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 November 26th, 2016, 02:39 PM #11 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 If $x$, $y \in \mathbb{N}$, dividing $9(x + y) + 7$ by 9 gives $x + y$ with remainder 7, so [$9(x + y) + 7$] and [7] are the same equivalence class. Thanks from beesee
November 26th, 2016, 04:17 PM   #12
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Quote:
 Originally Posted by skipjack If $x$, $y \in \mathbb{N}$, dividing $9(x + y) + 7$ by 9 gives $x + y$ with remainder 7, so [$9(x + y) + 7$] and [7] are the same equivalence class.
so for example $x = 1, y =1$
$9(1+1) + 7 = 9(2) + 7 = 18 + 7 = 25$
that expression $9(x+y)+7$ becomes $25$, i.e. a number that if divided by $9$ we obtain a quotient $2$, in fact $(x+y = 1+1 = 2)$, with remainder $7$. So $[25]$ and $[7]$ are the same equivalence class.

 November 26th, 2016, 04:54 PM #13 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Yes, that's correct.
 November 27th, 2016, 03:27 AM #14 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 Can that result be written in the form of a congruence? Last edited by skipjack; November 27th, 2016 at 05:14 AM.
 November 27th, 2016, 05:11 AM #15 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Yes, 25 ≡ 7 (mod 9). Thanks from beesee

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