Practice Problems for Chapter 7

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Section 7.3 Practice Problems for Chapter 7

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Exercises

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1.

Let $R^{+} = {y \in R | y > 0} .$ Define

$f : R \to R$ by $f (x) = e^{- x},$ for each $x \in R,$ and

$g : R \to R^{+}$ by $g (x) = e^{- x},$ for each $x \in R$

Determine if each of these functions is an injection or a surjection. Justify your conclusions

Note: Before writing proofs, it might be helpful to draw the graph of $y = e^{- x} .$ A reasonable graph can be obtained using $- 3 \leq x \leq 3$ and $- 2 \leq y \leq 10 .$ Please keep in mind that the graph does not prove any conclusion, but may help us arrive at the correct conclusions, which will still need proof.

Solution.

The function $f$ is an injection but not a surjection. To see that it is an injection, let $a, b \in R$ and assume that $f (a) = f (b) .$ This implies that $e^{- a} = e^{- b} .$ Now use the natural logarithm function to prove that $a = b .$ Since $e^{- x} > 0$ for each real number $x,$ there is no $x \in R$ such that $f (x) = - 1 .$ So $f$ is not a surjection.

The function $g$ is an injection and is a surjection. The proof that $g$ is an injection is basically the same as the proof that $f$ is an injection. To prove that $g$ is a surjection, let $b \in R + .$ To construct the real number $a$ such that $g (a) = b,$ solve the equation $e^{- a} = b$ for $a .$ The solution is $a = - \ln b .$ It can then be verified that $g (a) = b .$

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2.

For each of the following functions, determine if the function is an injection or a surjection (or both, and hence, a bijection). Justify all conclusions.

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(a)

$F : R \to R$ defined by $F (x) = 5 x + 3,$ for all $x \in R .$

Solution.

Let $F : R \to R$ be defined by $F (x) = 5 x + 3$ for all $x 2 R .$ Let $x 1, x 2 \in R$ and assume that $F (x_{1}) = F (x_{2}) .$ Then,

\begin{aligned} 5 x_{1} + 3 & = 5 x_{2} + 3 \\ 5 x_{1} & = 5_{x} 2 \\ x_{1} & = x_{2} \end{aligned}

Hence, $F$ is an injection. Now, let $y \in R .$ Then, $\frac{y - 3}{5} \in R$ and

\begin{aligned} F (\frac{y - 3}{5}) & = 5 (\frac{y - 3}{5}) + 3 \\ = (y - 3) + 3 \\ = y . \end{aligned}

Thus, $F$ is a surjection and hence $F$ is a bijection.

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(b)

$G : Z \to Z$ defined by $G (x) = 5 x + 3,$ for all $x \in Z .$

Solution.

The proof that $G$ is an injection is similar to the proof in Task 7.3.2.a that $F$ is an injection. Now, for each $x \in Z,$ $5 x + 3 \equiv 3 (\mod 5),$ and hence $G (x) \equiv 3 (\mod 5) .$ This means that there is no integer $x$ such that $G (x) = 0 .$ Therefore, $G$ is not a surjection.

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(c)

$f : (R - {4}) \to R)$ defined by $f (x) = \frac{3 x}{x - 4},$ for all $x \in (R - {4}) .$

Solution.

Let $a, b \in R - {4}$ and assume that $f (a) = f (b) .$ Then,

\begin{aligned} \frac{3 a}{a - 4} & = \frac{3 b}{b - 4} \\ 3 a (b - 4) & = 3 b (a - 4 b) \\ 3 a b - 12 a & = 3 a b - 12 b \\ - 12 a & = - 12 b \\ a & = b \end{aligned}

So $f$ is an injection.

Use a proof by contradiction to show there is no $a \in R - {4}$ such that $f (a) = 3 .$ Assume such an $a$ exists. Then

\begin{aligned} \frac{3 a}{a - 4} & = 3 \\ 3 a & = 3 a - 12 \\ 0 & = - 12 \end{aligned}

and this is a contradiction. Therefore, for all $x \in R - {4},$ $f (x) \neq 3$ and $f$ is not a surjection.

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(d)

$g : (R - {4}) \to (R - {3})$ defined by $g (x) = \frac{3 x}{x - 4},$ for all $x \in (R - {4}) .$

Solution.

