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Section 5.4 Converse and Contrapositive

In this section, we continue our discussion of conditional statements, including the converse and contrapositive of a conditional statement. The contrapositive provides an alternative proof format for proving that a conditional is true. We also introduce biconditional (if and only if) statements and the proof format for biconditional statements.

Subsection 5.4.A The Converse and Biconditional Statements

The statement \(P \to Q\) promises that if \(P\) is true, then \(Q\) is also true. It does not say the reverse.

Definition 5.4.1 Converse and inverse.

 
  1. For statements \(P\) and \(Q\text{,}\) the converse of \(P \to Q\) is \(Q \to P\text{.}\) For example, for the statement “for all integers \(s\text{,}\) if \(s\) is even, then \(s^2\) is even”. the converse is “for all integers \(s\text{,}\) if \(s^2\) is even, then \(s\) is even”.
  2. In this context, we can translate the conditional \(P \to Q\) as “\(Q\) if \(P\)” and we can translate the converse \(Q \to P\) as “\(Q\) only if \(P\)”.
  3. For statements \(P\) and \(Q\text{,}\) the inverse of \(P \to Q\) is \(\neg P \to \neg Q\text{.}\) For example, for the same statement “for all integers \(s\text{,}\) if \(s\) is even, then \(s^2\) is even”, the inverse is “for all integers \(s\text{,}\) if \(s\) is odd, then \(s^2\) is odd”.
Let’s compare the truth table for a conditional statement to its converse.

Example 5.4.2 The converse.

 
  1. Construct a truth table for a conditional statement and its converse.
    Solution.
    We constructed the truth table for the converse in Example 5.3.7. We repeat that here alongside the conditional for comparison in Table 5.3.8. Notice that both a statement and its converse can be true or either can be true with the other false.
    Table 5.4.3. Truth table for a conditional and its converse
    \(P\) \(Q\) \(P \to Q\) \(Q \to P\)
    \(\str{T}\) \(\str{T}\) \(\str{T}\) \(\str{T}\)
    \(\str{T}\) \(\str{F}\) \(\str{F}\) \(\str{T}\)
    \(\str{F}\) \(\str{T}\) \(\str{T}\) \(\str{F}\)
    \(\str{F}\) \(\str{F}\) \(\str{T}\) \(\str{T}\)
  2. Give an example of a true conditional statement that has a false converse.
    Solution.
    Consider the conditional statement If \(k=3\text{,}\) then \(k^2 = 9\text{.}\) This statement is true because if \(k=3\text{,}\) then \(k^2=3^2=9\text{.}\) The converse is if \(k^2=9\text{,}\) then \(k=3\text{.}\) The converse is false because if \(k=-3\text{,}\) then \(k^2=(-3)^2=9\) is true but \(k=-3=3\) is false. That is, we have a broken promise Theorem 5.3.17.
  3. Give an example of a true conditional statement that has a true converse.
    Solution.
    Consider the conditional statement If \(n\) is odd, then \(n+1\) is even. The converse is If \(n+1\) is even, then \(n\) is odd. We proved that both statements are true in Section 4.1 Exercise 8 and again in Section 5.3 Exercise 25.
When both a conditional statement, \(P \to Q\text{,}\) and its converse, \(Q \to P\text{,}\) are true, we say “\(P\) if and only if \(Q\)”. We give a formal definition of the biconditional.

Definition 5.4.4 Biconditional statements.

 
  1. For statements \(P\) and \(Q\text{,}\) the biconditional statement, denoted \(P \leftrightarrow Q\text{,}\) is the following compound statement:
    \begin{equation*} (P \to Q) \land (Q \to P)\text{.} \end{equation*}
    For example, we saw in Example 5.4.2 (c) that if \(P\) is the statement “\(n\) is odd” and \(Q\) is the statement “\(n+1\) is even”, then \(P \to Q\) and \(Q \to P\) are true, so the biconditional statement \(P \leftrightarrow Q\) is also true.
  2. The biconditional \(P \leftrightarrow Q\) is often translated as \(P\) if and only if \(Q\) and may also be abbreviated \(P\) iff \(Q\text{.}\) For example, when \(P\) is the statement “\(n\) is odd” and \(Q\) is the statement “\(n+1\) is even”, then the biconditional \(P \leftrightarrow Q\) is the statement “for any integer \(n\text{,}\) \(n\) is odd if and only if \(n+1\) is even”.
Let’s construct a truth table for a biconditional statement.

