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?


  • $\begingroup$ You say 'consider' these statements, ... but do you have to prove these statements? $\endgroup$ – Bram28 Nov 11 '18 at 16:40
  • $\begingroup$ no just whther they are true or false $\endgroup$ – Carlos Bacca Nov 11 '18 at 16:41
  • $\begingroup$ Ah! OK, can you make that more clear in your post? $\endgroup$ – Bram28 Nov 11 '18 at 16:43
  • $\begingroup$ 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 $4\not\mid2$. $\endgroup$ – Barry Cipra Nov 11 '18 at 16:48
  • 1
    $\begingroup$ 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 $ac\mid b^2$. $\endgroup$ – 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.

  • $\begingroup$ Thanks, is what i ahve done a proof in the cases when they are true? $\endgroup$ – Carlos Bacca Nov 11 '18 at 16:51
  • $\begingroup$ Sure, that is what you do. $\endgroup$ – Mohammad Riazi-Kermani Nov 11 '18 at 16:53
  • $\begingroup$ does a not divide c in that example $\endgroup$ – Carlos Bacca Nov 11 '18 at 16:54
  • $\begingroup$ Yes that is proof for true ones $\endgroup$ – Mohammad Riazi-Kermani Nov 11 '18 at 16:54
  • $\begingroup$ 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 $\endgroup$ – 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 $\,c\mid a\mid b)$


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