Let $f$ be the subset of $Z \times Z$ defined by $f = \{(ab, a+b) : a, b \in Z\}$. Is $f$ a function from $Z$ to $Z$? Justify your answer.
(N/A) The relation $f$ is defined as $f = \{(ab, a+b) : a, b \in Z\}$.
$A$ relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ has a unique image in $B$.
Consider the elements $a=2, b=6$ and $a=-2, b=-6$ in $Z$.
For $a=2, b=6$,we have $(ab, a+b) = (2 \times 6, 2+6) = (12, 8) \in f$.
For $a=-2, b=-6$,we have $(ab, a+b) = (-2 \times -6, -2-6) = (12, -8) \in f$.
Since the same first element $12$ corresponds to two different images $8$ and $-8$,the relation $f$ is not a function.
$A$ relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ has a unique image in $B$.
Consider the elements $a=2, b=6$ and $a=-2, b=-6$ in $Z$.
For $a=2, b=6$,we have $(ab, a+b) = (2 \times 6, 2+6) = (12, 8) \in f$.
For $a=-2, b=-6$,we have $(ab, a+b) = (-2 \times -6, -2-6) = (12, -8) \in f$.
Since the same first element $12$ corresponds to two different images $8$ and $-8$,the relation $f$ is not a function.
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