A conditional and its contrapositive
An if-then statement can be turned around in three ways. Which of them is guaranteed to say exactly what the original said?
Where you are 3 in Logical equivalence
Why this exists
“If x > 4, then x^2 > 16.” That is true for every real number. Now turn it around: “if x^2 > 16, then x > 4.” That is false, since x = -5 has (-5)^2 = 25 > 16. Turning a true statement around can break it.
Yet there is one way of turning it around that never breaks anything: “if x^2 \le 16, then x \le 4.” Swap the two parts and negate both, and the new statement is true exactly when the old one is. This is the contrapositive, and it matters because it is often far easier to check or prove than the statement you started with. Knowing which turnaround is safe, and why, lets you make that trade without changing the question.
Definition
Start from a conditional P \to Q, “if P, then Q”, where P is the hypothesis and Q the conclusion. Three other conditionals are built from the same two parts:
| Name | Statement | In words |
|---|---|---|
| original | P \to Q | if P, then Q |
| converse | Q \to P | swap the parts |
| inverse | \neg P \to \neg Q | negate both parts |
| contrapositive | \neg Q \to \neg P | swap the parts and negate both |
Two statements are logically equivalent, written \equiv, when their truth tables agree on every row: whatever P and Q are, the two statements are true together and false together.
The claim of this lesson is that a conditional is logically equivalent to its contrapositive, and the converse is logically equivalent to the inverse, but the original is not equivalent to the converse.
The lesson on conditionals, converse, inverse and contrapositive introduces these four forms and proves the same two equivalences with truth tables. This lesson picks the contrapositive out on its own and puts it to work: deciding and proving claims through it, and counting how many of the four can be true.
Intuition
By the truth table for the conditional, an if-then statement is false in exactly one situation: its hypothesis true and its conclusion false. So each of the four statements rules out exactly one combination of P and Q. Read off which one:
| Statement | It is false only when |
|---|---|
| P \to Q | P is true and Q is false |
| \neg Q \to \neg P | \neg Q is true and \neg P is false, so Q is false and P is true |
| Q \to P | Q is true and P is false |
| \neg P \to \neg Q | \neg P is true and \neg Q is false, so P is false and Q is true |
The original and the contrapositive rule out the same case. They are two sentences for one promise: “you will never find P without Q.” The converse and the inverse also rule out a shared case, but a different one: “you will never find Q without P.” That is a separate promise, and keeping the first says nothing about the second.
Results
Contrapositive law. For any statements P and Q,
Converse and inverse. The inverse is the contrapositive of the converse, so
Not the converse. P \to Q and Q \to P are not logically equivalent: they disagree when exactly one of P and Q is true.
So the four related conditionals fall into two pairs, and the members of a pair always share a truth value. Taking the contrapositive twice gets you back where you started, since \neg\neg P is P.
Why it is true
Build all four columns. The \neg columns simply flip each truth value, and each arrow column is false only where its left side is true and its right side is false.
| Row | P | Q | P \to Q | Q \to P | \neg P \to \neg Q | \neg Q \to \neg P |
|---|---|---|---|---|---|---|
| 1 | T | T | T | T | T | T |
| 2 | T | F | F | T | T | F |
| 3 | F | T | T | F | F | T |
| 4 | F | F | T | T | T | T |
The columns for P \to Q and \neg Q \to \neg P agree on all four rows, so they are logically equivalent. The columns for Q \to P and \neg P \to \neg Q agree on all four rows too. But P \to Q and Q \to P disagree on rows 2 and 3, so the original and its converse are not equivalent. This is a check of every possible case, so it holds whatever P and Q say.
There is also a one-line argument, using the fact that a conditional says the same as an or: P \to Q \equiv \neg P \lor Q. Then
and Q \lor \neg P is \neg P \lor Q written in the other order.
Worked examples
All four for an inequality
For real numbers x, take the statement “if x > 4, then x^2 > 16.” Write its converse, inverse and contrapositive, and decide which of the four are true.
SolutionThe hypothesis P is x > 4 and the conclusion Q is x^2 > 16. The negation of x > 4 is x \le 4, not x < 4, since x = 4 must land on one side or the other. Likewise \neg Q is x^2 \le 16.
