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Section 4.4 Practice Problems for ChapterĀ 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 \(-6\text{;}\) \(-2\text{;}\) 2; 6; 10. None of these integers are congruent to 3 modulo 6. For example, \(\notmodulo{10}{3}{6}\) since \(10 - 3 = 7\) and 6 does not divide 7.

(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 \(n\text{,}\) if \(\modulo{n}{2}{4}\text{,}\) then \(\notmodulo{n}{3}{6}\text{.}\)

Solution.
Proof.

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

\begin{equation} n - 2 = 4k\tag{3} \end{equation}

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

\begin{equation} n - 3 = 6m\tag{4} \end{equation}

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

\begin{equation*} 4k + 2 = 6m + 3 \end{equation*}

However, this equation can be rewritten as

\begin{equation*} 2 (2k + 1) = 2(3m + 1) + 1\text{.} \end{equation*}

Since \(2k + 1\) is an integer and \(3m + 1\) 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 \(\modulo{n}{2}{4}\text{,}\) then \(\notmodulo{n}{3}{6}\text{.}\)

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 \(y\text{,}\) if \(x\) is rational and \(x \ne 0\) and \(y\) is irrational, then \(x + y\) is irrational.

Solution.
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{align*} (-x) + (x + y) \amp = ((-x) + x) + y\\ \amp = 0 + y\\ \amp = y \end{align*}

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 \(y\text{,}\) if \(x\) is rational and \(y\) is irrational, then \(x + y\) is irrational.

3.

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

Solution.
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 \(\log_{2} 3 = a\text{,}\) then \(2^a = 3\text{.}\) This means that \(a\) is a positive rational number, and hence, there exist natural numbers \(m\) and \(n\) such that \(2^{\frac{m}{n}} = 3\text{.}\) Hence,

\begin{equation*} \left( 2^{\frac{m}{n}} \right)^n = 3^n\text{,} \end{equation*}

From this, we conclude that \(2^m = 3^n \text{.}\) 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.

4.

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

Solution.
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 \(\sqrt{2} + \sqrt{3} = r\text{,}\) where \(r\) is a rational number and \(r \ne 0\text{.}\) We now rewrite this equation and then square both sides of the resulting equation to obtain

\begin{align*} \sqrt{3} \amp = r - \sqrt{2}\\ 3 \amp = r^2 - 2r \sqrt{2} + 2 \end{align*}

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

\begin{align*} 2r \sqrt{2} \amp = r^2 - 1\\ \sqrt{2} \amp = \frac{r^2 - 1}{2r} \end{align*}

Since \(r \ne 0\text{,}\) \(2r \ne 0\text{,}\) 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.