Q.Prove that the Greatest Integer Function $f: \mathbf{R} \to \mathbf{R}$, given by $f(x) = [x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

One-one and onto. To disprove one-one, exhibit two distinct inputs with the same output; to disprove onto, exhibit a codomain value that is never an output. We apply both tests to $f(x) = [x]$.

Step 1 (not one-one): Take $x_1 = 1.2$ and $x_2 = 1.5$. Then $f(1.2) = [1.2] = 1$ and $f(1.5) = [1.5] = 1$. Two different inputs give the same output, so $f$ is not injective. (In general, every $x \in [n, n+1)$ maps to the same integer $n$.)

Step 2 (not onto): For every $x$, $[x]$ is an integer, so the range is $\mathbb{Z}$. Pick a non-integer such as $0.5 \in \mathbb{R}$: no real $x$ satisfies $[x] = 0.5$. So $0.5$ has no preimage and $f$ is not surjective.

Why this approach works

To prove a function is not one-one, we just need to find two different inputs that give the same output. For $[x]$, any two numbers in the same integer interval $[n, n+1)$ map to the same $n$ — so that's immediate.

To prove it's not onto, we need to show there's some real number that never appears as $f(x)$. Since $[x]$ always spits out an integer, any non-integer (like $0.5$) can never be the output. That's the whole idea.

Step-by-step proof

1. Not one-one

Take any integer $n$. For any $x$ in the interval $[n, n+1)$, the definition says $[x] = n$. So pick two distinct numbers in that interval, say:

Both are in $[n, n+1)$, so:

Thus $x_1 \neq x_2$ but $f(x_1) = f(x_2)$. Hence $f$ is not injective.

2. Not onto

The codomain is $\mathbf{R}$ (all real numbers). But what values does $f$ actually take? For any $x \in \mathbf{R}$, $[x]$ is always an integer. So:

Pick any non-integer real number, say $y = 0.5$. Is there any $x$ such that $[x] = 0.5$? No — because $[x]$ is always an integer. So $0.5$ has no preimage.

Thus $f$ is not surjective.

Method: Proving a function is neither one-one nor onto

To DISPROVE a property you need only one counterexample — far quicker than a general proof.

Steps

Step 1: Disprove one-one with two distinct inputs sharing an output

Find explicit $x_1\neq x_2$ with $f(x_1)=f(x_2)$. For a step function like $[x]$, any two numbers in the same interval $[n,n+1)$ work, e.g. $[1.2]=[1.5]=1$.

Step 2: Disprove onto by exhibiting an unreached codomain value

Identify the actual range of $f$ and pick one codomain element outside it. Since $[x]$ only outputs integers, any non-integer such as $0.5$ has no preimage.

Step 3: State both counterexamples explicitly — "constant on intervals" or "the range is small" is not a proof on its own; name the specific values that break each property.

Common Mistakes

Mistake 1: Saying "not one-one because it is constant on intervals" without a concrete counterexample.

Why it's wrong: a formal disproof needs two explicit distinct inputs with equal output. Correct approach: state e.g. $[1.2]=[1.5]=1$ while $1.2\neq 1.5$.

Mistake 2: Thinking the range of $[x]$ is all of $\mathbf{R}$.

Why it's wrong: $[x]$ only outputs integers, so its range is $\mathbb{Z}$, and non-integers like $0.5$ have no preimage. Correct approach: compare range $\mathbb{Z}$ with codomain $\mathbf{R}$ to see it is not onto.