Q.Check the injectivity and surjectivity of the following functions:

Recall: injective means different inputs give different outputs ($f(a)=f(b)\Rightarrow a=b$); surjective means every element of the codomain is hit.

(i) $f:\mathbf{N}\to\mathbf{N},\ f(x)=x^2$ — injective (on positive integers $a^2=b^2\Rightarrow a=b$); not surjective ($2$ is not a perfect square).

(ii) $f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^2$ — not injective ($f(-2)=f(2)=4$); not surjective ($-1$ has no preimage, squares are $\ge 0$).

(iii) $f:\mathbf{R}\to\mathbf{R},\ f(x)=x^2$ — not injective ($f(-a)=f(a)$); not surjective ($-1$ has no real preimage).

(iv) $f:\mathbf{N}\to\mathbf{N},\ f(x)=x^3$ — injective ($a^3=b^3\Rightarrow a=b$); not surjective ($2$ is not a perfect cube).

(v) $f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^3$ — injective ($x^3$ is strictly increasing, so $a^3=b^3\Rightarrow a=b$); not surjective, because $2$ has no integer cube root ($1^3=1,\ 2^3=8$, nothing gives $2$).

The whole question turns on one theme: the same formula behaves differently as we change the number system. Injectivity fails for $x^2$ whenever negatives are available (because $(-a)^2=a^2$); surjectivity fails whenever the codomain contains values the formula can never produce.

(i) $f:\mathbf{N}\to\mathbf{N},\ f(x)=x^2$

• Injective: on natural numbers there are no negatives, so $a^2=b^2\Rightarrow a=b$. Yes.
• Surjective: an output must be a perfect square, but $2,3,5,\dots$ are natural numbers that are not squares. No.

(ii) $f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^2$

• Injective: $f(-2)=4=f(2)$ with $-2\ne 2$. No.
• Surjective: squares are never negative, so $-1$ has no preimage. No.

(iii) $f:\mathbf{R}\to\mathbf{R},\ f(x)=x^2$

• Injective: $f(-a)=f(a)$ for any $a\ne 0$. No.
• Surjective: $x^2\ge 0$ always, so no negative real (e.g. $-1$) is an output. No.

(iv) $f:\mathbf{N}\to\mathbf{N},\ f(x)=x^3$

• Injective: $x^3$ is strictly increasing on $\mathbf{N}$, so $a^3=b^3\Rightarrow a=b$. Yes.
• Surjective: an output must be a perfect cube; $2$ is not a cube of any natural number. No.

(v) $f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^3$

• Injective: $x^3$ is strictly increasing on $\mathbf{Z}$ ($a<b\Rightarrow a^3<b^3$), so $a^3=b^3\Rightarrow a=b$. Yes.
• Surjective: we would need every integer to be a perfect cube. But $2$ is not: $1^3=1$ and $2^3=8$, so no integer cubes to $2$. No.

Method: Same formula, different number system — deciding one-one and onto

When one power formula is tested over $\mathbf{N}$, $\mathbf{Z}$, $\mathbf{R}$, the answer changes with the set. Two facts settle every case.

Steps

Step 1: Injectivity depends on whether negatives are available

An even power ($x^2$) satisfies $f(-a)=f(a)$, so it is NOT one-one on any set containing negatives ($\mathbf{Z}$, $\mathbf{R}$) but IS one-one on $\mathbf{N}$. An odd power ($x^3$) is strictly increasing, hence one-one on every set.

Step 2: Surjectivity depends on whether every codomain value is actually produced

Ask: is every element of the codomain an output? On $\mathbf{N}$ or $\mathbf{Z}$, $x^2$ and $x^3$ miss non-squares / non-cubes (e.g. $2$), so they are not onto; $x^2$ also misses all negatives.

Step 3: Report each case as the pair (injective?, surjective?)

Run Steps 1–2 for each domain/codomain separately; never assume the verdict carries over from one number system to another.

Common Mistakes

Mistake 1: Believing $x^2$ is one-one because "squaring is a normal function".

Why it's wrong: whenever negatives are available ($\mathbf{Z}$, $\mathbf{R}$), $f(-a)=f(a)$ breaks injectivity; it is one-one only on $\mathbf{N}$. Correct approach: check whether the domain contains negatives before deciding.

Mistake 2: Assuming $x^3$ is onto on $\mathbf{Z}$ because it is onto on $\mathbf{R}$.

Why it's wrong: over $\mathbf{R}$ every value has a real cube root, but the cube root of an integer need not be an integer (e.g. $2$ is not a perfect cube). Correct approach: judge surjectivity in the actual codomain, not by analogy to $\mathbf{R}$.