Science:Math Exam Resources/Courses/MATH220/April 2011/Question 05

MATH220 April 2011

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Question 05
Prove that the function $f:\mathbb {R} -\{1\}\to \mathbb {R} -\{2\}$ given by $f(x)={\frac {2x}{x-1}}$ is bijective.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Remember that bijective is the same as being both injective (one-to-one) and surjective (onto).

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will first show that this function is injective. Suppose that $f(x)=f(y)$ for some $x,y\in \mathbb {R} -\{1\}$. First, we note that if $f(x)=0$ then $x=0$. So we can suppose that $x\neq 0\neq y$. Then we would have ${\begin{aligned}{\frac {2x}{x-1}}={\frac {2y}{y-1}}&\iff {\frac {y-1}{y}}={\frac {x-1}{x}}\\&\iff 1-{\frac {1}{y}}=1-{\frac {1}{x}}\\&\iff {\frac {1}{y}}={\frac {1}{x}}\end{aligned}}$ which is true if and only if $x=y$, and so the function is injective. To show that the function is surjective, we need to "invert" the function. That is, we want to show that for each $r\in \mathbb {R} -\{2\}$ that we can find some $x_{r}$ so that $f(x_{r})=r$. Expanding this yields ${\begin{aligned}{\frac {2x_{r}}{x_{r}-1}}&=r\\2x_{r}&=r(x_{r}-1)\\2x_{r}&=rx_{r}-r\\2x_{r}-rx_{r}&=-r\\x_{r}(2-r)&=-r\\x_{r}&={\frac {-r}{2-r}}\\x_{r}&={\frac {r}{r-2}}\end{aligned}}$ Hence we can choose $x_{r}={\frac {r}{r-2}}$. In such a case, we have ${\begin{aligned}f(x_{r})&={\frac {2{\frac {r}{r-2}}}{{\frac {r}{r-2}}-1}}\\&={\frac {2{\frac {r}{r-2}}}{\frac {r-(r-2)}{r-2}}}\\&={\frac {2r}{r-(r-2)}}=r\end{aligned}}$ as desired. (Note that this final justification is included only as a sanity check and is not needed in a solution to obtain full marks).

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that a function is bijective if and only if it has an inverse function, that is, a function $\displaystyle g(x)$ such that both $\displaystyle f(g(x))=x$ and that $\displaystyle g(f(x))=x$ I claim that for $\displaystyle f(x)={\frac {2x}{x-1}}$ the function ${\begin{aligned}g:\mathbb {R} -\{2\}&\to \mathbb {R} -\{1\}\\x&\mapsto {\frac {x}{x-2}}\end{aligned}}$ is an inverse function for $\displaystyle f(x)$. To prove this, we proceed directly. Notice that ${\begin{aligned}f(g(x))&={\frac {2({\tfrac {x}{x-2}})}{({\tfrac {x}{x-2}})-1}}\\&={\frac {2({\tfrac {x}{x-2}})}{({\tfrac {x-x+2}{x-2}})}}\\&={\frac {2x}{x-2}}\cdot {\frac {x-2}{2}}\\&=x\end{aligned}}$ and that ${\begin{aligned}g(f(x))&={\frac {\tfrac {2x}{x-1}}{({\tfrac {2x}{x-1}})-2}}\\&={\frac {\tfrac {2x}{x-1}}{({\tfrac {2x-2(x-1)}{x-1}})}}\\&={\frac {2x}{x-1}}\cdot {\frac {x-1}{2}}\\&=x\end{aligned}}$ and this completes the proof. To see where $\displaystyle g(x)$ came from, check out solution 1.

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