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 August 5th, 2017, 05:23 AM #1 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 Help with modules Please help me prove the following theorem: This theorem apparently holds when $\displaystyle p$ is prime, and $\displaystyle a$ is NOT a multiple of $\displaystyle p$. If, $\displaystyle ab = ac$ $\displaystyle mod(p)$ Then $\displaystyle b = c$ $\displaystyle mod(p)$
 August 5th, 2017, 06:04 AM #2 Senior Member   Joined: Oct 2009 Posts: 627 Thanks: 190 When is $a$ invertible? I.e. can you find an $x$ such that $ax=1$ mod(p)?
 August 5th, 2017, 09:52 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,950 Thanks: 1842 The standard notation, for example, $\displaystyle ab \equiv ac \pmod p$, is preferred.
August 5th, 2017, 05:55 PM   #4
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Joined: Oct 2009

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Thanks: 190

Quote:
 Originally Posted by skipjack The standard notation, for example, $\displaystyle ab \equiv ac \pmod p$, is preferred.
Not really. Many books have different notations.

 August 5th, 2017, 05:58 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,950 Thanks: 1842 What I meant was "preferred on this website".

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