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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 for more on the different meanings of "mod". – Bill Dubuque Jan 16 '20 at 04:30

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