Swapping and Negating Change the Claim
Begin with a true conditional statement:
If an integer is divisible by \(4\), then it is even.
Let
\[ p:\text{ the integer is divisible by }4 \]and
\[ q:\text{ the integer is even}. \]The statement has the form
\[ p\rightarrow q. \]Now compare:
- If an integer is even, then it is divisible by \(4\).
- If an integer is not divisible by \(4\), then it is not even.
- If an integer is not even, then it is not divisible by \(4\).
These sentences sound similar, but they make different claims. The first two are false; the third is true because it is logically equivalent to the original statement.
\[ \boxed{p\rightarrow q\quad\equiv\quad\neg q\rightarrow\neg p} \]Reviewing the Conditional Statement
A conditional statement has the form
\[ \text{if }p,\text{ then }q, \]or symbolically,
\[ p\rightarrow q. \]| Part | Name | Role |
|---|---|---|
| \(p\) | Hypothesis or antecedent | The condition introduced by “if” |
| \(q\) | Conclusion or consequent | What must follow when the hypothesis is true |
The conditional \(p\rightarrow q\) makes one promise:
Whenever \(p\) is true, \(q\) must also be true.
It does not claim that \(p\) is the only way for \(q\) to occur. It also does not claim that \(q\) forces \(p\).
The Four Related Forms
| Name | Symbolic Form | Construction |
|---|---|---|
| Conditional | \(p\rightarrow q\) | Use the original order and signs |
| Converse | \(q\rightarrow p\) | Swap \(p\) and \(q\) |
| Inverse | \(\neg p\rightarrow\neg q\) | Negate both parts without swapping |
| Contrapositive | \(\neg q\rightarrow\neg p\) | Swap and negate both parts |
A compact memory aid is:
\[ \begin{aligned} \text{converse}&:\text{ swap},\\ \text{inverse}&:\text{ negate},\\ \text{contrapositive}&:\text{ swap and negate}. \end{aligned} \]In this lesson, inverse means the logical inverse of a conditional statement. In other settings, “inverse” may refer to an inverse operation, number, relation, matrix, or function. Context distinguishes these meanings.
A Reliable Formation Method
- Rewrite the statement in if-then form. Make the hypothesis and conclusion explicit.
- Label the parts. Assign \(p\) to the hypothesis and \(q\) to the conclusion.
- Write the symbolic template. Use \(q\rightarrow p\), \(\neg p\rightarrow\neg q\), or \(\neg q\rightarrow\neg p\).
- Substitute complete statements. Move or negate the whole proposition, not isolated words.
- Rewrite naturally. Remove awkward double negatives without changing the logic.
- Test the truth value separately. Forming a related statement does not establish that it is true.
Suppose the original statement is:
If \(x>5\), then \(x>2\).
Then
\[ p:x>5, \qquad q:x>2. \]The negations are
\[ \neg p:x\le5, \qquad \neg q:x\le2. \]Notice that the negation of \(x>5\) is \(x\le5\), not merely \(x<5\). Equality must be included.
Truth-Table Relationships
A conditional is false only when its hypothesis is true and its conclusion is false.
| \(p\) | \(q\) | \(p\rightarrow q\) | \(q\rightarrow p\) | \(\neg p\rightarrow\neg q\) | \(\neg q\rightarrow\neg p\) |
|---|---|---|---|---|---|
| True | True | True | True | True | True |
| True | False | False | True | True | False |
| False | True | True | False | False | True |
| False | False | True | True | True | True |
Compare the columns:
- \(p\rightarrow q\) and \(\neg q\rightarrow\neg p\) match in every row.
- \(q\rightarrow p\) and \(\neg p\rightarrow\neg q\) match in every row.
Matching truth values in every possible case establish logical equivalence.
Which Forms Are Equivalent?
\[ \boxed{p\rightarrow q\equiv\neg q\rightarrow\neg p} \]The conditional and its contrapositive are logically equivalent.
\[ \boxed{q\rightarrow p\equiv\neg p\rightarrow\neg q} \]The converse and inverse are logically equivalent to each other.
In general,
\[ p\rightarrow q\not\equiv q\rightarrow p \]and
\[ p\rightarrow q\not\equiv\neg p\rightarrow\neg q. \]Therefore:
- if the original conditional is true, its contrapositive is true;
- if the converse is false, the inverse is false;
- the truth of the original alone does not determine the truth of its converse; and
- the truth of the original alone does not determine the truth of its inverse.
Recognizing Conditionals in Natural Language
Conditional relationships do not always contain the phrase “if-then.”
| Natural-Language Form | Conditional Reading |
|---|---|
| All squares are rectangles. | If a figure is a square, then it is a rectangle. |
| Being divisible by \(6\) guarantees being divisible by \(3\). | If an integer is divisible by \(6\), then it is divisible by \(3\). |
| A number is even only if it is an integer. | If a number is even, then it is an integer. |
| Whenever it rains, the sensor activates. | If it rains, then the sensor activates. |
| Passing the inspection requires every safety test to pass. | If the item passes inspection, then every safety test passes. |
Words such as only if and requires often introduce a necessary condition, so their direction deserves special attention.