- Original: if x > 4, then x^2 > 16. True: multiplying x > 4 by the positive number x gives x^2 > 4x, and 4x > 16, so x^2 > 16.
- Converse: if x^2 > 16, then x > 4. False: x = -5 gives 25 > 16 but -5 \le 4.
- Inverse: if x \le 4, then x^2 \le 16. False: x = -5 again, since -5 \le 4 but 25 > 16.
- Contrapositive: if x^2 \le 16, then x \le 4. True: x^2 \le 16 means -4 \le x \le 4.
The single value x = -5 breaks both the converse and the inverse. That is no accident: they rule out the same case, so a counterexample to one is a counterexample to the other.
Negations that cancel
A whole number is claimed to satisfy: “if it is not a multiple of 3, then its digit sum is not a multiple of 3.” Write the contrapositive, and use it on 471.
SolutionHere P is “n is not a multiple of 3” and Q is “the digit sum of n is not a multiple of 3.” Negating a “not” removes it, so \neg Q is “the digit sum is a multiple of 3” and \neg P is “n is a multiple of 3.” The contrapositive \neg Q \to \neg P is if the digit sum of n is a multiple of 3, then n is a multiple of 3. That is the familiar test for divisibility by 3, which the original was hiding.
For 471 the digit sum is 4 + 7 + 1 = 12, a multiple of 3, so 471 is a multiple of 3. Indeed 471 = 3 \times 157.
Here the converse is true as well: “if the digit sum is not a multiple of 3, then n is not.” Its contrapositive, “if n is a multiple of 3, then so is its digit sum”, is the other half of the same test, so for this statement all four are true.
A statement about a whole class of shapes
“Every rhombus has perpendicular diagonals.” Write this as a conditional, then write its contrapositive and its converse. Decide whether each is true.
Solution“Every rhombus has...” means: for every quadrilateral, if it is a rhombus, then its diagonals are perpendicular. The words “for every quadrilateral” set the scene and are not negated; only the two parts are.
- Original: if a quadrilateral is a rhombus, then its diagonals are perpendicular. True, a known property of the rhombus.
- Contrapositive: if a quadrilateral’s diagonals are not perpendicular, then it is not a rhombus. True, being equivalent to the original, and the useful form for ruling a shape out.
- Converse: if a quadrilateral’s diagonals are perpendicular, then it is a rhombus. False. The kite with vertices (0, 3), (2, 0), (0, -1) and (-2, 0) has diagonals along the two axes, which are perpendicular, but its sides are \displaystyle \sqrt{13}, \displaystyle \sqrt{5}, \displaystyle \sqrt{5} and \displaystyle \sqrt{13}, so it is not a rhombus.
The inverse, “if a quadrilateral is not a rhombus, then its diagonals are not perpendicular”, is false too, and the same kite shows it.
Deciding a claim through its contrapositive
For real numbers x and y: “if x + y \ge 20, then x \ge 10 or y \ge 10.” Is it true?
SolutionDirectly, knowing only the sum, it is not obvious how to reach either number. Try the contrapositive. The negation of x + y \ge 20 is x + y < 20. The negation of x \ge 10 \text{ or } y \ge 10 is x < 10 \text{ and } y < 10, because an or is false only when both parts are false (De Morgan’s laws). So the contrapositive is
and that is plainly true: adding the two inequalities gives x + y < 20. The contrapositive is true, so the original is true.
Its converse, “if x \ge 10 or y \ge 10, then x + y \ge 20”, is false: x = 10 and y = 0 give a sum of 10.
Proving something by proving its contrapositive
Show that for every integer n: if 5n + 3 is even, then n is odd.
SolutionA direct start, “5n + 3 = 2k”, says little about n itself. The contrapositive is “if n is not odd, then 5n + 3 is not even”, that is, “if n is even, then 5n + 3 is odd”, and that one starts from something concrete.
Suppose n is even, so n = 2k for an integer k. Then
which is one more than an even number, so it is odd. The contrapositive holds for every integer, so the original does too. Proving a statement this way is its own technique, proof by contraposition.
Check. n = 1 gives 8, even, and 1 is odd. n = 4 gives 23, odd, so the statement makes no claim about it.