Example 5.4.5 Truth table of biconditional statement.

   
  1. Construct a truth table for a biconditional statement.
    Solution.
    We construct the truth table in Table 5.4.6.
    Table 5.4.6. Truth table for the biconditional \(P \leftrightarrow Q\)
    \(P\) \(Q\) \(P \to Q\) \(Q \to P\) \(P \leftrightarrow Q\)
    \(\str{T}\) \(\str{T}\) \(\str{T}\) \(\str{T}\) \(\str{T}\)
    \(\str{T}\) \(\str{F}\) \(\str{F}\) \(\str{T}\) \(\str{F}\)
    \(\str{F}\) \(\str{T}\) \(\str{T}\) \(\str{F}\) \(\str{F}\)
    \(\str{F}\) \(\str{F}\) \(\str{T}\) \(\str{T}\) \(\str{T}\)
  2. Use this truth table to explain when the biconditional \(P \leftrightarrow Q\) is true and when it is false.
    Solution.
    The biconditional \(P \leftrightarrow Q\) is true when \(P\) and \(Q\) are both true (row 1) or \(P\) and \(Q\) are both false (row 4). That is, \(P \leftrightarrow Q\) is true when \(P\) and \(Q\) have the same truth value meaning \(P \equiv Q\text{.}\) On the other hand, the biconditional \(P \leftrightarrow Q\) is false when \(P\) is true but \(Q\) is false (row 2) or when \(P\) is false but \(Q\) is true (row 3). That is, \(P \leftrightarrow Q\) is false when \(P\) and \(Q\) have the opposite truth value meaning \(Q \equiv \neg P\text{.}\)
Let’s practice working with the converse and biconditional statements.

Activity 5.4.A The converse and biconditional statements.

 
  1. State the converse of the statement “if the graph \(G\) has a vertex of degree five, then the graph \(G\) has at least six vertices”. Is the statement true? Is the converse true?
  2. State the converse of the statement “for any integer \(k\text{,}\) if \(k >10\text{,}\) then \(k > 5\)”. Is the statement true? Is the converse true?
  3. State the converse of the statement “for any integer \(t\text{,}\) if \(t \ge 1\text{,}\) then \(t >0\)”. Is the statement true? Is the converse true? Hint: \(t\) is an integer.
  4. True or false? For any integer \(m\text{,}\) \(4 \mid m^2\) if and only if \(4 \mid m\text{.}\)
  5. True or false? For any integer \(n\text{,}\) \(n\) is even if and only if \(3n\) is even.

Subsection 5.4.B Proof Format: Biconditional

Since the biconditional is defined as
\begin{equation*} P \leftrightarrow Q \equiv (P \to Q) \land (Q \to P)\text{,} \end{equation*}
we can combine the proof formats for “and” Theorem 5.1.11 with the proof format for a conditional statement Theorem 5.3.13 to get a new proof format to prove a biconditional statement. Normally in “and” proofs we begin the two parts with the words “First” and “Second” (or “First” and “Next”). In the proof of a biconditional statement, the second part proves the converse, so we begin with the word “Conversely” instead.
As usual, when we introduce a new proof format, we provide an example for you to copy and edit.

Activity 5.4.B Biconditional proof.

