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Section 3.4 Practice Problems for ChapterĀ 3

Exercises

1.

Is the following proposition true or false?

For all integers \(a\) and \(b\text{,}\) if \(ab\) is even, then \(a\) is even or \(b\) is even.
Justify your conclusion by writing a proof if the proposition is true or by providing a counterexample if it is false.

Solution.
Proof.

We will prove that for all integers \(a\) and \(b\text{,}\) if \(ab\) is even, then \(a\) is even or \(b\) is even by proving its contrapositive, which is:

For all integers \(a\) and \(b\text{,}\) if \(a\) is odd and \(b\) is odd, then \(ab\) is odd.
So we assume that both \(a\) and \(b\) are odd integers and will prove that \(ab\) is an odd integer. Since \(a\) and \(b\) are odd, there exist integers \(k\) and \(m\) such that \(a = 2k + 1\) and \(b = 2m + 1\text{.}\) Using substitution and algebra, we then see that

\begin{align*} ab \amp = (2k + 1)(2m +1)\\ \amp = 4km + 2k + 2m + 1\\ \amp = 2 (2km + k + m) + 1 \end{align*}

Since \(k\) and \(m\) are integers, the closure properties of the integers allow us to conclude that \((2km + k + m)\) is an integer. This means that \(ab\) has been written as two times an integer plus 1, and hence \(ab\) is an odd integer. This proves that for all integers \(a\) and \(b\text{,}\) if \(a\) is odd and \(b\) is odd, then \(ab\) is odd, which is the contrapositive of the proposition. So we have proved that For all integers \(a\) and \(b\text{,}\) if \(ab\) is even, then \(a\) is even or \(b\) is even.

2.

Are the following statements true or false? Justify your conclusion.

(a)

For each \(a \in \Z\text{,}\) if \(\modulo{a}{2}{5}\text{,}\) then \(\modulo{a^2}{4}{5}\text{.}\)

Solution.
Proof.

We assume that \(a\) is an integer and that \(\modulo{a}{2}{5}\) and will prove that \(\modulo{a^2}{4}{5}\text{.}\) Since \(\modulo{a}{2}{5}\text{,}\) then there exists an integer \(k\) such that \(a = 2 = 5k\) and so \(a = 2 + 5k\text{.}\) Then

\begin{align*} a^2 - 4 \amp = (2 + 5k)^2 - 4\\ \amp = 20k + 25k^2\\ \amp = 5(4k + 5k^2) \end{align*}

Since the integers are closed under addition and multiplication, \((4k + 5k^2 )\) is an integer, and so the last equation proves that 5 divides \(a^2 - 4\text{.}\) Hence, \(\modulo{a^2}{4}{5}\text{,}\) and this proves that for each integer \(a\text{,}\) if \(\modulo{a}{2}{5}\text{,}\) then \(\modulo{a^2}{4}{5}\text{.}\)

(b)

For each \(a \in \Z\text{,}\) if \(\modulo{a^2}{4}{5}\text{,}\) then \(\modulo{a}{2}{5}\text{.}\)

Solution.

This statement is false. A counterexample is \(a = 3\) since \(\modulo{3^2}{4}{5}\) and \(\notmodulo{3}{2}{5}\text{.}\)

(c)

For each \(a \in \Z\text{,}\) \(\modulo{a}{2}{5}\) if and only if \(\modulo{a^2}{4}{5}\text{.}\)

Solution.

This statement is false since the statement in TaskĀ 3.4.2.b is false.

3.

A real number x is defined to be a rational number provided

there exist integers \(m\) and \(n\) with \(n \ne 0\) such that \(x = \frac{m}{n}\text{.}\)
A real number that is not a rational number is called an irrational number.

It is known that if \(x\) is a positive rational number, then there exist positive integers \(m\) and \(n\) with \(n Ā¤\ne 0\) such that \(x = \frac{m}{n}\text{.}\)

Is the following proposition true or false? Explain.

Proposition. For each positive real number \(x\text{,}\) if \(x\) is irrational, then \(\sqrt{x}\) is irrational.

Solution.
Proof.

We will prove the contrapositive of this statement, which is

For each positive real number \(x\text{,}\) if \(\sqrt{x}\) is rational, then \(x\) is rational.
So we assume that \(x\) is a rational number and that \(\sqrt{x}\) is rational, and will prove that \(x\) is rational. Since \(\sqrt{x}\) is rational, there exist positive integers \(m\) and \(n\) such that \(\sqrt{x} = \frac{m}{n}\text{,}\) then \(x = \frac{m^2}{n^2}\text{.}\) Since \(m\) and \(n\) are positive integers, \(m^2\) and \(n^2\) are positive integers and we can conclude that \(x\) is a rational number. This proves the contrapositive of the statement and so we have proved that for each positive real number \(x\text{,}\) if \(x\) is irrational, then \(\sqrt{x}\) is irrational.