Understanding Functions and Relations

A relation $f$ from a set $A$ to a set $B$ is a subset of the Cartesian product \(A \times B\). For a relation to be a function from $A$ to $B$, two conditions must be met:

• Every element in the domain set $A$ must be related to an element in the codomain set $B$. (In the context of the given statements, the domain is the set of possible first elements of the pairs in the relation, and this set must be equal to \(\mathbb{Z}\) in statement 1 and \(\mathbb{N}\) in statement 2. The codomain is the set to which the second element belongs, which is \(\mathbb{Z}\) or \(\mathbb{N}\).)
• Each element in the domain set $A$ must be related to exactly one element in the codomain set $B$. This is the crucial condition we will check. If a single input value maps to more than one output value, the relation is not a function.

Let's examine each statement.

Analysis of Statement 1: Relation on Z × Z

Statement 1 defines the relation \(f = \{(xy, x - y); x, y \in \mathbb{Z}\}\) as a subset of \(\mathbb{Z} \times \mathbb{Z}\). The potential input values are of the form $xy$, and the potential output values are of the form $x-y$, where $x, y$ are integers.

For $f$ to be a function from \(\mathbb{Z}\) to \(\mathbb{Z}\), every integer $z$ in the domain (the set of all possible $xy$ values) must map to exactly one integer $w$ (an $x-y$ value).

Let's pick an input value, say $0$. Can we get $xy=0$ in more than one way such that the corresponding $x-y$ values are different?

• Consider $x=0$ and $y=1$. Then \(xy = 0 \times 1 = 0\), and $x-y = 0 - 1 = -1$. This gives the pair $(0, -1)$ in the relation $f$.
• Consider $x=0$ and $y=2$. Then \(xy = 0 \times 2 = 0\), and $x-y = 0 - 2 = -2$. This gives the pair $(0, -2)$ in the relation $f$.

We have found that the input value $0$ is mapped to two different output values, $-1$ and $-2$. According to the definition of a function, each input must have a unique output.

Therefore, the relation defined in Statement 1 is not a function from \(\mathbb{Z}\) to \(\mathbb{Z}\).

Analysis of Statement 2: Relation on N × N

Statement 2 defines the relation \(f = \{(xy, x + y); x, y \in \mathbb{N}\}\) as a subset of \(\mathbb{N} \times \mathbb{N}\). The potential input values are of the form $xy$, and the potential output values are of the form $x+y$, where $x, y$ are natural numbers (\(\mathbb{N} = \{1, 2, 3, \dots\}\)).

For $f$ to be a function from \(\mathbb{N}\) to \(\mathbb{N}\), every natural number $z$ in the domain (the set of all possible $xy$ values) must map to exactly one natural number $w$ (an $x+y$ value).

Let's pick an input value, say $6$. Can we get $xy=6$ using different pairs $(x, y)$ from \(\mathbb{N}\) such that the corresponding $x+y$ values are different?

• Consider $x=1$ and $y=6$. Then \(xy = 1 \times 6 = 6\), and $x+y = 1 + 6 = 7$. This gives the pair $(6, 7)$ in the relation $f$.
• Consider $x=2$ and $y=3$. Then \(xy = 2 \times 3 = 6\), and $x+y = 2 + 3 = 5$. This gives the pair $(6, 5)$ in the relation $f$.

We have found that the input value $6$ is mapped to two different output values, $7$ and $5$. According to the definition of a function, each input must have a unique output.

Therefore, the relation defined in Statement 2 is not a function from \(\mathbb{N}\) to \(\mathbb{N}\).

Conclusion

Based on the analysis of both statements:

• Statement 1 does not define a function because we found an input (0) that maps to multiple outputs (-1 and -2).
• Statement 2 does not define a function because we found an input (6) that maps to multiple outputs (7 and 5).

Neither of the statements given is correct in defining a function.

Revision Table: Function Definition Check

Additional Information: Relations and Functions

A relation is a very general concept simply showing a relationship between elements of sets. A function is a special type of relation with stricter rules.

• Relation: Any subset of a Cartesian product \(A \times B\). It just pairs elements from $A$ with elements from $B$.
• Function: A relation $f$ from $A$ to $B$ is a function if every element in $A$ is the first element of exactly one ordered pair in $f$. This means:
  • Every element in $A$ is used as an input.
  • No element in $A$ is used as an input more than once with different outputs.

The domain of a relation is the set of all first elements of the ordered pairs. The range of a relation is the set of all second elements of the ordered pairs. For a relation to be a function from $A$ to $B$, its domain must be exactly $A$, and its range must be a subset of $B$. The key functional property is the unique output for each input.