 
  1. Copy the following proof. Prove for any integer \(a\) that \(a \fn{ mod } 3 = 1\) if and only if \(3 \mid (a-1)\text{.}\)

    Proof.
    Let \(a\) be an integer.
    First, assume that \(a \fn{ mod } 3 = 1\text{.}\) By definition of \(\fn{mod}\text{,}\) it follows that \(a = 3q +1\) for some integer \(q\text{.}\) Subtracting one from each side of this equation we get \(a-1 = 3q\text{.}\) By definition of divides, it follows that \(3 \mid (a-1)\text{.}\)
    Conversely, assume that \(3 \mid (a-1)\text{.}\) By definition of divides, we can write \(a-1=3k\) for some integer \(k\text{.}\) Adding one to each side of this equation we get \(a = 3k+1\text{.}\) By definition of \(\fn{mod}\text{,}\) it follows that \(a \fn{ mod } 3 = 1\text{.}\)
  2. Edit the proof to prove for any integers \(a\) and \(n\) that \(a \fn{ mod } n = 1\) if and only if \(n \mid (a-1)\text{.}\)
We can now prove the connection between divides and fractions.

Example 5.4.10 Fractions and Divisibility.

Let \(d\) and \(n\) be integers with \(d \neq 0\text{.}\) Prove that \(d \mid n\) exactly when \(\frac{n}{d}\) is an integer.
Solution.
Proof. Let \(d\) and \(n\) be integers with \(d \neq 0\text{.}\)
First, assume \(d \mid n\text{.}\) By definition of divides, \(n=dk\) for some integer \(k\text{.}\) By Theorem 4.1.14, it follows that
\begin{equation*} \frac{n}{d} = \frac{dk}{d(1)} = \frac{k}{1} = k \end{equation*}
is an integer.
Conversely, assume \(\frac{n}{d}\) is an integer. Say,
\begin{equation*} \frac{n}{d} = k = \frac{k}{1}\text{.} \end{equation*}
By Theorem 4.1.14, it follows that \(n(1) = k(d)\) or, equivalently, \(n=dk\text{.}\) By the definition of divides, \(d \mid n\text{.}\) \(\Box\)

Subsection 5.4.C The Contrapositive

While a conditional statement and its converse are not logically equivalent, there is a related conditional that is logically equivalent to a conditional.

Definition 5.4.11 Contrapositive.

For statements \(P\) and \(Q\text{,}\) the contrapositive of the conditional statement \(P \to Q\) is the conditional statement \(\neg Q \to \neg P\text{.}\) For example, consider the true statement “if \(k=3\text{,}\) then \(k^2=9\)”. The contrapositive is the statement “if \(k^2 \neq 9\text{,}\) then \(k \neq 3\)”. which is also true.
Let’s explore the logical equivalence of a conditional and its contrapositive.

Example 5.4.12 Equivalence of statement and its contrapositive.

 
  1. Construct a truth table of conditional and its contrapositive. What do you notice?
    Solution.
    The truth table is shown in Table 5.4.13. Notice that the columns for the conditional and its contrapositive are identical. That means that a conditional and its contrapositive are logically equivalent.
    Table 5.4.13. Truth table for the contrapositive
    \(P\) \(Q\) \(P \to Q\) \(\neg Q\) \(\neg P\) \(\neg Q \to \neg P\)
    \(\str{T}\) \(\str{T}\) \(\str{T}\) \(\str{F}\) \(\str{F}\) \(\str{T}\)
    \(\str{T}\) \(\str{F}\) \(\str{F}\) \(\str{T}\) \(\str{F}\) \(\str{F}\)
    \(\str{F}\) \(\str{T}\) \(\str{T}\) \(\str{F}\) \(\str{T}\) \(\str{T}\)
    \(\str{F}\) \(\str{F}\) \(\str{T}\) \(\str{T}\) \(\str{T}\) \(\str{T}\)
  2. Explain in words why the contrapositive is logically equivalent to the original conditional.
    Solution.
    Think about what makes the conditional \(P \to Q\) is true. We are promising that if \(P\) is true, then \(Q\) is true. What happens if \(\neg Q\) is true? That means \(Q\) is false and so \(P\) cannot be true (or we would break our promise) and so \(P\) must be false which means \(\neg P\) is true. That is, if \(\neg Q\text{,}\) then \(\neg P\text{,}\) which is the contrapositive.
We add the contrapositive to our list of logical equivalences.
Let’s practice working with the contrapositive.

