Solutions to Problem Set 26

Math 310-01/02

11-17-2017

1. Define $f: \real - \{2\} \to
   \real$ by

$$f(x) = \dfrac{x}{x - 2}.$$

Prove that $\im f = \real - \{1\}
   = \{y \in \real \mid y \ne 1\}$ .

First, suppose $f(x) \in \im f$ , where $x \in \real - \{2\}$ . I need to show that $f(x) \in \{y \in \real \mid y \ne
   1\}$ . That is, I must show $f(x) \ne 1$ .

Suppose on the contrary that $f(x) = 1$ . Then

$$\eqalign{ \dfrac{x}{x - 2} & = 1 \cr \noalign{\vskip2pt} x & = x - 2 \cr 0 & = -2 \cr}$$

This contradiction shows that $f(x) \ne 1$ , so $f(x) \in \{y \in \real \mid y \ne
   1\}$ .

Conversely, suppose $y \in \{y
   \in \real \mid y \ne 1\}$ , so $y \ne 1$ .

First, $\dfrac{2 y}{y - 1} \ne
   2$ . For if $\dfrac{2 y}{y - 1} = 2$ , then

$$\eqalign{ \dfrac{2 y}{y - 1} & = 2 \cr \noalign{\vskip2pt} 2 y & = 2(y - 1) \cr 2 y & = 2 y - 2 \cr 0 & = -2 \cr}$$

This contradiction shows $\dfrac{2 y}{y - 1} \ne 2$ , so $\dfrac{2 y}{y - 1} \in \real
   - \{2\}$ .

Now

$$f\left(\dfrac{2 y}{y - 1}\right) = \dfrac{\dfrac{2 y}{y - 1}}{\dfrac{2 y}{y - 1} - 2} = \dfrac{2 y}{2 y - 2(y - 1)} = \dfrac{2 y}{2} = y.$$

This proves that $y \in \im f$ .

Hence, $\im f = \real - \{1\} =
   \{y \in \real \mid y \ne 1\}$ .


2. Define $f: \real^2 \to
   \real^2$ by

$$f(x, y) = (e^{x-y}, x + y).$$

Prove that f is injective, but f is not surjective.

If $f(x, y) = (-1, 0)$ , then $(e^{x-y}, x + y) = (-1, 0)$ , so $e^{x-y} = -1$ . Since $e^{x-y}
   > 0$ for all x and y, this is a contradiction. Therefore, f is not surjective.

Suppose $f(a, b) = f(c, d)$ . Then

$$(e^{a-b}, a + b) = (e^{c-d}, c + d), \quad\hbox{so}\quad e^{a-b} = e^{c-d} \quad\hbox{and}\quad a + b = c + d.$$

Taking logs on both sides of $e^{a-b} = e^{c-d}$ , I get $a - b = c - d$ . Adding this equation to $a + b = c + d$ , I get $2 a =
   2c$ , so $a = c$ . This allows me to cancel a and c from $a + b = c + d$ to get $b = d$ . Therefore, $(a, b) = (c, d)$ , and f is injective.


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