Question

If the function \(f:N\to N\) is defined as \(f(x)=x^2\), then \(f\) is:

• A. One-one and onto
• B. One-one but not onto
• C. Neither one-one nor onto
• D. Many-one and onto

Hint:

Squaring is injective on naturals but its range misses non-perfect-squares like 2.

Answer 1

The Correct Option is B

Step 1: Check one-one:
For \(x_1,x_2\in N\), if \(f(x_1)=f(x_2)\) then \(x_1^2=x_2^2\). Since \(x_1,x_2\) are natural numbers (positive), this forces \(x_1=x_2\). So \(f\) is one-one.

Step 2: Check onto:
For \(f\) to be onto, every \(y\in N\) must have a pre-image \(x\in N\) with \(x^2=y\), i.e. \(x=\sqrt y\). Take \(y=2\): \(\sqrt2\) is not a natural number, so \(2\) has no pre-image in \(N\).

Step 3: Why the other options are wrong:
Option A and D claim \(f\) is onto, which fails as shown. Option C claims \(f\) is not one-one, which is false since we proved injectivity above.

Final Answer:
\(f\) is one-one but not onto. \[ \boxed{\text{One-one but not onto}} \]