Let \(g(x) = 3x + 1\). Here R, Z, Q and N denote the sets of real numbers, integers, rational numbers and natural numbers. Consider the following statements:
(A) \(g : R \to R\) is one-one and onto.
(B) \(g : Z \to Z\) is onto.
(C) \(g : Q \to Q\) is one-one and onto.
(D) \(g : N \to N\) is one-one.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
On every one of these sets g is one-one:
\(3p + 1 = 3q + 1\)
\(p = q\)
For onto, solve \(g(x) = y\):
\( \displaystyle x = \frac{y - 1}{3} \)
If y is real, this x is real. If y is rational, this x is rational. If y is an integer, x need not be an integer: for \(y = 2\),
\( \displaystyle x = \frac{1}{3} \)
Also, for every natural number n, \(3n + 1\) is a natural number, so \(g : N \to N\) is a function, and it is one-one.
So g is one-one and onto on R and on Q, one-one but not onto on Z, and one-one on N.
Checking the statements
(A) \(g : R \to R\) is one-one and onto. — Correct. \(x = (y - 1)/3\) is real for every real y.
(B) \(g : Z \to Z\) is onto. — Incorrect. \(3x + 1 = 2\) gives \(x = 1/3\), not an integer.
(C) \(g : Q \to Q\) is one-one and onto. — Correct. \(x = (y - 1)/3\) is rational for every rational y.
(D) \(g : N \to N\) is one-one. — Correct. \(3p + 1 = 3q + 1\) gives \(p = q\).