-1

Attempt:

Prove that if $A|B$ and $B|A$, then $A=B$? Both $A$ AND $B$ are positive integers.

I tried to substitute with the fact that $A|B$ means $\exists c$ s.t $Ac_2=B$ and same $Bc_1=A$ when $B|A$, but I did not get anywhere.

Avv
  • 1,159
  • 2
    But A=-B is possible, right? You should specify more details about your setting – Benjamin Wang Mar 24 '21 at 00:38
  • @BenjaminWang. Thank you. Only $A=B$ is specified. – Avv Mar 24 '21 at 00:39
  • 1
    It depends on how you define divisibility.

    If $x \mid y$ means "$\exists k \in \Bbb Z^+$ such that $y = kx$", then yes this is true.

    If $x \mid y$ means "$\exists k \in \Bbb Z$ such that $y = kx$", then no. (In this case, for instance, $x \mid -x$ and $-x \mid x$, each with $k=-1$.) (Of course, then, this also assumes $x,y$ can be negative.)

     – PrincessEev Mar 24 '21 at 00:39
  • This is false: $2 \mid -2$ and $-2 \mid 2$ but $2 \ne -2$. If you don't add hypothesis about the sign of the numbers you won't be able to prove $A = B$ – jjagmath Mar 24 '21 at 00:43
  • @jjagmath. Positive numbers only. – Avv Mar 24 '21 at 00:51
  • Don't write it in a comment, edit your question – jjagmath Mar 24 '21 at 00:58

1 Answers1

3

If $A=c_1 B$ and $B= c_2A$, then we get $A = c_1B = c_1 c_2 A$, and if all these are non negative integers, it follows that $c_1 =c_2 =1$.

S.Farr
  • 1,190