Adding Equivalence Classes?

[cired2002]

Is (m,n) + (p,q) = (m+p , n+q) or = (mq + np , nq)

I am very confused on how the addition works

The question is
:- Let Q = {(m,n) | m,n is a subset of integers such that n is not equal to 0}. Let us define addition (+) and multiplication (*) on this set as follow. For (m,n), (p,q) belonging to Q

1) (m,n) + (p,q) = (mq+np,nq)
2) (m,n)*(p,q) = (mp + nq)

Show that Q with these operations is not a field

 

[Harry_the_cat]

The addition is defined as in 1), So that's what it is.

 

[Dr.Peterson]

Is (m,n) + (p,q) = (m+p , n+q) or = (mq + np , nq)

I am very confused on how the addition works

The question is
:- Let Q = {(m,n) | m,n is a subset of integers such that n is not equal to 0}. Let us define addition (+) and multiplication (*) on this set as follow. For (m,n), (p,q) belonging to Q

1) (m,n) + (p,q) = (mq+np,nq)
2) (m,n)*(p,q) = (mp + nq)

Show that Q with these operations is not a field

I think you've copied some things wrong; an image of the actual problem may be needed.

First, "m,n is a subset of integers" makes no sense. It probably says [imath]m,n\in \mathbb{Z}[/imath], which means "m and n are elements of the integers".

Second, if multiplication is defined by (m,n)*(p,q) = (mp + nq), then it is not a binary operation at all, since the result is not an ordered pair. If you didn't miscopy, then this is the answer to the question!

It's also worth noting that the problem doesn't involve equivalence classes. I imagine that will come later, as equivalence classes will come to the rescue.