Q.Show that the function $f: \mathbb{N} \to \mathbb{N}$, given by $f(x) = 2x$, is one-one but not onto.

Concept: One-One (Injective) Function — A function is one-one if distinct inputs always give distinct outputs: $f(a) = f(b) \implies a = b$.

Step 1 — Proving one-one:

Take any $a, b \in \mathbb{N}$ such that $f(a) = f(b)$.

Then $2a = 2b \implies a = b$. Hence $f$ is one-one.

Step 2 — Proving not onto:

For $f$ to be onto, every natural number must be of the form $2x$ for some $x \in \mathbb{N}$.

But odd numbers like $1, 3, 5, \dots$ are never reached because $2x$ is always even.

Step 3 — Conclusion:

Since $1 \in \mathbb{N}$ has no pre-image, $f$ is not onto.

The core idea is simple: we need to check two properties of the function $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = 2x$.

One-one (injective) means: if $f(a) = f(b)$, then $a = b$. In other words, no two different inputs produce the same output.

Onto (surjective) means: every element in the codomain $\mathbb{N}$ must be the image of some input. That is, for every $y \in \mathbb{N}$, there exists some $x \in \mathbb{N}$ such that $f(x) = y$.

Let’s test both.

1. Proving $f$ is one-one.

   Suppose $f(a) = f(b)$ for some $a, b \in \mathbb{N}$.

   Then $2a = 2b$. Dividing both sides by $2$ (which is valid in $\mathbb{N}$), we get $a = b$.

   So the condition holds: equal outputs force equal inputs. Hence $f$ is injective.

   Tip

   The key here is that multiplication by $2$ is a strictly increasing operation on $\mathbb{N}$. If $a \neq b$, then $2a \neq 2b$ automatically. So one-oneness is immediate from the monotonicity.

2. Proving $f$ is NOT onto.

   The codomain is $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$.

   The range of $f$ is $\{2, 4, 6, 8, \dots\}$ — all even natural numbers.

   Consider $y = 1$. Is there any $x \in \mathbb{N}$ such that $2x = 1$?

   Solving gives $x = \frac{1}{2}$, which is not a natural number.

   So $1$ has no preimage. Similarly, every odd number ($3, 5, 7, \dots$) is missing from the range.

   Since there exists at least one element in $\mathbb{N}$ (in fact, infinitely many) that is not the image of any input, $f$ is not onto.

   Watch out

   A common mistake is to think "onto" means the function covers all numbers it can produce. But onto is about covering the entire codomain, not just the range. Here the codomain is all naturals, but the range is only evens — so it fails.

3. Visualising the result.

   Think of $\mathbb{N}$ as two interleaved sets: odds and evens.

   $f$ sends every natural to an even number, but never to an odd.

   So the evens get "hit" exactly once each (one-one), but the odds are completely untouched (not onto).

   Input $x$	Output $f(x)=2x$
   1	2
   2	4
   3	6
   4	8
   5	10
   …	…

   Notice: $1, 3, 5, 7, \dots$ never appear in the output column.

Method: Proving a function is one-one but not onto

Use this when a rule injects a set into itself or a larger set but skips some target values.

Steps

Step 1: Prove one-one (injective)

Assume $f(x_1)=f(x_2)$ and simplify to reach $x_1=x_2$. For a linear rule $f(x)=ax\ (a\ne0)$, $ax_1=ax_2$ gives $x_1=x_2$ directly.

Step 2: Prove not onto (not surjective)

Find at least one element of the codomain that no input can produce. Describe the range and show it is a proper subset of the codomain — for $f(x)=2x$ on $\mathbb{N}$, the range is the even numbers, so every odd number is missed.

Step 3: Conclude

State both results: injective, not surjective. The transferable idea is that "onto" is about covering the entire codomain, so exhibiting one unreached element is enough to disprove it.

Common Mistakes

Mistake 1: Thinking $f(x)=2x$ is onto because "it produces lots of numbers"

Why it's wrong: the codomain is all of $\mathbb{N}$, but the range is only the even numbers, so every odd number (like $1$) has no preimage. Correct approach: to be onto, $2x=y$ must have a solution in $\mathbb{N}$ for every $y$; $y=1$ gives $x=\tfrac12\notin\mathbb{N}$, so $f$ is not onto.

Mistake 2: Proving injectivity by testing a couple of values

Why it's wrong: $f(1)=2,\ f(2)=4$ being different is not a proof. Correct approach: assume $f(x_1)=f(x_2)$, i.e. $2x_1=2x_2$, and cancel the $2$ to get $x_1=x_2$ for all inputs.