Activity 5.4.C Contrapositive.

 
  1. Write the contrapositive of the true statement “for all integers \(n\text{,}\) if \(n>10\text{,}\) then \(n>5\)”.
  2. Write the contrapositive of the true statement “for all graphs \(G\text{,}\) if \(G\) has at most five vertices, then \(G\) has at most 10 edges”. Hint: To negate “at most” use “more than”.
  3. Write the contrapositive of the true statement “for all integers \(n\text{,}\) if \(n^2\) is odd, then \(n\) is odd”.
  4. Which statement do you think would be easier to prove: the statement from (c) or its contrapositive? Explain.

Subsection 5.4.D Proof Format: Conditional Using the Contrapositive

Since a conditional and its contrapositive are logically equivalent, we have a new way to prove \(P \to Q\text{:}\) we can prove its contrapositive \(\neg Q \to \neg P\) instead.
Let’s do an example with this new proof format.

Activity 5.4.D Contrapositive proof if \(n^2\) is odd, then \(n\) is also odd.

 
  1. Perhaps you noticed in Activity 5.4.C (d) that the contrapositive would be easier to prove. Copy the following proof. Prove for any integer \(n\text{,}\) if \(n^2\) is odd then \(n\) is odd.

    Proof.
    Let \(n\) be an integer. We will prove the contrapositive, namely that if \(n\) is even, then \(n^2\) is even.
    Assume that \(n\) is even. By the definition of even, we can write \(n=2k\) for some integer \(k\text{.}\) Then
    \begin{equation*} n^2=(2k)^2=(2k)(2k) = (2)(2)kk = 4k^2 = 2(2k^2)\text{.} \end{equation*}
    Since \(2k^2\) is an integer, it follows from the definition of even that \(n^2\) is even.
  2. Edit the proof to prove for any integer \(n\) if \(5n\) is odd, then \(n\) is odd.
When a proof by contradiction does not use the supposed hypothesis until the ending contradiction, we can often rewrite the proof using the contrapositive, avoiding the proof by contradiction.

Example 5.4.18 Rewriting a proof by contradiction using the contrapositive.

 
  1. Read the following proof by contradiction. Prove that for any integers \(a\) and \(b\) if \(ab\) is even, then \(a\) is even or \(b\) is even.

    Proof.
    Suppose not. Then for some integers \(a\) and \(b\) we would have that \(ab\) is even but neither \(a\) nor \(b\) is even.
    First, since neither \(a\) nor \(b\) is even, that means \(a\) and \(b\) are odd. By the definition of odd, we can write \(a = 2k_1+1\) and \(b = 2k_2+1\) for some integers \(k_1\) and \(k_2\text{.}\) Multiplying \(a\) and \(b\) we get
    \begin{equation*} ab=(2k_1+1)(2k_2+1)=4k_1k_2+2k_1+2k_2+1 \end{equation*}
    \begin{equation*} = 2(2k_1k_2 + k_1 + k_2) + 1 \end{equation*}
    and so \(ab\) is odd.
    On the other hand, we said \(ab\) was even. This is a contradiction.
    What are the contradictory statements \(Q\) and \(\neg Q\text{?}\)
    Solution.
    The contradictory statements are \(Q\) “\(ab\) is odd” and \(\neg Q\) “\(ab\) is even”.
  2. State the contrapositive of the statement “For any integers \(a\) and \(b\) if \(ab\) is even, then \(a\) is even or \(b\) is even.”
    Solution.
    By DeMorgan’s Laws Theorem 5.1.23, the negation of the conclusion “\(a\) is even or \(b\) is even”> is the statement “\(a\) and \(b\) are odd”. The contrapositive is “for any integers \(a\) and \(b\text{,}\) if \(a\) and \(b\) are odd, then \(ab\) is odd”. Notice that we keep the universal quantifier.
  3. Adapt the proof from (a) to prove for any integers \(a\) and \(b\) if \(ab\) is even, then either \(a\) is even or \(b\) is even using the contrapositive Theorem 5.4.16. Hint: you should not have to change much of the proof.
    Solution.
    Prove: for any integers \(a\) and \(b\) if \(ab\) is even, then either \(a\) is even or \(b\) is even.
    Proof. We will prove the contrapositive, namely for any integers \(a\) and \(b\text{,}\) if \(a\) and \(b\) are odd, then \(ab\) is odd.
    Let \(a\) and \(b\) be odd integers. By the definition of odd, we can write \(a = 2k_1+1\) and \(b = 2k_2+1\) for some integers \(k_1\) and \(k_2\text{.}\) Multiplying \(a\) and \(b\) we get
    \begin{equation*} ab=(2k_1+1)(2k_2+1)=4k_1k_2+2k_1+2k_2+1 \end{equation*}
    \begin{equation*} = 2(2k_1k_2 + k_1 + k_2) + 1 \end{equation*}
    and so \(ab\) is odd.

