Relation ≤ on Z = ℕ x ℕ given by (a,b) ≤ (c,d) iff a + d ≤ b + c is linear ordering Prove the relation ≤ on Z = ℕ x ℕ given by (a,b) ≤ (c,d) iff a + d ≤ b + c is a linear ordering.

Note: The set we are referring to here is not the familiar set ℤ but the set Z that is in bijection with this set (when we construct the integers from natural numbers).
I understand that in the construction of integers, we define a relation ~ on Z which is defined by (a,b) ~ (c,d) iff a + d = b + c. It seems intuitive that for x,y ∈ ℤ where x=(a,b) and y =(c,d), then x ≤ y <==> a + d ≤ b + c. Hence, I'm not quite sure what sort of proof I need to give in order to establish this. Do I need to run the antisymmetry, transitivity, totality check?
Hope someone can give me a nudge in the right direction. 
 A: I'll give a geometric interpretation; hopefully this will guide you in the right direction. Since the equivalence relation is defined on $\mathbb{N} \times \mathbb{N}$ we can imagine it as a plane. Let's look at some point $(a,b)$. We'll consider this before "quotienting out" by the equivalence relation you mentioned (the one that defines $\mathbb{Z}$ from $\mathbb{N}$), i.e. identifying elements if they are equivalent. The line $y = -(x-a) + b$ cuts the plane in half precisely diagonally (with negative slope). We'll work with these two halves from here on out. If you think about it (it's not too hard to see), the points less than or equal to $(a,b)$ are the ones to the left (really, top-left) of that line, and the ones greater than or equal to $(a,b)$ are the ones to the right (really, bottom-right). This is an if and only if, so this gives you totality for free; any point in the plane must lie to the left of the line or to the right of the line (or both, but I'll get to that in a second). The only way $(a,b) \leq (x,y)$ and $(a,b) \geq (x,y)$ is if $(x,y)$ lies in both half-planes, i.e. precisely on that line, and vice-versa. But the equivalence relation you gave precisely says that all of the elements on that line are treated as the same! So this interpretation gives us antisymmetry as well. For transitivity: in our interpretation, we want to show that if $(a,b)$ lies to the left of $(c,d)$ and $(c,d)$ lies to the left of $(e,f)$, then $(a,b)$ lies to the left of $(e,f)$. If you draw out the diagonal lines on a piece of paper this is entirely obvious.
Of course you need this in algebraic form. Let's look at $(a,b), (c,d)$. If $(a,b) \leq (c,d)$ and $(a,b) \geq (c,d)$ then we have $a + d = b + c$. Use your equivalence relation to conclude. Transitivity is almost exactly the same. For totality, note that the ordering on the natural numbers is total and consider the quantities $a + d, b + c$ and get some kind of relation between them.
A: I believe one possibility is to check the conditions by hand. This can be done by a reasonable amount of (rather boring) work. Because the claim also follows from a reasoning like "there exists a thing called the integers, there is a bijection between them and $Z$, and the order defined here works just as the standard order on the integers", I suppose you are not supposed to do anything more fancy.
Assymetry: Suppose $(a,b) \leq (c,d) \leq (a,b)$. Then, $a+d = b+c$. But this means that $(a,b) \sim (c,d)$, so they represent the same element of $Z$. 
Totality: Take $(a,b)$ and $(c,d)$. Then one of the values $a+b$, $c+d$ is greater than or equal to the other. Hence, $(a,b)$ and $(c,d)$ can be compared.
Transivity: Suppose $(a,b) \leq (c,d) \leq (e,f)$. Then, $a + d \leq c + b$ and $c+f \leq d+e$. It follows that $a+d+f \leq c+b+f \leq d+e+b$, so $a+f \leq e+b$.
