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.

November 26th, 2016, 04:17 PM  #12  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Quote:
$9(1+1) + 7 = 9(2) + 7 = 18 + 7 = 25$ 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).


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