Science:Math Exam Resources/Courses/MATH220/December 2010/Question 05 (b)

MATH220 December 2010

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

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
Recall that a function is bijective if it is one to one and onto.

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Solution

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 first show our function is one to one (injective). Let $\displaystyle x,y\in \mathbb {R} -\{-2\}$ be such that $f(x)=f(y)$, that is, $\displaystyle {\frac {x+1}{x+2}}={\frac {y+1}{y+2}}$ Cross multiplying gives $\displaystyle (x+1)(y+2)=(y+1)(x+2).$ Simplifying yields $\displaystyle xy+2x+y+2=xy+2y+x+2$ and cancelling like terms shows that $\displaystyle x=y.$ To show our function is onto, suppose that $\displaystyle y\in \mathbb {R} -\{1\}$. We want to know if there is a $\displaystyle x\in \mathbb {R} -\{-2\}$ such that $f(x)=y$, that is, $\displaystyle {\frac {x+1}{x+2}}=y$ If we isolate for our x, we see that ${\begin{aligned}(x+1)&=y(x+2)\\x+1&=xy+2y\\1-2y&=xy-x\\{\frac {1-2y}{y-1}}&=x\end{aligned}}$ Notice in the last step dividing by $\displaystyle y-1$ is allowed since $\displaystyle y\neq 1$. Defining $x={\frac {1-2y}{y-1}}$ implies that $f(x)=y$. It remains to show that $x\in \mathbb {R} -\{-2\}$, that is, $x\neq -2$. Using a proof by contradiction we assume that $x=-2$, which implies ${\begin{aligned}{\frac {1-2y}{y-1}}&=-2\\1-2y&=-2y+2\\1&=2\end{aligned}}$ A contradiction! Hence, indeed, $x\in \mathbb {R} -\{-2\}$ and the proof is complete.

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