Teacher Guide: Proof by Contrapositive

Prove an if-then statement by assuming its conclusion is false and showing that its hypothesis must be false too.

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10 questions on Mathematical Proofs. Students join with a name, you see everyone's score.

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Learning Objectives

Explain why a statement and its contrapositive are logically equivalent
Decide when a proof by contrapositive is easier than a direct proof
Negate conclusions that contain "and", "or" and inequalities correctly
Write complete proofs by contrapositive about even and odd integers and inequalities
Distinguish proof by contrapositive from proof by contradiction

Prerequisites

• Direct Proof: assume the hypothesis and derive the conclusion
• Contrapositive: forming $\neg q \to \neg p$ from an if-then statement
• Introduction to Mathematical Proofs: hypothesis, conclusion and Q.E.D.
• Definitions of even ( $2k$ ) and odd ( $2k + 1$ ) integers

Discussion Starters

1. "If it rains, the street is wet." What can you conclude if the street is dry? What if the street is wet?
2. Why is the hypothesis " $n^2$ is even" harder to use than " $n$ is odd"?
3. What is the opposite of "at least one of the numbers is 5 or more"?
4. How would you explain to a friend the difference between a contrapositive proof and a contradiction proof?

Common Misconceptions

The contrapositive and the converse are the same thing

Remediation: Use a true statement with a false converse: "if a number is divisible by 4, it is even". Its converse fails ( $6$ is even but not divisible by 4), while its contrapositive ("if a number is odd, it is not divisible by 4") is true.

Negating just means putting "not" in front of each word

Remediation: Practice with " $x$ and $y$ are odd": students test pairs like $(2, 3)$ and see that the statement already fails when only one number is even, which leads to " $x$ is even or $y$ is even".

A proof by contrapositive must end with a contradiction

Remediation: Write both proofs of "if $n^2$ is even, then $n$ is even" side by side and highlight that the contrapositive version simply ends at " $n^2$ is odd".

Differentiation Ideas

For Struggling Students:

• Give a template: Contrapositive: ... / Assume: ... / Then: ... / So: ... / Therefore the original statement is true
• Start with everyday sentences before switching to numbers
• Check each step with concrete numbers such as $n = 7$ or $n = 10$

For On-Level Students:

• Prove the classic even and odd statements without a template
• Negate statements with and, or and inequalities, then prove them
• Decide for each statement whether a direct or a contrapositive proof is shorter

For Advanced Students:

• Prove: if $n^2$ is divisible by 3, then $n$ is divisible by 3 (cases $n = 3k + 1$ and $n = 3k + 2$ )
• Use the result " $n^2$ even implies $n$ even" to follow the proof that $\sqrt{2}$ is irrational
• Rewrite a proof by contrapositive as a proof by contradiction and compare their lengths

Standards Alignment

MP3 (CCSS.MATH.PRACTICE.MP3)
Construct viable arguments and critique the reasoning of others
MP6 (CCSS.MATH.PRACTICE.MP6)
Attend to precision, including the precise use of logical connectives and negations
HSA-REI.A.1 (CCSS.MATH.CONTENT.HSA.REI.A.1)
Explain each step in solving an equation as following from the equality of numbers asserted at the previous step

Lesson Resources

visualTruth Table: Two Equal Columns
Students build the columns $p \to q$ and $\neg q \to \neg p$ and see they match
activityDirect or Contrapositive?
Sort 8 statements by which method gives the shorter proof, then prove two of them
worksheetNegation Drill
Negate statements with and, or, at least one and inequalities before writing proofs

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

A proof by contrapositive proves an if-then statement

p \to q

by proving its contrapositive

\neg q \to \neg p

instead. You assume the conclusion is false and show that the hypothesis must then be false.

Why this is allowed (a recap of the lesson Contrapositive): a statement and its contrapositive are logically equivalent. Their columns in the truth table are identical in every row:

\begin{array}{c|c|c|c|c|c} p & q & p \to q & \neg q & \neg p & \neg q \to \neg p \\ \hline F & F & T & T & T & T \\ F & T & T & F & T & T \\ T & F & F & T & F & F \\ T & T & T & F & F & T \end{array}

So once

\neg q \to \neg p

is proven,

p \to q

is proven too.

