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Let $r_1=(a+b)(mod n )$ , $r_2=c(mod n) $ and $r_3=a(mod n) $.

Where $0\le r_1,r_2,r_3<n$

L. H. S. =$(a+b)(mod n) + c(mod n) $.

R. H. S. =$(a+(b+c)(mod n) )(mod n) $.

L. H. S. =$r_1 + r_2$.

$r_1=(a+b)(mod n) . r_2=c(mod n). r_1+r_2=(a+b+c)(mod n) $.

Now $n|((a+b+c)-a) (mod n). Now (a+b+c)-a =nk1+(r_1+r_2)-nk2-r_3$.

$R. H. S. =(a +(a+b+c-a)(mod n) )(mod n) =( nk3+r_3+(r_2+r_1-r_3))(mod n) =(r_1+r_2).$

Is my attempt correct?

Guria Sona
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    it follows from associativity of addition for real numbers – J. W. Tanner Jan 15 '20 at 13:34
  • So is my proof wrong? – Guria Sona Jan 15 '20 at 13:35
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    What is your definition of "addition modulo $n$? In particular, is it an operation on the *ring* $\,\Bbb Z_n?\ \ $ – Bill Dubuque Jan 15 '20 at 16:26
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    Note that $a=b\bmod n$ means something different from $a=b\pmod{n}$. One is the binary modulo operator, the other is the relation symbol. Please say **explicitly** which one you mean and do not confuse them. They are not interchangeable. – Arturo Magidin Jan 15 '20 at 20:31
  • @Bill Dubuque. Yes .i meant addition modulo n – Guria Sona Jan 16 '20 at 04:24
  • Please (1) give your definition of addition modulo $n$. (ii) Explain what you mean by “$\bmod n$” vs. $\pmod{n}$. (iii) Use proper MathJax; for the binary operator, use `\bmod`: so `a\bmod n` produces $a\bmod n$. For the parenthetical equivalence relation, use `\pmod{n}`. So `a\equiv b\pmod{n}` produces $a\equiv b\pmod{n}$. And again: **they are not the same thing**. – Arturo Magidin Jan 16 '20 at 04:24
  • @GuriaSona: The question is what is your **definition** of “addition modulo $n$”. Your **precise** definition. Is it defined on equivalence classes of integers? Is it defined on integers themselves? Is it defined on the set $\{0,1,\ldots n-1\}$? Is it defined on the set $\{1,\ldots,n\}$? Is it defined on some other set? And **how** is it defined? – Arturo Magidin Jan 16 '20 at 04:25
  • @GuriaSona Your reply did not answer either of my questions. As Arturo explained, you need to be more precise so we can determine what you mean. See [this answer](https://math.stackexchange.com/a/614944/242) for more on the different meanings of "mod". – Bill Dubuque Jan 16 '20 at 04:30

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