If (A={1,2,3}), (B={1,2,3}), and (R={(x,y):x+y\text{ is even}}), why is (R) not a function?

Correct answer: Because (x=1) has images (1) and (3)

A relation is a function only when every input has exactly one output. Here, a pair (x,y) belongs to R when x+y is even. For x=1, the sum is even when y is odd. Both y=1 and y=3 are in B, and 1+1=2 and 1+3=4 are even. Thus the same input x=1 is related to two different outputs, 1 and 3.

This violates the defining rule of a function, which permits only one image for each input. Therefore option A is correct. Notice that x=2 does have images, namely 2, and the set B is not empty. The fact that A and B are equal also has no bearing on whether the relation is a function; the problem is specifically the multiple outputs for x=1.