Necessary and Sufficient Conditions
For the conditional
\[ p\rightarrow q, \]- \(p\) is a sufficient condition for \(q\), and
- \(q\) is a necessary condition for \(p\).
Return to:
If an integer is divisible by \(4\), then it is even.
- Divisibility by \(4\) is sufficient for evenness.
- Evenness is necessary for divisibility by \(4\).
- Evenness is not sufficient for divisibility by \(4\), because \(6\) is even but not divisible by \(4\).
The converse \(q\rightarrow p\) asks whether the necessary condition is also sufficient. Sometimes it is; often it is not.
Testing with Counterexamples
A universal conditional is disproved by one case in which the hypothesis is true and the conclusion is false.
Consider the converse:
If an integer is even, then it is divisible by \(4\).
The integer \(6\) gives
\[ \text{even: true}, \qquad \text{divisible by }4:\text{ false}. \]Therefore, \(6\) is a counterexample and the converse is false.
The same value tests the inverse:
If an integer is not divisible by \(4\), then it is not even.
For \(6\), the hypothesis is true and the conclusion is false, so the inverse is also false—as expected because the converse and inverse are equivalent.
A useful counterexample must satisfy the hypothesis. A case where the hypothesis is false cannot disprove the conditional.
Valid and Invalid Inference
A conditional can support two standard valid argument forms.
Modus Ponens
\[ \begin{aligned} p&\rightarrow q,\\ p&\text{ is true},\\ \therefore\ q&\text{ is true}. \end{aligned} \]If divisibility by \(4\) implies evenness and \(20\) is divisible by \(4\), then \(20\) is even.
Modus Tollens
\[ \begin{aligned} p&\rightarrow q,\\ q&\text{ is false},\\ \therefore\ p&\text{ is false}. \end{aligned} \]If divisibility by \(4\) implies evenness and \(15\) is not even, then \(15\) is not divisible by \(4\). This inference uses the contrapositive.
Affirming the Consequent
\[ \begin{aligned} p&\rightarrow q,\\ q&\text{ is true},\\ \therefore\ p&\text{ is true}. \end{aligned} \]This is invalid because it assumes the converse. Knowing that \(6\) is even does not show that \(6\) is divisible by \(4\).
Denying the Antecedent
\[ \begin{aligned} p&\rightarrow q,\\ p&\text{ is false},\\ \therefore\ q&\text{ is false}. \end{aligned} \]This is invalid because it assumes the inverse. An integer can fail to be divisible by \(4\) and still be even.
Proof by Contrapositive
Because a conditional and its contrapositive are equivalent, proving the contrapositive proves the original statement.
Consider the claim:
If \(n^2\) is odd, then \(n\) is odd.
A direct proof begins with \(n^2\) odd. The contrapositive begins with the simpler condition:
If \(n\) is not odd, then \(n^2\) is not odd.
For an integer, “not odd” means even. Let
\[ n=2k \]for some integer \(k\). Then
\[ n^2=(2k)^2=4k^2=2(2k^2). \]Since \(2k^2\) is an integer, \(n^2\) is even and therefore not odd. The contrapositive is true, so the original conditional is true.
Proof by contrapositive is especially useful when negating the conclusion produces a more workable starting point.
When the Converse Is Also True
If both a conditional and its converse are true, then the relationship is biconditional:
\[ p\leftrightarrow q. \]This is read “\(p\) if and only if \(q\).”
For an integer \(n\):
\(n\) is even if and only if \(n^2\) is even.
This combines:
\[ n\text{ even}\rightarrow n^2\text{ even} \]and
\[ n^2\text{ even}\rightarrow n\text{ even}. \]A biconditional should not be written merely because the original conditional is true. The converse requires its own proof or established justification.
Worked Example: Divisibility
Start with:
If an integer is divisible by \(10\), then it is divisible by \(5\).
Let \(p\) mean “the integer is divisible by \(10\)” and \(q\) mean “the integer is divisible by \(5\).”
| Form | Statement | Truth |
|---|---|---|
| Conditional | If an integer is divisible by \(10\), then it is divisible by \(5\). | True |
| Converse | If an integer is divisible by \(5\), then it is divisible by \(10\). | False; \(15\) is a counterexample |
| Inverse | If an integer is not divisible by \(10\), then it is not divisible by \(5\). | False; \(15\) is a counterexample |
| Contrapositive | If an integer is not divisible by \(5\), then it is not divisible by \(10\). | True |
The matching truth values reflect the two equivalence pairs.
Worked Example: Geometry
Start with:
If a quadrilateral is a square, then it is a rectangle.
| Form | Statement |
|---|---|
| Conditional | If a quadrilateral is a square, then it is a rectangle. |
| Converse | If a quadrilateral is a rectangle, then it is a square. |
| Inverse | If a quadrilateral is not a square, then it is not a rectangle. |
| Contrapositive | If a quadrilateral is not a rectangle, then it is not a square. |
The conditional and contrapositive are true. The converse and inverse are false because a nonsquare rectangle is a counterexample.