The function $g$ is a bijection. The proof that is an injection is similar to the proof that $f$ is an injection in Task 7.3.2.c. To prove that it is a surjection let $y \in R - {3} .$ Then, $\frac{4 y}{y - 3} \in R - {4}$ and

\begin{aligned} g (\frac{4 y}{y - 3}) & = \frac{3 (\frac{4 y}{y - 3})}{(\frac{4 y}{y - 3}) - 4} \\ = \frac{12 y}{4 y - 4 (y - 3)} \\ = \frac{12 y}{12} \\ = y . \end{aligned}

This proves that $g$ is a surjection.

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3.

Let $s$ be the function that associates with each natural number the sum of its distinct natural number divisors. This is called the sum of the divisors function. For example, the natural number divisors of 6 are 1, 2, 3, and 6, and so

\begin{aligned} s (6) & = 1 + 2 + 3 + 6 \\ = 12 \end{aligned}

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(a)

Calculate $s (k)$ for each natural number $k$ from 1 through 15.

Solution.

\begin{aligned} s (1) & = 1 & s (5) & = 6 & s (9) & = 13 & s (13) & = 14 \\ s (2) & = 3 & s (6) & = 12 & s (10) & = 18 & s (14) & = 24 \\ s (3) & = 4 & s (7) & = 8 & s (11) & = 12 & s (15) & = 24 \\ s (4) & = 7 & s (8) & = 15 & s (12) & = 28 & s (16) & = 31 \end{aligned}

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(b)

Is the sum of the divisors function an injection? Is it a surjection? Justify your conclusions.

Solution.

The sum of the divisors function $s$ is not an injection. For example, $s (6) = s (11) .$ This function is also not a surjection. For example, for all $x \in N,$ $s (x) \neq 2$ and for all $x \in N,$ $s (x) \neq 5 .$

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4.

Let $M_{2} (R)$ represent the set of all 2 by 2 matricies over $R .$

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(a)

Define $det: M_{2} (R) \to R$ by

det [\begin{matrix} a & b \\ c & d \end{matrix}] = a b - b c .

This is the determinant function on the set of 2 by 2 matrices over the real numbers. Is the determinant function an injection? Is the determinant function a surjection? Justify your conclusions.

Solution.

The determinant function is not an injection. For example,

det [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = det [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] .

The determinant function is a surjection. To prove this, let $a \in R .$ Then

det [\begin{matrix} a & 0 \\ 0 & 1 \end{matrix}] = a .

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(b)

Define $tran: M_{2} (R) \to M_{2} (R)$ by

tran [\begin{matrix} a & b \\ c & d \end{matrix}] = A^{T} = [\begin{matrix} a & c \\ b & d \end{matrix}] .

This is the transpose function on the set of 2 by 2 matrices over the real numbers. Is the transpose function an injection? Is the transpose function a surjection? Justify your conclusions.

Solution.

The transpose function is a bijection. To prove it is an injection let $[\begin{matrix} a & b \\ c & d \end{matrix}], [\begin{matrix} p & q \\ r & s \end{matrix}] \in M_{2} (R)$ and assume that

tran [\begin{matrix} a & b \\ c & d \end{matrix}] = tran [\begin{matrix} p & q \\ r & s \end{matrix}] .

Then $[\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} p & q \\ r & s \end{matrix}] .$ Therefore, $a = p,$ $b = q,$ $c = r,$ and $d = s$ and hence, $[\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} p & q \\ r & s \end{matrix}] .$ To prove that the transpose function is a surjection, let $[\begin{matrix} a & b \\ c & d \end{matrix}] \in M_{2} (R) .$ Then,

tran [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] .

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(c)

Define $F : M_{2} (R) \to R$ by

F [\begin{matrix} a & b \\ c & d \end{matrix}] = a^{2} + d^{2} - b^{2} - c^{2}

Is the function $F$ an injection? Is the function $F$ a surjection? Justify your conclusions.

Solution.

The function $F$ is not an injection. For example

F [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 and F [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] = 0 .

The function $F$ is a surjection. To prove this, let $y \in R .$ Consider three cases.

If $y = 0,$ then $F [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 = y .$
If $y > 0,$ then $\sqrt{y} \in R$ and $F [\begin{matrix} \sqrt{y} & 0 \\ 0 & 0 \end{matrix}] = {(\sqrt{y})}^{2} = y .$
If $y < 0,$ then $\sqrt{- y} \in R$ and $F [\begin{matrix} 0 & \sqrt{- y} \\ 0 & 0 \end{matrix}] = - {({\sqrt{- y}}^{2})}^{2} = y .$

Constructing and Writing Mathematical Proofs: A Guide for Mathematics Students

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Section 7.3 Practice Problems for Chapter 7

Exercises

1.

2.

(a)

(b)

(c)

(d)

3.

(a)

(b)

4.

(a)

(b)

(c)