The plan of the proof:

1.Write the contrapositive: negate the hypothesis and the conclusion, then swap them.
2.Assume $\neg q$ (the conclusion is false).
3.Work forward with definitions and algebra, exactly as in a direct proof.
4.Arrive at $\neg p$ (the hypothesis is false).
5.Conclude: the contrapositive is true, so the original statement is true.

When to choose it: when the hypothesis

p

gives you little to work with ("

n^2

is even" says nothing simple about

n

), but the negated conclusion

\neg q

is concrete ("

n

is odd" means

n = 2k + 1

Not the same as proof by contradiction: there you assume

p

and

\neg q

together and look for *any* contradiction. Here you assume only

\neg q

and aim at one fixed target:

\neg p

Worked Examples

Prove: for every integer $n$ , if $n^2$ is even, then $n$ is even.

Write the contrapositive

Negate both parts and swap them: "if $n$ is not even, then $n^2$ is not even" → If $n$ is odd, then $n^2$ is odd

Assume the negated conclusion

Assume $n$ is odd, so $n = 2k + 1$ for some integer $k$ → $n = 2k + 1$

Square and expand

$n^2 = (2k + 1)^2 = 4k^2 + 4k + 1$ → $n^2 = 4k^2 + 4k + 1$

Bring it into the form 2m + 1

$n^2 = 2(2k^2 + 2k) + 1$ , and $m = 2k^2 + 2k$ is an integer → $n^2$ is odd

Conclude

We proved " $n$ odd $\Rightarrow$ $n^2$ odd", the contrapositive of the statement → If $n^2$ is even, then $n$ is even

Answer: The contrapositive (if $n$ is odd, then $n^2$ is odd) is true, so the original statement is true. Q.E.D.

Common Mistakes

Proving the converse instead of the contrapositive: for "if $n^2$ is even, then $n$ is even", assuming $n$ is even and showing $n^2$ is even

Why it's wrong: That proves $q \to p$ , the converse, which is a different statement. A statement and its converse are not equivalent.

Correct: Negate both parts and swap them: assume $n$ is odd ( $\neg q$ ) and show $n^2$ is odd ( $\neg p$ ).

Negating " $x$ and $y$ are odd" as " $x$ and $y$ are even"

Why it's wrong: The conclusion fails as soon as one of the numbers is even. "Both even" covers only part of the cases, so the proof would be incomplete.

Correct: "Not ( $a$ and $b$ )" is "not $a$ or not $b$ ": $x$ is even or $y$ is even. Then prove every case.

Negating $x \geq 5$ as $x \leq 5$

Why it's wrong: The value $x = 5$ satisfies both $x \geq 5$ and $x \leq 5$ , so $x \leq 5$ is not the opposite of $x \geq 5$ .

Correct: The negation of $x \geq 5$ is $x < 5$ , and the negation of $x > 5$ is $x \leq 5$ .

Assuming $\neg q$ and then hunting for "some contradiction"

Why it's wrong: That mixes two methods. In a proof by contrapositive there is no contradiction: you have one clear target.

Correct: Assume $\neg q$ and work until you reach exactly $\neg p$ . The search for any contradiction belongs to proof by contradiction.

Why It Matters

Proof by contrapositive is one of the most useful tools in a mathematician's kit:

It turns vague hypotheses into concrete ones: " $n^2$ is even" is hard to use, " $n$ is odd, so $n = 2k + 1$ " is easy to use
It handles "or" conclusions neatly: the negation of " $x \geq 5$ or $y \geq 5$ " is the friendly " $x < 5$ and $y < 5$ "
It powers classic results: the proof that $\sqrt{2}$ is irrational starts from the fact that $n^2$ even forces $n$ even, which is proven by contrapositive
It is everyday reasoning: "if the alarm works, it beeps in a test" means "no beep, so the alarm is broken"
It prepares you for proof by contradiction, the next lesson in this topic, by training you to negate statements correctly

Once you can negate statements and switch viewpoints, many proofs that look impossible directly become a few lines long.

Real World Applications

Quality Control

Inspectors use contrapositive reasoning to rule out cases without weighing every item twice.

Example:

Rule: if a pack of 4 boxes weighs at least 2000 g, then at least one box weighs at least 500 g. Contrapositive: if every box weighs less than 500 g, the pack weighs less than 2000 g.

1Try It Yourself

A machine fills 4 boxes and none of them can hold more than 495 g.

