – Mohammad Riazi-Kermani Nov 11 '18 at 16:53
• does a not divide c in that example – Carlos Bacca Nov 11 '18 at 16:54
• Yes that is proof for true ones – Mohammad Riazi-Kermani Nov 11 '18 at 16:54
• The counter example is an example which shows the statement is false . In my example $3$ does not divide $5$ and $5$ divides $15$ but $3$ divides $15$ which is against the claim of statement – Mohammad Riazi-Kermani Nov 11 '18 at 17:02

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 $$\,c\mid a\mid b)$$