Subsection 5.4.E Exercises

Subsubsection Exercises for 5.4.A The Converse and Biconditional Statements

1. Practice.

Consider the statement “for any real number \(x\text{,}\) if \(x^2 \le 100\text{,}\) then \(x \le 10\)”.
  1. Is the statement true or false?
  2. State the converse.
  3. Give a counterexample to show that the converse is false.
Check.
  1. True. Note: In fact, \(-10 \le x \le 10\text{.}\)
  2. Hint: The converse should still include \(\le\text{.}\)
  3. Hint: Broken promise.

2. Practice.

Consider the statement “for all integers \(n\text{,}\) if \(6 \mid n\text{,}\) then \(3 \mid n\)”.
  1. Is the statement true or false?
  2. State the converse.
  3. Give a counterexample to show that the converse is false.

3. Practice.

Consider the statement “for any integers \(a\) and \(b\text{,}\) if \(a \fn{ mod } 5 = b \fn{ mod } 5\text{,}\) then \(a^2 \fn{ mod } 5 = b^2 \fn{ mod } 5\)”
  1. Is the statement true or false?
  2. State the converse.
  3. Give a counterexample to show that the converse is false.
Check.
  1. True
  2. Hint: The converse should also involve \(=\text{.}\)
  3. Hint: Broken promise.

4. Practice.

Use a truth table to show that the converse \(Q \to P\) and the inverse \(\neg P \to \neg Q\) are logically equivalent. Include intermediate columns for \(\neg P\) and \(\neg Q\text{.}\) This result helps explain why we rarely discuss the inverse.

5. Practice.

Decide if each biconditional statement from Activity 5.4.A is true or false. Explain briefly.
  1. For any integers \(k\text{,}\) \(k>10\) if and only if \(k>5\text{.}\)
  2. For any integer \(n\text{,}\) \(n\) is even if and only if \(3n\) is even.
  3. For any integer \(m\text{,}\) \(4 \mid m^2\) if and only if \(4 \mid m\text{.}\)
Check.
  1. Hint: One direction is true, but the other direction is false. Find a counterexample.
  2. Hint: It is true. Check both directions.
  3. Hint: One direction is true, but the other direction is false. Find a counterexample.

6. Understand.

 
  1. Construct a truth table for \(\neg(P \leftrightarrow Q)\text{.}\) Hint: Copy the truth table from Example 5.4.5 and add one additional column.
  2. The negation \(\neg(P \leftrightarrow Q)\) is logically equivalent to a simpler statement. What is it? Hint: Table 5.1.6.

7. Recap.

Do you know …
  1. What the converse of \(P \to Q\) is?
  2. How to find examples to show that \(P \to Q\) and its converse are not equivalent?
  3. How the biconditional statement \(P \leftrightarrow Q\) is defined?
  4. When a biconditional statement is true and when it is false?

