if a|b and b|a, then a=b

am=b and bn=a but how do I show a=b?

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- Jun 17th 2010, 01:00 PMdwsmithif a|b and b|a, then a=b
if a|b and b|a, then a=b

am=b and bn=a but how do I show a=b? - Jun 17th 2010, 01:01 PMMoo
a/b=m and a/b=n, then...

- Jun 17th 2010, 01:05 PMdwsmith
- Jun 17th 2010, 01:09 PMAckbeet
am = b and bn = a implies a = (am)n = a(mn). What does mn have to be?

- Jun 17th 2010, 01:12 PMdwsmith
- Jun 17th 2010, 01:14 PMAckbeet
I think you can know more than that: a = a(mn) implies a (mn-1) = 0. Since a is not zero, mn - 1 = 0, which implies...

- Jun 17th 2010, 01:22 PMundefined
I'm assuming a and b are positive integers? Otherwise consider

a = 2

b = -2

If a and b are positive integers, then a|b implies a \le b and b|a implies b \le a, therefore a = b. - Jun 17th 2010, 02:20 PMmelese

Modifying your question: Let $\displaystyle a$ and $\displaystyle b$ be integers. If $\displaystyle a|b $ and $\displaystyle b|a $, then $\displaystyle a=\pm b $ ($\displaystyle b $ or $\displaystyle -b $).

$\displaystyle a=bn$ and $\displaystyle b=am$, where $\displaystyle n $ and $\displaystyle m $ are integers , then substitute for $\displaystyle b $ to get $\displaystyle a=(am)n $ or $\displaystyle a=a(mn) $. Divide both sides by $\displaystyle a(\neq0)$, and so both $\displaystyle m$ and $\displaystyle n$ equal $\displaystyle 1$ or both equal $\displaystyle -1$. - Jun 17th 2010, 02:37 PMundefined
Hmm, I'm not sure if this will display properly, I keep getting LaTeX errors on my previews.

Another way to get melese's result above is

$\displaystyle a|b \Rightarrow |a| \le |b|$

$\displaystyle b|a \Rightarrow |b| \le |a|$

$\displaystyle |a| = |b|$

$\displaystyle a = \pm b$

Since this isn't rendering properly, here is an image I got using this site:

Attachment 17884 - Jun 19th 2010, 07:39 AMBruno J.