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Section 2.3 Practice Problems for ChapterĀ 2

Exercises

1.

Use a counterexample to prove the following statement is false.

For all integers \(a\) and \(b\text{,}\) if 5 divides \(a\) or 5 divides \(b\text{,}\) then 5 divides \((5a + b)\text{.}\)

Solution.

A counterexample for this statement will be values of \(a\) and \(b\) for which 5 divides \(a\) or 5 divides \(b\text{,}\) and 5 does not divide \(5a + b\text{.}\) One counterexample for the statement is \(a = 5\) and \(b = 1\text{.}\) For these values, the hypothesis is true since 5 divides \(a\) and the conclusion is false since \(5a + b = 26\) and 5 does not divide 26.

2.

Construct a table of values for \(3m^2 + 4m + 6\) using at least six different integers for \(m\text{.}\) Make one-half of the values for \(m\) even integers and the other half odd integers. Is the following proposition true or false?

If \(m\) is an odd integer, then \(3m^2 + 4m + 6\) is an odd integer.
Justify your conclusion. This means that if the proposition is true, then you should write a proof of the proposition. If the proposition is false, you need to provide an example of an odd integer for which \(3m^2 + 4m + 6\) is an even integer.

Solution.

All examples should indicate the proposition is true. Following is a proof.

Proof.

We assume that \(m\) is an odd integer and will prove that \(( 3m^2 + 4m + 6 )\) . Since \(m\) is an odd integer, there exists an integer \(k\) such that \(m = 2k + 1\text{.}\) Substituting this into the expression \(( 3m^2 + 4m + 6 )\) and using algebra, we obtain

\begin{align*} 3m^2 + 4m + 6 \amp = 3 (2k + 1)^2 + 4 (2k+1) + 6\\ \amp = (12k^2 + 12k + 3 ) + (8k + 4) + 6\\ \amp = 12k^2 + 20k + 13\\ \amp = 12k^2 + 20k + 12 + 1\\ \amp = 2 (6k^2 + 10k + 6) + 1 \end{align*}

By the closure properties of the integers, \((6k^2 + 10k + 6)\) is an integer, and hence, the last equation shows that \(( 3m^2 + 4m + 6 )\) is an odd integer.

This proves that if \(m\) is an odd integer, then \(( 3m^2 + 4m + 6 )\) is an odd integer.

3.

The Pythagorean Theorem for right triangles states that if \(a\) and \(b\) are the lengths of the legs of a right triangle and \(c\) is the length of the hypotenuse, then \(a^2 + b^2 = c^2\) . For example, if \(a = 5\) and \(b = 12\) are the lengths of the two sides of a right triangle and if \(c\) is the length of the hypotenuse, then the \(c^2 = 5^2 + 12^2\) and so \(c^2 = 169\text{.}\) Since \(c\) is a length and must be positive, we conclude that \(c = 13\text{.}\)

Construct and provide a well-written proof for the following proposition.

Proposition. If \(m\) is a real number and \(m\text{,}\) \(m + 1\text{,}\) and \(m + 2\) are the lengths of the three sides of a right triangle, then \(m = 3\text{.}\)

Solution.
Proof.

We let \(m\) be a real number and assume that \(m\text{,}\) \(m + 1\text{,}\) and \(m + 2\) are the lengths of the three sides of a right triangle. We will use the Pythagorean Theorem to prove that \(m = 3\text{.}\) Since the hypotenuse is the longest of the three sides, the Pythagorean Theorem implies that \(m^2 + (m + 1)^2 = (m+ 2)^2\text{.}\) We will now use algebra to rewrite both sides of this equation as follows

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

The last equation is a quadratic equation. To solve for \(m\text{,}\) we rewrite the equation in standard form and then factor the left side. This gives

\begin{align*} m^2 - 2m - 3 \amp = 0\\ (m-3)(m + 1) \amp = 0 \end{align*}

The two solutions of this equation are \(m = 3\) and \(m = 1\text{.}\) However, since m is the length of a side of a right triangle, \(m\) must be positive and we conclude that \(m = 3\text{.}\) This proves that if \(m\text{,}\) \(m + 1\text{,}\) and \(m + 2\) are the lengths of the three sides of a right triangle, then \(m =3\text{.}\)

4.

Let \(n\) be a natural number and let \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) be integers. Prove each of the following.

(a)

If \(\modulo{a}{b}{n}\) and \(\modulo{c}{d}{n}\text{,}\) then \(\modulo{(a + c)}{(b + d)}{n}\text{.}\)

Solution.

