Let `A=R-{2}` and `B=R-{1}` . If `f: A->B` is a mapping defined by `f(x)=(x-1)/(x-2)` , show that `f` is bijective.

Explanation

To show that the function \( f: A \to B \) defined by \( f(x) = \frac{x-1}{x-2} \) is bijective, we need to prove that it is both one-to-one (injective) and onto (surjective).

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

To prove that \( f \) is one-to-one, we assume that \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in A \). This means:

\[ \frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \]

Cross-multiplying gives:

\[ (x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2) \]

Expanding both sides:

\[ x_1 x_2 - 2x_1 - x_2 + 2 = x_2 x_1 - 2x_2 - x_1 + 2 \]

Now, simplifying both sides, we can cancel \( x_1 x_2 \) and \( 2 \):

\[ -2x_1 - x_2 = -2x_2 - x_1 \]

Rearranging gives:

\[ -2x_1 + x_1 = -2x_2 + x_2 \]

This simplifies to:

\[ -x_1 = -x_2 \]

Thus, we have:

\[ x_1 = x_2 \]

Since \( x_1 \) and \( x_2 \) are equal, we conclude that \( f \) is one-to-one.

### Step 2: Proving that \( f \) is onto (surjective)

To prove that \( f \) is onto, we need to show that for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

Starting with the equation:

\[ y = \frac{x - 1}{x - 2} \]

We can rearrange this to solve for \( x \):

\[ y(x - 2) = x - 1 \]

Expanding gives:

\[ yx - 2y = x - 1 \]

Rearranging terms yields:

\[ yx - x = 2y - 1 \]

Factoring out \( x \):

\[ x(y - 1) = 2y - 1 \]

Thus, we can express \( x \) as:

\[ x = \frac{2y - 1}{y - 1} \]

Now, we need to ensure that \( x \) is in \( A \) (which is \( \mathbb{R} - \{2\} \)). The expression \( \frac{2y - 1}{y - 1} \) will not equal 2 for any \( y \in B \) (which is \( \mathbb{R} - \{1\} \)).

To check when \( x = 2 \):

\[ \frac{2y - 1}{y - 1} = 2 \]

Cross-multiplying gives:

\[ 2y - 1 = 2(y - 1) \]

This simplifies to:

\[ 2y - 1 = 2y - 2 \]

Which leads to:

\[ -1 = -2 \]

This is a contradiction, meaning \( x \) can never be 2. Thus, for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

### Conclusion

Since \( f \) is both one-to-one and onto, we conclude that \( f \) is bijective.

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