Solutions to Problem Set 24

Math 310-01/02

11-15-2017

1. (a) Find $5^{-1} \mod{29}$ (that is, find the number n in $\{0, 1, \ldots, 28\}$ which satisfies the equation $5 n = 1 \mod{29}$ ).

(b) Use the result of (a) to solve the equation $5 x = 17 \mod{29}$ . Your answer should be a number in the range $\{0, 1, \ldots, 28\}$ .

(a) $5 \cdot 6 = 30 = 1 =
   \mod{29}$ , so $5^{-1} = 6 \mod{29}$ .

(b)

$$\eqalign{ 5 x & = 17 \mod{29} \cr 6 \cdot 5 x & = 6 \cdot 17 \mod{29} \cr x & = 15 \mod{29} \cr} \quad\halmos$$


2. Prove that there is no number x such that $9 x = 1 \mod{99}$ .

(This shows that 9 does not have a multiplicative inverse mod 99.)

Suppose $9 x = 1 \mod{99}$ . Multiplying by 11, I get

$$\eqalign{ 99 x & = 11 \mod{99} \cr 0 & = 11 \mod{99} \cr}$$

This contradiction shows that there is no such x.


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

$$f(x, y) = (e^x + y, 3 x).$$

Prove that f is bijective by constructing an inverse $f^{-1}(a, b)$ .

Demonstrate that your inverse works by showing that $f(f^{-1}(a, b)) = (a, b)$ and $f^{-1}(f(x, y)) = (x, y)$ .

First, I'll work backwards to "guess" the inverse. If $f(x, y) = (a, b)$ , then $(x, y) = f^{-1}(a, b)$ . Substitute the definition of $f(x, y)$ in $f(x, y) = (a, b)$ :

$$(e^x + y, 3 x) = (a, b), \quad\hbox{so}\quad e^x + y = a \quad\hbox{and}\quad 3 x = b.$$

The second equation gives $x =
   \dfrac{b}{3}$ . Then

$$e^{b/3} + y = a, \quad\hbox{so}\quad y = a - e^{b/3}.$$

So

$$f^{-1}(a, b) = \left(\dfrac{b}{3}, a - e^{b/3}\right).$$

Next, I'll check that the guess works:

$$f(f^{-1}(a, b)) = f\left(\dfrac{b}{3}, a - e^{b/3}\right) = \left(e^{b/3} + a - e^{b/3}, 3\cdot \dfrac{b}{3}\right) = (a, b).$$

$$f^{-1}(f(x, y)) = f^{-1}(e^x + y, 3 x) = \left(\dfrac{3 x}{3}, e^x + y - e^{3 x/3}\right) = (x, e^x + y - e^x) = (x, y).$$

This proves that $f^{-1}$ is the inverse of f, so f is bijective.


He who has not lost his head over some things has no head to lose. - Jean-Paul Richter


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