Practice Problems for Chapter 4

Section 4.4 Practice Problems for Chapter 4

Exercises

1.

(a)

Determine at least five different integers that are congruent to 2 modulo 4. Are any of these integers congruent to 3 modulo 6?

Solution.

Some integers that are congruent to 2 modulo 4 are $- 6;$ $- 2;$ 2; 6; 10. None of these integers are congruent to 3 modulo 6. For example, $10 ≢ 3 (\mod 6)$ since $10 - 3 = 7$ and 6 does not divide 7.

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(b)

Is the following proposition true or false? Justify your conclusion with a counterexample (if it is false) or a proof (if it is true).

Propostion. For each integer $n,$ if $n \equiv 2 (\mod 4),$ then $n ≢ 3 (\mod 6) .$

Solution.

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Proof.

We will use a proof by contradiction. Let $n \in Z$ and assume that $n \equiv 2 (\mod 4)$ and that $n \equiv 3 (\mod 6) .$ Since $n \equiv 2 (\mod 4),$ we know that 4 divides $n - 2 .$ Hence, there exists an integer $k$ such that

\begin{matrix} (3) & n - 2 = 4 k \end{matrix}

We can also use the assumption that $n \equiv 3 (\mod 6)$ to conclude that 6 divides $n - 3$ and that there exists an integer $m$ such that

\begin{matrix} (4) & n - 3 = 6 m \end{matrix}

If we now solve equations (3) and (4) for $n$ and set the two expressions equal to each other, we obtain

4 k + 2 = 6 m + 3

However, this equation can be rewritten as

2 (2 k + 1) = 2 (3 m + 1) + 1 .

Since $2 k + 1$ is an integer and $3 m + 1$ is an integer, this last equation is a contradiction since the left side is an even integer and the right side is an odd integer. Hence, we have proven that if $n \equiv 2 (\mod 4),$ then $n ≢ 3 (\mod 6) .$

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2.

For the following, it may be useful to use the facts that the set of rational numbers $Q$ is closed under addition, subtraction, multiplication, and division by nonzero rational numbers.

Prove the following proposition:

Proposition. For all real numbers $x$ and $y,$ if $x$ is rational and $x \neq 0$ and $y$ is irrational, then $x + y$ is irrational.

Solution.

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Proof.

We will use a proof by contradiction. So we assume that there exist real numbers $x$ and $y$ such that $x$ is rational, $y$ is irrational, and $x + y$ is rational. Since $x$ is rational, we know that $- x$ is rational. Since the rational numbers are closed under addition, we know that $(- x) + (x + y)$ is rational, and we see that

\begin{aligned} (- x) + (x + y) & = ((- x) + x) + y \\ = 0 + y \\ = y \end{aligned}

However, this shows that $y$ must be a rational number, but we have also assumed that $y$ is irrational. Since a real number cannot be both rational and irrational, this is a contradiction. We have therefore proved that for all real numbers $x$ and $y,$ if $x$ is rational and $y$ is irrational, then $x + y$ is irrational.

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3.

Is the base 2 logarithm of 3, $\log_{2} 3,$ a rational or irrational number? Justify your conclusion.

Solution.

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Proof.

We will use a proof by contradiction to prove that $\log_{2} 3$ is an irrational number.

So we assume that $\log_{2} 3$ is a rational number. So, if $\log_{2} 3 = a,$ then $2^{a} = 3 .$ This means that $a$ is a positive rational number, and hence, there exist natural numbers $m$ and $n$ such that $2^{\frac{m}{n}} = 3 .$ Hence,

{(2^{\frac{m}{n}})}^{n} = 3^{n},

From this, we conclude that $2^{m} = 3^{n} .$ However, $2^{m}$ is an even integer and $3^{n}$ is an odd integer. This is a contradiction, and so we have proved that $\log_{2} 3$ is an irrational number.

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4.

Is the real number $\sqrt{2} + \sqrt{3}$ a rational or irrational number? Justify your conclusion.

Solution.

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Proof.

We will use a proof by contradiction to prove that $\sqrt{2} + \sqrt{3}$ is an irrational number. So we assume that $\sqrt{2} + \sqrt{3}$ is a rational number and so we can write $\sqrt{2} + \sqrt{3} = r,$ where $r$ is a rational number and $r \neq 0 .$ We now rewrite this equation and then square both sides of the resulting equation to obtain

\begin{aligned} \sqrt{3} & = r - \sqrt{2} \\ 3 & = r^{2} - 2 r \sqrt{2} + 2 \end{aligned}

We continue and rewrite this equation to isolate $\sqrt{2}$ on one side of the equation.

\begin{aligned} 2 r \sqrt{2} & = r^{2} - 1 \\ \sqrt{2} & = \frac{r^{2} - 1}{2 r} \end{aligned}

Since $r \neq 0,$ $2 r \neq 0,$ and since the rational numbers are closed under division by a nonzero rational number, the last equation shows that $\sqrt{2}$ is a rational number. This is a contradiction since it is known that $\sqrt{2}$ is irrational. This proves that $\sqrt{2} + \sqrt{3}$ is an irrational number.

Constructing and Writing Mathematical Proofs: A Guide for Mathematics Students

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Section 4.4 Practice Problems for Chapter 4

Exercises

1.

(a)

(b)

Proof.

2.

Proof.

3.

Proof.

4.

Proof.