If $A|B$ and $B|A$ then prove $A=\pm B$

Question

If $A|B$ and $B|A$ then prove $A=\pm B$

So far I have $A|B \iff AX=B$ and $B|A \iff BY=A$ with $X,Y \in \mathbb{Z}$

Not sure how to finish, any help.

http://math.stackexchange.com/questions/1200244/prove-divisibility-of-c-mid-d-and-d-mid-c#comment2442957_1200244 — user26486, May 21 '15 at 18:33

score 2 · Answer 1 · answered May 21 '15 at 18:30

2

You're almost there. Note that if you plug $A = BY$ into the first expression, you have

$$(BY)X = B.$$

Since we are discussing divisibility, $B\neq 0$ so we can divide both sides by $B$ to get

$$XY = 1.$$

What integers multiply together to give $1$?

answered May 21 '15 at 18:30

Cameron Williams

29,432

$\pm 1$ then. What is the last move? Think I got it thanks, $X=1, Y=1$ or $X=Y=-1$ so sub into find $A=\pm B$ – Ryan May 21 '15 at 18:32
Yep exactly right! – Cameron Williams May 21 '15 at 18:33
Brilliant thanks mate. – Ryan May 21 '15 at 18:34

score 1 · Answer 2 · answered May 21 '15 at 18:30

1

You get that $AXY=A$, so unles $A=0$ we may conlcude that $XY=1$, whihc has only two solutions. If onthe other hand $A=0$ then the given considitions im ply $B=0$ anyway.

answered May 21 '15 at 18:30

Hagen von Eitzen

374,180

score 1 · Answer 3 · answered May 21 '15 at 18:31

If $A|B$ then $B=AX$ as you said, but if $B|A$ we have as you write $A=BY$, substitute in our initial we get that $B=AX=BYX$, this can only be possible if $XY=1$, so we have two numbers times one another equals 1, which is $X=Y=1$ or $X=Y=-1$, hence it can be either of those.

If $A|B$ and $B|A$ then prove $A=\pm B$

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