There is a relation $\sim$ on $\mathbb{Z}\times\mathbb{Z}$ such that $(a,b)\sim(c,d)$ when $a+b=c+d$. Let $R={[(a,b)]:(a,b)\in\mathbb{Z}\times\mathbb{Z}}$ (i.e $R$ is the set of all equivalence classes of $\mathbb{Z}\times\mathbb{Z}$ under the equivalence relation $\sim$). Is $[(x,y)]*[(w,z)]=[(x+w,y+z)]$ well defined? ----After several examples, I think it is well defined. $[(2,3)]*[(4,5)]=[(6,8)]$, $[(1,4)]*[(2,7)]=[(3,11)]$ and $(6,8)\sim(3,11)$... But I am not sure how to prove it in a general way. Please help!
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2You don't need to write $(a,b)$~$(c,d)$. You can write $(a,b)\sim(c,d)$, entirely within MathJax. It has proper spacing that way, and the fonts match. – Michael Hardy Oct 18 '13 at 03:46
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Let $ (a,b)\sim(c,d)$ and $(e,f)\sim(g,h)$. Thus $a+b=c+d, e+f=g+h$. This implies $a+b+e+f=c+d+g+h$. Thus $(a+e)+(b+f)=(c+g)+(d+h)$. Hence $(a+e,b+f)\sim (c+d,g+h)$.
Anupam
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You have to prove that for any $a,b,c,d,e,f,g,h$ such that $(a,b)\sim(c,d)$ and $(e,f)\sim(g,h)$ then $(a,b)*(e,f)\sim(c,d)*(g,h)$.
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So you mean for example, If $(2,3)$~$(4,5)$, $(1,4)$~$(2,7)$, then, $(3,7)$~$(6,12)$?? – Wes Oct 18 '13 at 01:42
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That is what I mean if you do it algebraically. Prove that if $(a,b),\text{~},(c,d)$ and $(e,f),\text{~},(g,h)$, then, $(a+e,b+f),\text{~},(c+g,d+h)$. – Carlos Eugenio Thompson Pinzón Oct 18 '13 at 01:47
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1In your example neither $(2,3),\text{~},(4,5)$. So if the premise is false don't expect the conclusion to be true. – Carlos Eugenio Thompson Pinzón Oct 18 '13 at 02:04
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Wait,,,, Why $(2,3)$~$(4,5)$???They are not necessarily relation, are they? I made those examples with my prof. Oh, I am so confused. – Wes Oct 18 '13 at 02:06
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2The code said ,\text{~},. It looks as if attention to proper spacing was done deliberately. I changed it to \sim. Thus: $(a,b)\sim(c,d)$ is coded as (a,b)\sim(c,d). Proper spacing and matching of fonts is left to the software. TeX was invented for tihngs like this. – Michael Hardy Oct 18 '13 at 03:49