Justin T. answered • 04/22/15

Experienced Math, SAT Math, SAT Math Subject Tests tutor

Let a = (x,y), b = (w,z) such that a,b ∈

**N**x**N**and f(a),f(b) ∈**N**x**N**.f(a) = f(b) implies (x+y, 2x-3y) = (w+z, 2w-3z) implies x+y = w+z, 2x-3y = 2w-3z by the definition of equality of elements in

**N**x**N**.x+y = w+z

**⇒**x-w = -(y-z)2x-3y = 2w-3z 2x-2w = 3y-3z

**⇒**x-w = -(y-z)

**⇒**-2(y-z) = 3(y-z)

2(x-w) = 3(y-z)

The last equation holds if and only if y-z = 0 implies also x-w = 0 implies y = z and x = w, which implies, if a = (x,y), b = (w,z), a = b.

Note that the domain of ƒ, D =

**{**(x,y)∈

**N**x

**N**| f((x,y)) = (x+y, 2x-3y)∈

**N**x

**N }**=

**{**(x,y)| 2x > 3y, x,y ∈

**N }**⊂

**N**x

**N**(note ⊂ as opposed to ⊆), but amazingly though as was just shown, these restrictions played no part in determining whether ƒ was an injective function or not.