This example also exposes a classification relationship:
\[ \{\text{squares}\}\subseteq\{\text{rectangles}\}. \]Membership in the smaller class guarantees membership in the larger class, but membership in the larger class does not guarantee membership in the smaller one.
Interpreting Practical Rules
Suppose a workplace policy states:
If an employee completes the required safety training, then the employee receives equipment access.
The rule has the form
\[ \text{training completed}\rightarrow\text{access received}. \]The contrapositive is:
If the employee does not receive equipment access, then the employee has not completed the required safety training.
That conclusion follows only if the original policy is accurate and has no unstated exceptions.
The converse—“if the employee receives access, then the employee completed training”—does not automatically follow. Access might also be granted by a temporary authorization, prior certification, or administrative error.
In practical settings, formal logic is only as reliable as the rule's definitions, assumptions, and completeness. Before applying a contrapositive to policy or data, confirm that the original conditional genuinely holds.
Common Logic Errors
Swapping Without Renaming the Statement
Swapping the hypothesis and conclusion produces the converse, not an equivalent restatement in general.
Negating Only One Part
The inverse negates both \(p\) and \(q\). The contrapositive swaps and negates both.
Negating an Inequality Incorrectly
The negation of \(x>a\) is \(x\le a\); the equality boundary must be included.
Assuming the Converse Is True
Truth does not generally survive reversal. Test the converse separately.
Confusing the Inverse with the Contrapositive
The inverse keeps the original order; the contrapositive reverses the order.
Using a Counterexample That Fails the Hypothesis
To disprove \(p\rightarrow q\), find a case where \(p\) is true and \(q\) is false.
Affirming the Consequent
From \(p\rightarrow q\) and \(q\), we cannot conclude \(p\) unless the converse is also known.
Denying the Antecedent
From \(p\rightarrow q\) and \(\neg p\), we cannot conclude \(\neg q\) unless the inverse is also known.
Practice
- For “If a number is divisible by \(8\), then it is even,” write the converse, inverse, and contrapositive.
- Determine whether the converse in Question 1 is true. Give a counterexample if it is false.
- Negate the proposition \(x\ge7\).
- For “All equilateral triangles are isosceles,” write the statement in if-then form and write its contrapositive.
- Explain why the inverse and converse of one conditional always have the same truth values.
- Identify the error: “If a figure is a square, then it has four sides. This figure has four sides, so it is a square.”
- Identify the valid inference: If \(n\) is divisible by \(6\), then \(n\) is divisible by \(3\). The integer \(14\) is not divisible by \(3\). What follows?
- Prove by contrapositive: If \(n^2\) is even, then \(n\) is even.
- Write a biconditional joining “an integer is divisible by \(2\)” and “the integer is even.”
- A policy says, “Applicants with a valid license may operate the vehicle.” Explain why “Anyone operating the vehicle has a valid license” may require an additional assumption.
Selected Answers
- Converse: If a number is even, then it is divisible by \(8\). Inverse: If a number is not divisible by \(8\), then it is not even. Contrapositive: If a number is not even, then it is not divisible by \(8\).
- False; \(6\) is even but not divisible by \(8\).
- \(x<7\).
- If a triangle is equilateral, then it is isosceles. Contrapositive: If a triangle is not isosceles, then it is not equilateral.
- The inverse \(\neg p\rightarrow\neg q\) is the contrapositive of the converse \(q\rightarrow p\), so they are logically equivalent.
- The argument affirms the consequent and assumes the converse. Many four-sided figures are not squares.
- \(14\) is not divisible by \(6\), by modus tollens.
- Contrapositive: if \(n\) is odd, then \(n^2\) is odd. Write \(n=2k+1\); then \(n^2=2(2k^2+2k)+1\), which is odd.
- An integer is divisible by \(2\) if and only if it is even.
- The rule explicitly gives license \(\rightarrow\) permission. The reverse requires a separate rule that operation is permitted only for licensed applicants and that no exceptions or violations occur.
Summary
- A conditional has the form \(p\rightarrow q\), with hypothesis \(p\) and conclusion \(q\).
- The converse swaps the two parts: \(q\rightarrow p\).
- The inverse negates both parts without swapping: \(\neg p\rightarrow\neg q\).
- The contrapositive swaps and negates both parts: \(\neg q\rightarrow\neg p\).
- The conditional is logically equivalent to its contrapositive.
- The converse is logically equivalent to the inverse.
- The truth of a conditional does not automatically establish the truth of its converse or inverse.
- A counterexample to \(p\rightarrow q\) must make \(p\) true and \(q\) false.
- Modus ponens and modus tollens are valid; affirming the consequent and denying the antecedent are invalid.
- If both a conditional and its converse are true, they form a biconditional.