Subsubsection Exercises for 5.4.B Proof Format: Biconditional

8. Practice.

Use Theorem 5.4.8 to prove that \(n\) is even if and only if \(n+3\) is odd, for any integer \(n\text{.}\)
Check.
In the converse, write \(n+3=2k+1 \) for some integer \(k\text{.}\) Solve for \(n \) and express it the form \(n=(\underline{\hspace{.5in}})+1\)

9. Practice.

Use Theorem 5.4.8 to prove that \(2n-7=5\) if and only if \(n =6\) for any integer \(n\text{.}\)
Check.
Hint: Here is an outline of the proof.
Proof. Let \(n\) be an integer.
First, assume \(2n-7=5\text{.}\) Then, \(\ldots\) and so \(n=6\text{.}\)
Next, assume \(n=6\text{.}\) Then, \(2n-7 = 2(6)-7\ldots\text{.}\)

10. Recap.

Do you know …
  1. How to format the proof of a biconditional statement?
  2. Why the second part of the proof of a biconditional statement begins with the word “Conversely”?

Subsubsection Exercises for 5.4.C The Contrapositive

11. Practice.

Write the contrapositive of each statement.
  1. For any real number \(x\text{,}\) if \(3x \le 6\text{,}\) then \(x \le 2\text{.}\)
  2. For any integer \(s\text{,}\) if \(s^2\) is even, then \(s\) is even.
  3. For all graphs \(G\text{,}\) if \(G\) has at least 5 vertices, then \(G\) must have at least 5 edges.
Check.
  1. For any real number \(x\text{,}\) if \(x > 2\text{,}\) then \(3x > 6\text{.}\)
  2. Hint: The negation of even is odd.
  3. Hint: The negation of “at least” is “fewer than”.

12. Understand.

Let \(H\) be a subgraph of a graph \(G\text{.}\) Consider the statement If \(G\) is acyclic, then \(H\) is acyclic.
  1. State the contrapositive of this statement and explain why it’s true.
  2. Explain why it follows that this statement is true.
Check.
  1. Hint: The negation of “acyclic” is “contains a cycle.”
  2. If \(H\) contains a subgraph \(C\) that is a cycle, then \(C\) is a subgraph of \(G\) and so \(G\) contains a cycle.

13. Understand.

Write the contrapositive of each statement.
  1. For all real numbers \(x\) and \(y\text{,}\) if \(x\) is rational and \(y\) is irrational, then \(x+y\) is irrational. Note: from Definition 6.1.4 rational and irrational are negations.
  2. For all real numbers \(x\) and \(y\text{,}\) if \(x+y\) is irrational, then either \(x\) is irrational or \(y\) is irrational.
  3. For all real numbers \(x\) and \(y\text{,}\) if \(x+y\) is rational, then \(x\) is rational and \(y\) is rational.
Check.
Hint: Use DeMorgan’s laws any time you need to negate a statement involving “and” or “or”.
  1. Hint: The contrapositive begins “For all real numbers \(x\) and \(y\text{,}\) if \(x+y\) is rational, then …”
  2. Hint: Your answer should include the word “and”. It should not include the word “irrational”.
  3. Hint: Your answer should not include the word “rational”.

14. Recap.

Do you know …
  1. What the contrapositive of \(P \to Q\) is?
  2. Why a conditional statement and its contrapositive are equivalent?
  3. How to use a truth table to verify that a conditional statement and its contrapositive are logically equivalent?

Subsubsection Exercises for 5.4.D Proof Format: Conditional Using the Contrapositive

15. Practice.

Use the contrapositive Theorem 5.4.16 to prove that for any integer \(s\text{,}\) if \(s^2\) is even then \(s\) is even.