How many of the four can be true?
Show that of a conditional and its three relatives, the number that are true is always even, both when P and Q are single statements and when the conditional is a claim about every real x. Show that for single statements at least two are true, and find a claim about every real x for which none of the four is true.
SolutionThe four split into two equivalent pairs: the original with the contrapositive, and the converse with the inverse. For single statements this is the truth table above. For a claim about every x the pairing still holds, because at each single value of x the original and the contrapositive agree, so one fails somewhere exactly when the other does. Either way the members of a pair are true together or false together, so each pair adds 0 or 2 to the count. The total is 0, 2 or 4, never 1 or 3.
For single statements with fixed truth values, 0 cannot happen: if P \to Q is false then P is true and Q is false, and then Q \to P is true, since its hypothesis Q is false. So the count is 2 or 4.
A claim about every x escapes this, because different values of x can break different directions. Take, for every real x, “if x > 0, then x^2 > 1.”
- Original and contrapositive (“if x^2 \le 1, then x \le 0”): both false at x = \tfrac{1}{2}, since \tfrac{1}{2} > 0 but \tfrac{1}{4} \le 1.
- Converse (“if x^2 > 1, then x > 0”) and inverse (“if x \le 0, then x^2 \le 1”): both false at x = -2, since (-2)^2 = 4 > 1 but -2 \le 0.
All four are false, yet no single value of x breaks all four: x = \tfrac{1}{2} breaks one pair and x = -2 the other.
Boundaries
Only the contrapositive is safe. Its equivalence to the original tells you nothing about the converse or the inverse; they may be true (Example 2) or false (Example 1).
Statements about every member hold value by value. “For every x, if P(x) then Q(x)” is equivalent to “for every x, if \neg Q(x) then \neg P(x)”, because the two agree at each single x. The words “for every” stay as they are. Negating them as well produces a different statement, one about the negation of a quantified statement.
The contrapositive is not the negation. The negation of P \to Q is “P and \neg Q”, which is true exactly when P \to Q is false. The contrapositive has the same truth value as P \to Q; the negation always has the opposite one.
Negating a compound part takes care. When P or Q contains an and or an or, its negation flips the connective, as in Example 4.
It relies on every statement being true or false. In logics that drop that assumption, such as intuitionistic logic, P \to Q still gives \neg Q \to \neg P, but \neg Q \to \neg P no longer gives back P \to Q.
Where this goes next
The contrapositive turns straight into a way of proving things: proof by contraposition proves P \to Q by assuming \neg Q and reaching \neg P, as in Example 5. The same equivalence is why modus tollens is valid: it is modus ponens applied to the contrapositive.
A true theorem may have a false converse, so each direction needs its own proof (a theorem and its converse); when both hold, the parts are joined by a biconditional. Geometry puts the contrapositive to work in indirect proof.
Exercises
Writing the four
- Write the converse, inverse and contrapositive of “if x = 5, then x^2 = 25.” For real x, say which of the four are true.
- Write the converse, inverse and contrapositive of “if a quadrilateral is a rectangle, then its diagonals are equal in length.” Say which are true.
- In symbols, write the converse, inverse and contrapositive of \neg P \to Q.
- Write the contrapositive of the converse of P \to Q. Which of the four related conditionals is it?
True or false, all four
For each statement, decide which of the original, converse, inverse and contrapositive are true.
- For integers n: if n is odd, then n^3 is odd.
- For real x: if x^2 = x, then x = 1.
- For real x: if x < 0, then x^3 < 0.
- For integers n: if n > 10, then n is even.
Deciding through the contrapositive
Write the contrapositive, and use it to decide whether the statement is true.
- For integers a and b: if ab is odd, then a and b are both odd.
- For real x: if x^2 - 6x + 8 \ne 0, then x \ne 2.
- For real x: if x^2 - 6x + 8 \ne 0, then x \ne 5.
- If 30 whole numbers are chosen, then two of them have the same last digit.
Proving through the contrapositive
- Prove that for every integer n: if 7n - 4 is odd, then n is odd.
- Prove that for real numbers x and y: if x \ne y, then \dfrac{x + y}{2} \ne x.
