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January 5th, 2017, 03:58 AM  #1  
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  Each integer is congruent modulo n to precisely one of the integers 0,1,...,n1.
Hi, I have the following exercise: Choose n = 5, then $\displaystyle [3] \oplus [4] = [3+4] = [7] = [2]$ END. I don't understand what is the passage that makes $[7] = [2]$ Why the set that contains the numbers that have remainders equal to 7 when divided by 5, is the same (contains the same elements), of the set that contains the numbers that have remainders equal to 2 when divided by 5? Let's see what we have: for hypotesis we have $\displaystyle [3] \oplus [4]$, i.e.: [3] is the set of the elements that have the same remainder equal to 3 when divided by 5; [4] is the set of the elements that have the same remainder equal to 4 when divided by 5; if we consider $a \in [3], b \in [4]$ $a \equiv 3 \mbox{ (mod 5) } \\ b \equiv 4 \mbox{ (mod 5) }$ Considering this theorem: Quote:
$a+b \equiv 3+4 \mbox{ (mod 5)} \\ a+b \equiv 7 \mbox{ (mod 5)} \\ 5  a+b7 \\ a+b7 = 5q \\ a+b = 5q+7$ but, considering the division algorithm, the remainder must be within this interval: $0 \le r < m$, but $7 > m$ in what way can I "transform" it to obtain "congruent to 2"? I have read in another book this phrase: Quote:
Can you help me please? Thanks! Last edited by beesee; January 5th, 2017 at 04:09 AM.  
January 5th, 2017, 05:09 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,225 Thanks: 186 
Here ... $a + b = 5q + 7$ All you have to do is recognise that $7 = 5 + 2$ $a + b = 5q + 5 + 2$ $a + b = (5q + 5) + 2$ $a + b = 5(q +1) + 2$ $a + b = 5Q + 2$ $5q$ represents multiples of $5$ , so does $5Q$. As far as multiples of $5$ are concerned , $5$ is absorbed from $7$ leaving $2$ Also , If you expand this notation $[7]$ congruent to modulus $5$ you get $[7]$ = { $... 18 , 13 , 8 , 3 , 2 , 7 , 12 , 17 ...$ } All these numbers are in the same equivalence class , they all have remaider $2$ when divided by $5$ It is customary to pick the smallest positive integer within the equivalence class to represent the whole equivalence class , in this case $2$ That is why $[7] = [2]$ 

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congruent, integer, integers, modulo, precisely 
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