Is my answer correct for the following question?

Show that the function $f : \mathbb{R} − {3} \to \mathbb{R} − {2}$ defined by $f(x) = \frac{2x−3}{x-3}$ is a bijection, and find the inverse function. I don't necessarily have to prove anything.

To show it's a bijection, I need to show that it's one-to-one and onto.

A function is one-to-one iff. $f(a)=f(b)$ implies that $a=b$

$\frac{2a-3}{a-3} =\frac{2b-3}{b-3}$





When $f(a)=f(b)$, $b=a$, therefore, this function is one-to-one.

A function is onto iff. for every $b\in B$, there is an $a \in A$ such that $f(a)=b$

For any $b\in R$-{2}, there's $a\in R$-{3} such that $f(a)=b$

Since $f(x)$ is one-to-one and onto, it is a bijection.

Finding the inverse:



$x= \frac{2y-3}{y-3}$



y= $\frac{3x-3}{x-2}$


There's a couple parts in your answer that are confusing:

When proving $f$ is 1-1, you have $(2a−3)(b−3)=(a−3)(2b−3)$ implies $b-3 = a-3$, but I'm not sure how you're making that jump. I would instead note that you have: $$(2a-3)(b-3) = (a-3)(2b-3) \longrightarrow 2ab - 6a -3b + 9 = 2ab -3a -6b + 9$$ Then by canceling the $2ab$ and the 9, you will get $3b = 3a$, or $b=a$ as desired.

Then, when proving that $f$ is onto, we need to start with an output value $y$ and show that there is some $x$ in the domain such that $f(x) = y$. In your explanation, you've discussed for which $x$ values the function is well-defined; ie, the domain of the function.

Instead, we let $y \in \mathbb{R} - \{2\}$. Then $y = \frac{2x-3}{x-3}$, so $y(x-3) = 2x - 3$, or $3 - 3y = 2x - yx = (2-y)x$. Then $x = \frac{3-3y}{2-y}$, which is well defined for all $y \in \mathbb{R}-\{2\}$. Note too that this $x$ will never be 3, or we'd have $6=3$. Thus for all $y$ in the range, we can find $x \in \mathbb{R} - \{3\}$, $x = \frac{3-3y}{2-y}$ such that $f(x) = y$, so our function is onto.

Your inverse function looks correct to me.


For the injectivity part, you say $$(2a-3)(b-3)=(a-3)(2b-3),$$ and then jump to $$a-3=b-3.$$ I do not think that the first follows immediately from the second. Or if it does in some way I am not seeing, you should probably add some intermediate steps.

For surjectivity, you need to prove that for every $b\in\mathbb R-\{2\}$, there is an $a\in\mathbb R-\{3\}$ so that $f(a)=b$. What you proved (or rather, stated) is that for all $a\in\mathbb R-\{3\}$, $f(a)\in\mathbb R$. Notice the difference? Examine the definition of surjectivity again.

  • $\begingroup$ Thanks! For some reason I thought I could cancel out the two terms but they weren't the same. Is my surjectivity part correct now? $\endgroup$ – Spectre Nov 8 '17 at 3:10
  • $\begingroup$ No, you are still trying to prove the wrong thing with surjectivity. Now you are stating that for any $a\in\mathbb R-\{3\}$, $f(a)\in\mathbb R-\{2\}$. You need to show that for any $b\in\mathbb R-\{2\}$, there is some value of $a\in\mathbb R-\{3\}$ so that $f(a)=b$. These are two very different concepts. For example, the function $g:\mathbb R\to\mathbb R$ defined by $g(x)=6$ for all $x$ has the property that for all $x\in\mathbb R$, $g(x)\in\mathbb R$, but it is certainly not surjective. $\endgroup$ – Alex S Nov 8 '17 at 3:17
  • $\begingroup$ Ok I changed it again. $\endgroup$ – Spectre Nov 8 '17 at 3:27
  • 2
    $\begingroup$ What you have there now is simply a statement of what surjectivity means. To prove that the function is surjective, you need to prove that the statement is true. Emma's answer shows how to do this. $\endgroup$ – Alex S Nov 8 '17 at 3:32

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