Suppose that $(a_1, b_1) + (c_1, d_1) = (a_1d_1 + b_1c_1, b_1d_1)$. Prove this addition is well-defined.

Question

I know that if $(a_1, b_1) + (c_1, d_1) = (a_1 + c_1, b_1 + d_1)$, there can be $(a_1, b_1) \equiv (a_2, b_2)$ and $(c_1, d_1) \equiv (c_2, d_2)$ such that $(a_1, b_1) + (c_1, d_1) \not \equiv (a_2, b_2) + (c_2, d_2)$.

So this is not well-defined. But why would the new addition be well-defined ?

Thank you!

Answer 1

We have an equivalence relation defined on pairs of numbers (assume for simplicity: naturals):

$(x,y) \equiv (z,w) \text { iff } xw=yz$,

where the product $xw$ is simply the product of two numbers (as well as $x+y$ is the sum of numbers).

We define the sum of pairs as follows (in terms of sum and product of numbers):

$(x,y)+(a,b)=(xb+ya,yb)$.

This product is again a pair, but we have to check that the operation is well-defined, i.e. that:

if $(x,y) \equiv (z,w) \text { and } (a,b) \equiv (c,d)$,

then the result does not change, i.e. that:

$(x,y)+(a,b)=(z,w)+(c,d)$.

In order to do this, we have to use the definition of sum of pairs, and verify that:

$(xb+ya,yb)=(zd+wc,wd)$.

But this in turn amounts to verify that:

$[xb+ya]wd=[zd+wc]yb$

where now we have no more pairs but numbers and operations with numbers.