16. Practice.

Use the contrapositive Theorem 5.4.16 to prove for any integer \(m\text{,}\) if \(3m\) is odd, then \(m\) is odd. (We proved the converse in Activity 5.3.B (a).)
Check.
Here is an outline of the proof.
Proof. Let \(m\) be an integer.
We will prove the contrapositive, namely \(\ldots\text{.}\)
Assume, \(\ldots\text{.}\)
Then, \(\ldots\text{.}\)
Thus, \(\ldots\text{.}\)

17. Understand.

In this exercise we prove Theorem 4.4.4 (a).
Use the contrapositive Theorem 5.4.16 to prove that for any integers \(n\text{,}\) \(p\text{,}\) and \(m\text{,}\) if \(p \nmid n\text{,}\) then \(mp \nmid n\text{.}\) Note that this statement can be rephrased as if \(p\) is not a divisor of \(n\text{,}\) then no multiple of \(p\) is a divisor of \(n\text{.}\)
Check.
Here is an outline of the proof.
Proof. Let \(n\text{,}\) \(p\text{,}\) and \(m\) be integers.
We will prove the contrapositive, namely if \(mp \mid n\text{,}\) then \(p \mid n\text{.}\)
Assume, \(mp \mid n\text{.}\) Then, by definition of divides we can write \(n= \ldots = pk\) where \(k=\ldots\) is an integer. Thus, \(p \mid n\text{.}\)

18. Understand.

In this exercise we prove Theorem 4.4.4 (b).
Prove for any integers \(a\text{,}\) \(b\text{,}\) and \(n\text{,}\) if \(ab=n\text{,}\) then \(a \le \sqrt{n}\) or \(b \le \sqrt{n}\text{.}\) Hint: Prove the contrapositive Theorem 5.4.16.
Check.
Hint: Use DeMorgan’s laws. Recall that \(\sqrt{n}\sqrt{n} = n\text{.}\)

19. Recap.

Do you know …
  1. How to prove a conditional statement by proving the contrapositive instead?
  2. When we want to prove the contrapositive instead of the original conditional statement?
  3. What we can write in a proof using the contrapositive so that the reader knows we are using the contrapositive?

20. Explore.

Use Theorem 5.4.8 to prove that \(m\) is odd if and only if \(m^3\) is odd. Hint: Use the contrapositive Theorem 5.4.16 for the converse (second part).
Check.
Hint: First, prove if \(m\) is odd, then \(m^3\) is odd. Conversely, prove the contrapositive of the converse, namely if \(m\) is even, then \(m^3\) is even. Another hint:
\begin{equation*} (2k+1)^3 = (2k+1)(2k+1)(2k+1) = (4k^2+4k+1)(2k+1)=\ldots \end{equation*}

21. Explore.

Copy the following proof by contradiction that \(\sqrt{2}\) is irrational, being sure to figure out why each statement is true.
Proof.
Suppose not. Then \(\sqrt{2}\) is rational.
By the definition of rational Definition 6.1.4, we can write
\begin{equation*} \sqrt{2} = \frac{s}{t} \end{equation*}
for some positive integers \(s\) and \(t\) with \(t \neq 0\text{.}\)
By Theorem 4.1.14, we may assume that we have canceled any common factors so that \(s\) and \(t\) are coprime.
On the other hand, multiply each side of this equation by \(t\) to get
\begin{equation*} s = \sqrt{2}t\text{.} \end{equation*}
Square each side of this equation to get
\begin{equation*} s^2=2t^2\text{.} \end{equation*}
Thus, \(s^2\) is even.
By Exercise 15, it follows that \(s\) is even.
By definition of even, we can write \(s = 2k\) for some integer \(k\text{.}\)
Substituting in we get
\begin{equation*} (2k)^2 = 2t^2\text{.} \end{equation*}
Simplifying we get \(t^2=2k^2\text{.}\) Thus, \(t^2\) is also even.
By Exercise 15, it follows that \(t\) is also even.
Since \(s\) and \(t\) are even, it follows that two is a common divisor of \(s\) and \(t\text{.}\)
Thus, \(s\) and \(t\) are not coprime.
This is a contradiction.
Thus \(\sqrt{2}\) is irrational.
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