Question Statement

If $f:A\to B$ is defined by $f(x)=\frac{x-2}{x-3}$ for all $x\in A$ where $A=R-\{3\}$ and $B=R-\{1\}$, show that the function $f$ is bijective.

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Background and Explanation

A function is bijective if it is both one-one (injective) and onto (surjective). To prove this, we must show that every unique input has a unique output, and every element in the codomain is mapped to by at least one element in the domain.

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Solution

Given the function: $f(x)=\frac{x-2}{x-3}$ With Domain $A=R-\{3\}$ and Codomain $B=R-\{1\}$.

Part 1: Proving $f$ is one-one (Injective)

To show the function is one-one, we assume $f(x_1)=f(x_2)$ and must prove that $x_1=x_2$.

Let $f(x_1)=f(x_2)$: $\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$

By cross-multiplying the denominators: $(x_1-2)(x_2-3)=(x_1-3)(x_2-2)$

Expanding both sides: $x_1{x}_2-3x_1-2x_2+6=x_1{x}_2-2x_1-3x_2+6$

Subtracting $x_1{x}_2$ and $6$ from both sides: $-3x_1-2x_2=-2x_1-3x_2$

Rearranging the terms to group $x_1$ and $x_2$: $-3x_1+2x_1=-3x_2+2x_2$ $-x_1=-x_2$ $x_1=x_2$

Since $f(x_1)=f(x_2)$ implies $x_1=x_2$, the function $f$ is one-one.

Part 2: Proving $f$ is onto (Surjective)

To show the function is onto, we must show that for every $y$ in the codomain $B$, there exists an $x$ in the domain $A$ such that $f(x)=y$.

Let $f(x)=y$: $y=\frac{x-2}{x-3}$

Now, we solve for $x$ in terms of $y$: $y(x-3)=x-2$ $xy-3y=x-2$ $xy-x=3y-2$ $x(y-1)=3y-2$ $x=\frac{3y-2}{y-1}$

For $x$ to be a real number, the denominator $y-1$ cannot be zero: $y-1\ne 0\Rightarrow y\ne 1$

This matches our codomain $B=R-\{1\}$. Since for every $y\in B$, there exists a corresponding $x=\frac{3y-2}{y-1}$, and we can verify that $x\ne 3$ for any $y\in B$: If $\frac{3y-2}{y-1}=3$, then $3y-2=3y-3$, which leads to $-2=-3$ (impossible).

Thus, Range $f=B=R-\{1\}$. Since the range equals the codomain, the function $f$ is onto.

Conclusion

Because the function $f$ is both one-one and onto, it is a bijective function.

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Key Formulas or Methods Used

• One-one (Injective) Definition: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$
• Onto (Surjective) Definition: Range of $f$ = Codomain of $f$
• Bijective Definition: A function that is both injective and surjective.

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Summary of Steps

1. Test for Injective property: Set $f(x_1)=f(x_2)$, cross-multiply, and simplify to see if $x_1=x_2$.
2. Test for Surjective property: Set $y=f(x)$ and rearrange the equation to express $x$ in terms of $y$.
3. Identify Restrictions: Determine the values of $y$ for which $x$ is defined to find the range.
4. Compare: Verify that the calculated range matches the given codomain $B$.
5. Conclude: If both conditions are met, the function is bijective.