Show that for every integeranot divisible by 11 there is another integerbwith

a*b == 1 (mod 11).

This problem has me stumped...

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- Sep 22nd 2010, 12:46 PMpaulrbExistence proof
Show that for every integer

*a*not divisible by 11 there is another integer*b*with

a*b == 1 (mod 11).

This problem has me stumped... - Sep 22nd 2010, 12:55 PMundefined
- Sep 22nd 2010, 01:29 PMpaulrb
Thanks...I should have specified though, I've only been in this number theory class for 3 weeks and we have only covered basic theorems about congruence / number theory. Nothing with fermat's little theorem / euler's theorem so I don't think it would be acceptable for me to use that in a proof.

- Sep 22nd 2010, 02:05 PMundefined
I think we can do as follows. It will be a stronger result than what you need to prove.

Let p be any prime, and let a be an integer not divisible by p.

Consider the set $\displaystyle S=\{0,a,2a,\,\ldots\,,(p-1)a\}$. Claim: each element of S is distinct modulo p.

Proof: Suppose the opposite. Then there exist integers x, y such that $\displaystyle x\not\equiv y\pmod{p}$ and $\displaystyle ax\equiv ay\pmod{p}$. The latter implies $\displaystyle a(x-y)\equiv 0\pmod{p}$. So we have a product of two non-multiples of p being divisible by p; contradiction.

Now consider that there are p distinct equivalence classes in S modulo p, thus it is the set of all equivalence classes modulo p, and $\displaystyle \overline{1}\in S$. - Sep 22nd 2010, 02:44 PMpaulrb
That's a great proof, thank you!

- Sep 23rd 2010, 09:12 AMroninpro
I'd just like to point out that a more standard approach would be to use Bezout's identity.

If 11 does not divide $\displaystyle a$, then $\displaystyle \gcd(a,11)=1$. By Bezout's identity, there exist integers $\displaystyle x,y$ such that $\displaystyle ax+11y=1$. Taking everything modulo 11, $\displaystyle ax+11y\equiv 1\pmod{11}$, and $\displaystyle ax\equiv 1\pmod{11}$. The term $\displaystyle x$ is what we are looking for. - Sep 23rd 2010, 09:18 AMundefined