Show with the contrapositive that the pack cannot reach 2000 g.

Step 1: Write the mathematical expression

Largest possible total of the 4 boxes:

Testing Software

Programmers test code with the contrapositive: if the code is correct, a test gives the right output, so a wrong output proves a bug.

Example:

A function should compute $f(n) = 3n + 2$ . We proved: if $f(n)$ is odd, then $n$ is odd. Contrapositive: an even $n$ must give an even result.

2Try It Yourself

A test calls the function with $n = 8$ and the program returns 27.

What should the correct output be, and what does 27 tell you?

Step 1: Write the mathematical expression

Compute the correct value $f(8)$ :

Check Digits in Codes

Many codes use a simple rule to catch typing errors. The contrapositive tells you when a code is certainly wrong.

Example:

Rule: if a code is valid, then the sum of its digits is divisible by 9. Contrapositive: if the digit sum is not divisible by 9, the code is not valid.

3Try It Yourself

Someone types the code 4726.

Use the contrapositive to decide whether the code can be valid.

Step 1: Write the mathematical expression

Sum of the digits:

Key Takeaways

1A proof by contrapositive proves $p \to q$ by proving $\neg q \to \neg p$
2It works because a statement and its contrapositive are logically equivalent (identical truth table columns)
3Plan: assume the conclusion is false, work forward, reach the hypothesis being false
4Choose it when the negated conclusion is concrete, for example " $n$ is odd" gives $n = 2k + 1$
5Negate carefully: "and" becomes "or", "or" becomes "and", $\geq$ becomes $<$
6Unlike proof by contradiction, you assume only $\neg q$ and aim at one target, $\neg p$

Frequently Asked Questions

Why does proving the contrapositive prove the original statement?

Because they are logically equivalent: in the truth table,

p \to q

and

\neg q \to \neg p

are false in exactly the same row (

p

true,

q

false) and true in all others. Proving one proves the other.

How do I know when to use contrapositive instead of a direct proof?

Look at the hypothesis. If it is hard to use (a statement about

n^2

, a product, a sum) while the negated conclusion gives you a formula (

n = 2k

n = 2k + 1

x < 5

and

y < 5

), the contrapositive is usually shorter.

What is the difference from proof by contradiction?

In proof by contradiction you assume

p

and

\neg q

and look for any impossible situation. In proof by contrapositive you assume only

\neg q

and must reach the specific statement

\neg p

. Every proof by contrapositive can be rewritten as a contradiction proof, but not every contradiction proof is a contrapositive.

If I prove the contrapositive, is the converse also true?

Not automatically. The converse

q \to p

is a separate statement. For example, "if

n^2

is even, then

n

is even" and its converse "if

n

is even, then

n^2

is even" are both true, but each needs its own proof.

Glossary

Contrapositive: The statement $\neg q \to \neg p$ formed from $p \to q$ by negating both parts and swapping them
Proof by contrapositive: A proof of $p \to q$ that assumes $\neg q$ and derives $\neg p$
Logically equivalent: Two statements that have the same truth value in every row of the truth table
Negation: The statement "not $p$ " ( $\neg p$ ), true exactly when $p$ is false
Converse: The statement $q \to p$ ; it is not equivalent to $p \to q$
Even / odd integer: An integer of the form $2k$ (even) or $2k + 1$ (odd), where $k$ is an integer

Formula Card

Contrapositive equivalence

p \to q \;\equiv\; \neg q \to \neg p

Proving the right side proves the left side

Negating "and"

\neg(a \land b) \equiv \neg a \lor \neg b

Not both: at least one fails

Negating "or"

\neg(a \lor b) \equiv \neg a \land \neg b

Neither: both fail

Even and odd

n = 2k

(even),

\; n = 2k + 1

(odd)

The concrete forms you get after negating "even" or "odd"

Teacher Guide: Proof by Contrapositive

Use this lesson with your class

For Teachers

For Struggling Students:

For On-Level Students:

For Advanced Students:

Lesson Content

Definition

Worked Examples

Common Mistakes

Why It Matters

Real World Applications

Quality Control

Testing Software

Check Digits in Codes

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Contrapositive equivalence

Negating "and"

Negating "or"

Even and odd

More in This Topic

Introduction to Mathematical Proofs

Direct Proof

Two-Column Proofs

Paragraph Proofs