Check if f(x) = x² is one-one and onto from R to R

One-one (Injective): A function $f$ is one-one if distinct inputs give distinct outputs. Formally: $f(a) = f(b) \implies a = b$ for all $a, b \in$ domain.

Equivalently: if $a \neq b$, then $f(a) \neq f(b)$.

Onto (Surjective): A function $f: A \rightarrow B$ is onto if every element of the codomain $B$ is the image of at least one element from $A$. Formally: For every $y \in B$, there exists $x \in A$ such that $f(x) = y$.

Take $f(a) = f(b)$:

So either $a = b$ OR $a = -b$.

Since $a = -b$ is possible (e.g., $a = 2, b = -2$), we can have $f(a) = f(b)$ with $a \neq b$.

Counterexample: $f(2) = 4$ and $f(-2) = 4$. So two different inputs (2 and -2) give the same output (4).

Conclusion: $f(x) = x^2$ is NOT one-one (not injective).

We need to check if every real number $y \in \mathbb{R}$ can be written as $y = x^2$ for some real $x$.

Consider $y = -1$ (a negative real number). Is there any $x \in \mathbb{R}$ such that $x^2 = -1$?

There is no real solution — the square of any real number is non-negative. So $x^2 \geq 0$ for all $x \in \mathbb{R}$.

This means no negative number is in the range of $f$. The range of $f$ is $[0, \infty)$, but the codomain is $\mathbb{R}$ (which includes negative numbers).

Since range $\subsetneq$ codomain, the function is NOT onto.

Conclusion: $f(x) = x^2$ is NOT onto (not surjective) as a function from $\mathbb{R}$ to $\mathbb{R}$.

$f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^2$:

• NOT one-one: Because $f(2) = f(-2) = 4$ (two distinct inputs map to the same output)
• NOT onto: Because negative real numbers (like $-1$) have no pre-image in $\mathbb{R}$ under $f$

Therefore, $f(x) = x^2$ is neither injective nor surjective as a function from $\mathbb{R}$ to $\mathbb{R}$.