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

MATH220 April 2011

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

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

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[show] Solution 1
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).

[show] Solution 2
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.