Reasoning
- Explain, using the one case each statement rules out, why the converse and the inverse are logically equivalent.
- The inverse of a certain conditional is true. Which of the other three must be true? Give one example where the original is true and one where it is false.
- Show that the contrapositive of the contrapositive of P \to Q is P \to Q.
- A conditional and its converse are both true. Which of the other two must be true, and what single statement says the same as the pair?
- Write both the negation and the contrapositive of P \to Q, and show with a truth table that they are never true at the same time.
Answers
- Converse: if x^2 = 25, then x = 5. Inverse: if x \ne 5, then x^2 \ne 25. Contrapositive: if x^2 \ne 25, then x \ne 5. The original and contrapositive are true; the converse and inverse are false, both broken by x = -5, since (-5)^2 = 25.
- Converse: if a quadrilateral’s diagonals are equal, it is a rectangle. Inverse: if it is not a rectangle, its diagonals are not equal. Contrapositive: if its diagonals are not equal, it is not a rectangle. The original and contrapositive are true. The converse and inverse are false: an isosceles trapezoid that is not a rectangle has equal diagonals.
- Converse Q \to \neg P; inverse P \to \neg Q (since \neg\neg P is P); contrapositive \neg Q \to P.
- The converse is Q \to P, whose contrapositive is \neg P \to \neg Q: the inverse.
- All four are true. The original holds since n = 2k + 1 gives n^3 = 2(4k^3 + 6k^2 + 3k) + 1. The converse holds by its contrapositive: if n = 2k is even, n^3 = 8k^3 is even. The inverse and contrapositive match these.
- The original and contrapositive are false: x = 0 has 0^2 = 0 but 0 \ne 1. The converse (if x = 1, then x^2 = x) and the inverse (if x^2 \ne x, then x \ne 1) are true, since 1^2 = 1.
- All four are true. A negative number cubed is negative, and if x \ge 0 then x^3 \ge 0, which is the inverse; the converse and contrapositive match them.
- None is true. The original and contrapositive (if n is odd, then n \le 10) both fail at n = 11. The converse (if n is even, then n > 10) and inverse (if n \le 10, then n is odd) both fail at n = 2.
- True. Contrapositive: if a is even or b is even, then ab is even. If a = 2k, then ab = 2(kb), and likewise for b.
- True. Contrapositive: if x = 2, then x^2 - 6x + 8 = 0. Indeed 4 - 12 + 8 = 0.
- False. Contrapositive: if x = 5, then x^2 - 6x + 8 = 0. But 25 - 30 + 8 = 3 \ne 0, so the contrapositive is false, and so is the original, at x = 5.
- True. Contrapositive: if no two of the numbers share a last digit, then fewer than 30 were chosen. There are only 10 last digits, so at most 10 numbers can have different ones.
- Contrapositive: if n is even, then 7n - 4 is even. With n = 2k, 7n - 4 = 14k - 4 = 2(7k - 2), which is even.
- Contrapositive: if \dfrac{x + y}{2} = x, then x = y. Multiply by 2: x + y = 2x, so y = x.
- The converse Q \to P is false only when Q is true and P is false. The inverse \neg P \to \neg Q is false only when \neg P is true and \neg Q is false, which is the same case. Two statements false in exactly the same case agree everywhere.
- The converse, which is equivalent to the inverse. The original is not settled. For real x, “if x = 1, then x^3 = 1” is true and so is its inverse, “if x \ne 1, then x^3 \ne 1”. But “if x > 3, then x > 4” is false at x = 3.5, while its inverse, “if x \le 3, then x \le 4”, is true.
- The contrapositive of P \to Q is \neg Q \to \neg P. Its contrapositive swaps and negates again: \neg\neg P \to \neg\neg Q, which is P \to Q.
- Both. The contrapositive is equivalent to the original and the inverse to the converse, so all four are true. Together P \to Q and Q \to P say that P and Q are true together or false together, which is the biconditional P \leftrightarrow Q.
- Negation: P \land \neg Q. Contrapositive: \neg Q \to \neg P. Down the four rows (TT, TF, FT, FF) the negation reads F, T, F, F and the contrapositive reads T, F, T, T. On every row exactly one of them is true, so they are never true together.