For both parts, we assume that n is a natural number and \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are integers and that \(\modulo{a}{b}{n}\) and \(\modulo{c}{d}{n}\text{.}\) We can then conclude that \(n\) divides \(a - b\) and \(n\) divides \(c - d\text{.}\) So there exist integers \(k\) and \(m\) such that

\begin{align*} a - b \amp = kn \amp \amp \text{and} \amp c - d \amp = mn\\ a \amp = b+ kn \amp \amp \text{and} \amp c \amp = d+ mn \end{align*}

We then see that

\begin{align*} ( a + c ) - ( b + d ) \amp = (a - b) + (c - d)\\ \amp = kn + mn\\ \amp = (k + m)n \end{align*}

Since the integers are closed under addition, \(k + m\) is an integer and the last equation proves that \(\modulo{a + c}{b + d}{n}\text{.}\)

(b)

If \(\modulo{a}{b}{n}\) and \(\modulo{c}{d}{n}\text{,}\) then \(\modulo{ac}{bd}{n}\text{.}\)

Solution.

For both parts, we assume that n is a natural number and \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are integers and that \(\modulo{a}{b}{n}\) and \(\modulo{c}{d}{n}\text{.}\) We can then conclude that \(n\) divides \(a - b\) and \(n\) divides \(c - d\text{.}\) So there exist integers \(k\) and \(m\) such that

\begin{align*} a - b \amp = kn \amp \amp \text{and} \amp c - d \amp = mn\\ a \amp = b+ kn \amp \amp \text{and} \amp c \amp = d+ mn \end{align*}

We then see that

\begin{align*} ac - bd \amp = (b + kn)(d + mn) - bd\\ \amp = ( bd + bmn + knd + kmn^2) - bd\\ \amp = (bm + kd + kmn)n \end{align*}

Since the integers are closed under addition and multiplication, \((bm + kd + kmn)\) is an integer and the last equation proves that \(\modulo{ac}{bd}{n}\text{.}\)

5.

One way to prove that two sets are equal is to prove that each one is a subset of the other one. Consider the following proposition:

Proposition. Let \(A\) and \(B\) be subsets of some universal set. Then \(A - (A - B) = A \cap B\text{.}\)
Prove this proposition is true or give a counterexample to prove it is false.

Solution.

The proposition is true. Use the choose-an-element method to prove that each set is a subset of the other set.

Proof.

Let \(A\) and \(B\) be subsets of some universal set. We will prove that \(A - (A - B) = A \cap B\) by proving that \(A - (A - B) \subseteq A \cap B\) and that \(A \cap B \subseteq A - (A - B)\text{.}\)

First, let \(x \in A - (A - B)\text{.}\) This means that

\begin{equation*} x \in A \text{ and } x \not \in (A - B)\text{.} \end{equation*}

We know that an element is in \((A - B)\) if and only if it is in \(A\) and not in \(B\text{.}\) Since \(x \not \in (A - B)\text{,}\) we conclude that \(x \not \in A\)\(x \in B\text{.}\) However, we also know that \(x \in A\) and so we conclude that \(x \in B\text{.}\) This proves that

\begin{equation*} x \in A \text{ and } x \in B \end{equation*}

This means that \(x \in A \cap B\text{,}\) and hence we have proved that \(A - (A - B) \subseteq A \cap B\text{.}\)

Now choose \(y \in A \cap B\text{.}\) This means that

\begin{equation*} y \in A \text{ and } y \in B\text{.} \end{equation*}

We note that \(y \in (A - B)\) if and only if \(y \in A\) and \(y \not \in B\) and hence, \(y \not \in (A - B)\) if and only if \(y \not \in A\) or \(y \in B\text{.}\) Since we have proved that \(y \in B\text{,}\) we can conclude that \(y \not \in (A - B)\text{,}\) and hence, we have established that \(y \in A\) and \(y \not \in (A - B)\text{.}\) So, \(y \in A - (A - B)\text{,}\) and this proves that if \(y \in A \cap B\text{,}\) then \(y \in A - (A - B)\) and hence, \(A \cap B \subseteq A - (A - B)\text{.}\)

Since we have proved that \(A - (A - B) \subseteq A \cap B\) and \(A \cap B \subseteq A - (A - B)\text{,}\) we conclude that \(A - (A - B) = A \cap B\text{.}\)