Show that the function `f:NtoN` defined by `f(x)=2x-1` is one-one but not onto.

Answer

To show that the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = 2x - 1 \) is one-one but not onto, we will follow these steps:

### Step 1: Show that \( f \) is one-one (injective)

To prove that \( f \) is one-one, we need to show that if \( f(a) = f(b) \), then \( a = b \).

Assume \( f(a) = f(b) \):
\[ 2a - 1 = 2b - 1 \]

Now, we can simplify this equation:
\[ 2a = 2b \]
Dividing both sides by 2:
\[ a = b \]

Since we have shown that \( f(a) = f(b) \) implies \( a = b \), the function \( f \) is one-one.

### Step 2: Show that \( f \) is not onto (surjective)

To show that \( f \) is not onto, we need to demonstrate that there exists at least one element in the codomain \( \mathbb{N} \) that is not the image of any element in the domain \( \mathbb{N} \).

The range of the function \( f(x) = 2x - 1 \) for \( x \in \mathbb{N} \) can be calculated as follows:
- For \( x = 1 \), \( f(1) = 2(1) - 1 = 1 \)
- For \( x = 2 \), \( f(2) = 2(2) - 1 = 3 \)
- For \( x = 3 \), \( f(3) = 2(3) - 1 = 5 \)
- For \( x = 4 \), \( f(4) = 2(4) - 1 = 7 \)
- For \( x = 5 \), \( f(5) = 2(5) - 1 = 9 \)

From this pattern, we can see that the outputs are all odd natural numbers: \( 1, 3, 5, 7, 9, \ldots \).

Now, consider the even natural number \( 2 \). There is no \( x \in \mathbb{N} \) such that \( f(x) = 2 \):
\[ 2x - 1 = 2 \implies 2x = 3 \implies x = \frac{3}{2} \]
Since \( \frac{3}{2} \) is not a natural number, \( 2 \) is not in the range of \( f \).

Thus, \( f \) is not onto.

### Conclusion

We have shown that the function \( f(x) = 2x - 1 \) is one-one (injective) and not onto (surjective).

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