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I'm having a bit of trouble on this proof. It's part of the construction of the integers. $R$ is the relation, $\mathbb{N}$ the natural numbers, $((x,y),(n,m)) \in (\mathbb{N}\times\mathbb{N})\times(\mathbb{N}\times\mathbb{N}) | x + m = y + n$.

$(x',y')\in[(x,y)] = x'+ y = y' + x$

$(n'm')\in[(n,m)] = n' + m = m' + n$

After that It gets messy when I try to multiply the two together, which isn't coming out correctly.

$x'n'+x'm+yn'+ym = y'm'+y'n+xm'+xn$

And also is addition and multiplication closed under $(\mathbb{N}\times\mathbb{N})/R$?

Another quick question to show the additive identity

$[(x,y)]+[(0,0)] = [(x+0,y+0)] = [(x,y)] = [(0+x,0+y)]$

Would this be correct?

Alex Ravsky
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HiPolyEraser
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1 Answers1

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To show that the multiplication of the equivalence classes is well-defined, we have to show that for any natural numbers $x,y,x’,y’,n,m,n’,m’$ if $(x',y')R(x,y)$ and $(n',m')R(n,m)$ then $$(xn + ym, xm + yn)R(x’n’ + y’m’, x’m’ + y’n’).$$ That is, if $x'+ y = y' + x$ and $ n' + m = m' + n $ then $xn + ym+ x’m’ + y’n’= xm + yn+ x’n’ + y’m’$. This is true, because $$xn + ym+ x’m’ + y’n’-(xm + yn+ x’n’ + y’m’)=(x-y)(n-m)-(x’-y’)(n’-m’)=0.$$

And also is addition and multiplication closed under $(\mathbb{N}\times\mathbb{N})/R$?

I assume that the sum of classes $[(x,y)]$ and $[(n,m)]$ is a class $[(x+y,n+m)]$, which does not depend on a choice of representatives from $[(x,y)]$ and $[(n,m)]$. Then the addition ad multiplication are well-defined on $(\mathbb{N}\times\mathbb{N})/R$ and the latter set is closed with respect to these operations.

Another quick question to show the additive identity $[(x,y)]+[(0,0)] = [(x+0,y+0)] = [(x,y)] = [(0+x,0+y)]$ Would this be correct?

Yes.

Alex Ravsky
  • 106,166