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

**dwsmith** · June 17th 2010, 01:00 PM

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

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

**Moo** · June 17th 2010, 01:01 PM

a/b=m and a/b=n, then...

**dwsmith** · June 17th 2010, 01:05 PM

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?

**Ackbeet** · June 17th 2010, 01:09 PM

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

**dwsmith** · June 17th 2010, 01:12 PM

Originally Posted by Ackbeet

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

An integer.

**Ackbeet** · June 17th 2010, 01:14 PM

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...

**undefined** · June 17th 2010, 01:22 PM

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.

**melese** · June 17th 2010, 02:20 PM

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 $a$ and $b$ be integers. If $a|b$ and $b|a$ , then $a=\pm b$ ( $b$ or $-b$ ).

$a=bn$ and $b=am$ , where $n$ and $m$ are integers , then substitute for $b$ to get $a=(am)n$ or $a=a(mn)$ . Divide both sides by $a(\neq0)$ , and so both $m$ and $n$ equal $1$ or both equal $-1$ .

**undefined** · June 17th 2010, 02:37 PM

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

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

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

$|a| = |b|$

$a = \pm b$

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

if a|b and b|a, then a=b-divides_abs_val.png

**Bruno J.** · June 19th 2010, 07:39 AM

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

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

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

$|a| = |b|$

$a = \pm b$

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

Click image for larger version.

Name: divides_abs_val.png
Views: 43
Size: 1.2 KB
ID: 17884

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

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