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I am new to discrete mathematics, and this was one of the question that the prof gave out. I am bit lost in this, since I never encountered discrete mathematics before. What do I need to do to prove that it is bijection, and find the inverse? Do I choose any number(integer) and put it in for the R and see if the corresponding question is bijection(both one-to-one and onto)?

Show that the function $f: \Bbb R \setminus \{-1\} \to \Bbb R \setminus \{2\}$ defined by $$ f(x) = \frac{4x + 3}{2x + 2} $$ is a bijection, and find the inverse function. (Hint: Pay attention to the domain and codomain.)

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  • $\begingroup$ wait, what does \ stand for? In the question it did say R - {-1} -> R - {2} $\endgroup$ – Dennis_Y Nov 6 '14 at 11:13
  • $\begingroup$ That is another way of writing the set difference. $\endgroup$ – N. F. Taussig Nov 6 '14 at 11:15
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A function is bijective if it is injective (one-to-one) and surjective (onto).

You can show $f$ is injective by showing that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.

You can show $f$ is surjective by showing that for each $y \in \mathbb{R} - \{2\}$, there exists $x \in \mathbb{R} - \{-1\}$ such that $f(x) = y$.

If $f(x_1) = f(x_2)$, then

\begin{align*} \frac{4x_1 + 3}{2x_1 + 2} & = \frac{4x_2 + 3}{2x_2 + 3}\\ (4x_1 + 3)(2x_2 + 2) & = (2x_1 + 2)(4x_2 + 3)\\ 8x_1x_2 + 8x_1 + 6x_2 + 6 & = 8x_1x_2 + 6x_1 + 8x_2 + 6\\ 8x_1 + 6x_2 & = 6x_1 + 8x_2\\ 2x_1 & = 2x_2\\ x_1 & = x_2 \end{align*} Thus, $f$ is injective.

Let $y \in \mathbb{R} - \{2\}$. We must show that there exists $x \in \mathbb{R} - \{-1\}$ such that $y = f(x)$. Suppose

$$y = \frac{4x + 3}{2x + 2}$$

Solving for $x$ yields \begin{align*} (2x + 2)y & = 4x + 3\\ 2xy + 2y & = 4x + 3\\ 2xy - 4x & = 3 - 2y\\ (2y - 4)x & = 3 - 2y\\ x & = \frac{3 - 2y}{2y - 4} \end{align*} which is defined for each $y \in \mathbb{R} - \{2\}$. Moreover, $x \in \mathbb{R} - \{-1\}$. To see this, suppose that $$-1 = \frac{3 - 2y}{2y - 4}$$ Then \begin{align*} -2y + 4 & = 3 - 2y\\ 4 & = 3 \end{align*} which is a contradiction.

The inverse function is found by interchanging the roles of $x$ and $y$. Hence, the inverse is $$y = \frac{3 - 2x}{2x - 4}$$ To verify the function $$g(x) = \frac{3 - 2x}{2x - 4}$$ is the inverse, you must demonstrate that \begin{align*} (g \circ f)(x) & = x && \text{for each $x \in \mathbb{R} - \{-1\}$}\\ (f \circ g)(x) & = x && \text{for each $x \in \mathbb{R} - \{2\}$} \end{align*}

\begin{align*} (g \circ f)(x) & = g\left(\frac{4x + 3}{2x + 2}\right)\\ & = \frac{3 - 2\left(\dfrac{4x + 3}{2x + 2}\right)}{2\left(\dfrac{4x + 3}{2x + 2}\right) - 4}\\ & = \frac{3(2x + 2) - 2(4x + 3)}{2(4x + 3) - 4(2x + 2)}\\ & = \frac{6x + 6 - 8x - 6}{8x + 6 - 8x - 8}\\ & = \frac{-2x}{-2}\\ & = x\\ (f \circ g)(x) & = f\left(\frac{3 - 2x}{2x - 4}\right)\\ & = \frac{4\left(\dfrac{3 - 2x}{2x - 4}\right) + 3}{2\left(\dfrac{3 - 2x}{2x - 4}\right) + 2}\\ & = \frac{4(3 - 2x) + 3(2x - 4)}{2(3 - 2x) + 2(2x - 4)}\\ & = \frac{12 - 8x + 6x - 12}{6 - 4x + 4x - 8}\\ & = \frac{-2x}{-2}\\ & = x \end{align*} Hence, $g = f^{-1}$, as claimed.

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  • $\begingroup$ (g∘f)(x)=x (f∘g)(x)=x for these two, at the last part I get integer/0, is it correct? $\endgroup$ – Dennis_Y Nov 7 '14 at 1:47
  • $\begingroup$ @Dennis_Y I have edited my answer to show how I obtained \begin{align*} (g \circ f)(x) & = x\\ (f \circ g)(x) & = x\end{align*} $\endgroup$ – N. F. Taussig Nov 7 '14 at 2:19
  • $\begingroup$ Oh, I see it now. Thank you :) $\endgroup$ – Dennis_Y Nov 7 '14 at 2:28
  • $\begingroup$ You're welcome. $\endgroup$ – N. F. Taussig Nov 7 '14 at 2:29
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\begin{align} y &= \frac{4x + 3}{2x + 2} \\ \implies(2x+2)y &= 4x + 3 \\\implies (2y)x+2y &= 4x + 3 \\ \cdots \end{align}

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To find the inverse $$x = \frac{4y+3}{2y+2} \Rightarrow 2xy + 2x = 4y + 3 \Rightarrow y (2x-4) = 3 - 2x \Rightarrow y = \frac{3 - 2x}{2x -4}$$

For injectivity let $$f(x) = f(y) \Rightarrow \frac{4x+3}{2x+2} = \frac{4y+3}{2y+2} \Rightarrow 8xy + 6y + 8x + 6 = 8xy + 6x + 8y + 6 \Rightarrow 2x = 2y \Rightarrow x= y$$

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