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

• Jun 17th 2010, 01:00 PM
dwsmith
if 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 PM
Moo
a/b=m and a/b=n, then...
• Jun 17th 2010, 01:05 PM
dwsmith
Quote:

Originally Posted by Moo
a/b=m and a/b=n, then...

Shouldn't it be \frac{b}{a}=m \ \mbox{and} \ \frac{a}{b}=n but I don't see how that helps.

Does Latex not work?
• Jun 17th 2010, 01:09 PM
Ackbeet
am = b and bn = a implies a = (am)n = a(mn). What does mn have to be?
• Jun 17th 2010, 01:12 PM
dwsmith
Quote:

Originally Posted by Ackbeet
am = b and bn = a implies a = (am)n = a(mn). What does mn have to be?

An integer.
• Jun 17th 2010, 01:14 PM
Ackbeet
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 PM
undefined
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 PM
melese
Quote:

Originally Posted by dwsmith
if a|b and b|a, then a=b

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

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 PM
undefined
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 AM
Bruno J.
Quote:

Originally Posted by undefined
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

This works, but it makes use of the total order on $\displaystyle \mathbb{Z}$, which you don't really need. Ackbeet's proof works if $\displaystyle \mathbb{Z}$ is replaced by the polynomial ring $\displaystyle \mathbb{Z}[x]$, for instance.