Let \(A = \{1, 2, 3, 4, 5, 6\}\) and let \(f : A \to A\) be defined by \(f(x)\) = the remainder when \(x^{2}\) is divided by 7. Then f is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → neither one-one nor onto
Working out each image:
\(1^{2} = 1\), remainder 1
\(2^{2} = 4\), remainder 4
\(3^{2} = 9 = 7 + 2\), remainder 2
\(4^{2} = 16 = 14 + 2\), remainder 2
\(5^{2} = 25 = 21 + 4\), remainder 4
\(6^{2} = 36 = 35 + 1\), remainder 1
Not one-one: \(f(3) = f(4) = 2\).
Not onto: the range is \(\{1, 2, 4\}\), so 3, 5 and 6 are not images.
So f is neither one-one nor onto.