Q.Let $f$ be the subset of $\mathbb{Z} \times \mathbb{Z}$ defined by $f = \{(ab, a + b) : a, b \in \mathbb{Z}\}$. Is $f$ a function from $\mathbb{Z}$ to $\mathbb{Z}$? Justify your answer.

Concept: A relation $f \subseteq A \times B$ is a function from $A$ to $B$ if every element in $A$ appears as the first coordinate of exactly one ordered pair in $f$.

Here $f = \{(ab, a+b) : a, b \in \mathbb{Z}\}$. The first coordinate is $ab$ (the product) and the second is $a+b$ (the sum). For $f$ to be a function from $\mathbb{Z}$ to $\mathbb{Z}$, each integer must map to exactly one integer.

Check uniqueness: Consider the integer $0$ as a first coordinate. We can write $0 = ab$ in multiple ways:

• Take $a=0, b=1$: then $ab=0$ and $a+b=1$, giving the pair $(0,1)$.
• Take $a=0, b=2$: then $ab=0$ and $a+b=2$, giving the pair $(0,2)$. …

Why this matters: the definition of a function

A subset $f \subseteq \mathbb{Z} \times \mathbb{Z}$ is a function from $\mathbb{Z}$ to $\mathbb{Z}$ if and only if every integer $x$ in the domain appears as the first coordinate of at most one ordered pair in $f$, and that pair is unique. In other words, if $(x, y_1) \in f$ and $(x, y_2) \in f$, we must have $y_1 = y_2$.

The relation $f$ is built by taking all possible products $ab$ as the "input" and pairing each with the sum $a+b$ as the "output." The question is whether a given product uniquely determines the sum.

Step-by-step analysis

1. Rewrite the relation in function notation.

   If $f$ were a function, we would write $f(ab) = a+b$. The domain would be the set of all products $ab$ (which is all of $\mathbb{Z}$, since every integer is a product), and the rule would assign to each product the corresponding sum.

2. Look for a counterexample.

   Suppose two different pairs of integers yield the same product but different sums. Pick $a=2, b=3$: then $ab = 6$ and $a+b=5$, so $(6,5) \in f$. Now pick $a=1, b=6$: then $ab=6$ and $a+b=7$, so $(6,7) \in f$. Both ordered pairs have first coordinate $6$, but second coordinates $5 \neq 7$.

3. Conclude that $f$ violates the definition.

   Because the element $6$ in the domain is paired with two distinct outputs, $f$ is not a function.

A more general perspective …