# Show that multiplication $[(x, y)] * [(n,m)] = [(xn + ym, xm + yn)]$ is also well-defined

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?

• "which isn't coming out correctly" - can you explain what you are expecting to happen? – Ted Nov 23 '15 at 6:11
• trying to get [(x'n'+y'm',x'm'+y'n')] = [(xn+ym,xm+yn)] – HiPolyEraser Nov 23 '15 at 6:15