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November 26th, 2016, 03:39 PM   #11
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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.
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November 26th, 2016, 05:17 PM   #12
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Quote:
Originally Posted by skipjack View Post
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.
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November 26th, 2016, 05:54 PM   #13
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Yes, that's correct.
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November 27th, 2016, 04:27 AM   #14
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Can that result be written in the form of a congruence?

Last edited by skipjack; November 27th, 2016 at 06:14 AM.
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November 27th, 2016, 06:11 AM   #15
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Yes, 25 ≡ 7 (mod 9).
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