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Year 11 Specialist (Unit 1 & 2) Introduction to proof

Proof by contrapositive

20 practice questions 0 video lessons Theory + worked examples

Master proof by contrapositive for Year 11 Specialist Mathematics in Queensland (QCAA). Rather than proving a statement directly, you prove its equivalent contrapositive, which swaps and negates the original.

You will learn to build the converse, inverse and contrapositive of a statement, see why only the contrapositive is equivalent to the original, and use it to prove results about integers and inequalities.

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Theory

Proof by contrapositive proves a statement of the form "if \(P\) then \(Q\)" by instead proving the logically equivalent contrapositive "if not \(Q\) then not \(P\)" in Year 11 Specialist Mathematics (QCAA, Queensland). This page shows how the contrapositive is built, how it differs from the converse and inverse, and how to use it in a proof.

A conditional statement has the form \(P\Rightarrow Q\), read "if \(P\), then \(Q\)". Here \(P\) is the hypothesis and \(Q\) is the conclusion. From it we can build three related statements.

The converse is \(Q\Rightarrow P\): you swap the hypothesis and conclusion. The inverse is \(\neg P\Rightarrow\neg Q\): you negate both parts but keep the order. The contrapositive is \(\neg Q\Rightarrow\neg P\): you negate both parts and swap them.

The key fact is that a statement is logically equivalent to its contrapositive: \(P\Rightarrow Q\) and \(\neg Q\Rightarrow\neg P\) are always true or false together. The converse and inverse are equivalent to each other, but neither is equivalent to the original.

To prove by contrapositive, assume \(\neg Q\) and reason directly until you reach \(\neg P\). Because the contrapositive is equivalent to the original, this proves \(P\Rightarrow Q\). This is useful when \(\neg Q\) is an easier place to start than \(P\).

The four related conditional forms Original P implies Q with Converse Q implies P on top; Contrapositive not-Q implies not-P with Inverse not-P implies not-Q on the bottom. Blue double arrows link the original to its contrapositive and the converse to the inverse as equivalent pairs; a red dashed arrow links the original to the converse as not equivalent. OriginalP ⇒ Q ConverseQ ⇒ P Contrapositive¬Q ⇒ ¬P Inverse¬P ⇒ ¬Q equivalent equivalent not equivalent
The four forms. The original and its contrapositive are equivalent; the converse and inverse are equivalent to each other but not to the original.
Layout of a proof by contrapositive Five stacked steps joined by downward arrows: prove if P then Q; state the contrapositive if not Q then not P; assume not Q; reason directly to not P; conclude the original is proved. Prove: if P then Q Contrapositive: if not Q then not P Assume not Q Reason directly to not P Original is proved
A contrapositive proof: state \(\neg Q\Rightarrow\neg P\), assume \(\neg Q\), reason to \(\neg P\), and the original follows.

Starting from the conditional \(P\Rightarrow Q\), the related forms are:

\[ \text{converse: } Q\Rightarrow P \qquad \text{inverse: } \neg P\Rightarrow\neg Q \qquad \text{contrapositive: } \neg Q\Rightarrow\neg P \]
QP,¬P¬Q,¬Q¬P

The equivalence that makes the method work is:

\[ (P\Rightarrow Q) \;\equiv\; (\neg Q\Rightarrow\neg P) \]
(PQ)(¬Q¬P)
Negate carefully. The negation of \(x>3\) is \(x\le 3\) (the boundary is now included); the negation of "\(n\) is odd" is "\(n\) is even"; and by De Morgan, the negation of "\(A\) and \(B\)" is "\(\neg A\) or \(\neg B\)".

How to prove a statement by contrapositive

  1. Identify \(P\) (the hypothesis) and \(Q\) (the conclusion) in the statement "if \(P\), then \(Q\)".
  2. Write the contrapositive \(\neg Q\Rightarrow\neg P\): negate both parts and swap them (do not just swap — that gives the converse).
  3. Assume \(\neg Q\) and reason directly, step by step, using algebra or definitions.
  4. Reach \(\neg P\). The contrapositive is proved, so the original statement is proved too.
Example 1 — Find the contrapositive
Write the contrapositive of "If it is snowing, then it is cold."
Solution

Identify the parts, negate each, then swap:

\(P\)\(:\)\(\text{it is snowing}\)
\(Q\)\(:\)\(\text{it is cold}\)
\(\neg Q\)\(:\)\(\text{it is not cold}\)
\(\neg P\)\(:\)\(\text{it is not snowing}\)
\(\text{contrapositive}\)\(:\)\(\neg Q \Rightarrow \neg P\)

The contrapositive is: "If it is not cold, then it is not snowing."

Example 2 — Converse vs contrapositive
For "If \(n\) is a multiple of \(9\), then \(n\) is a multiple of \(3\)", write the converse and the contrapositive.
Solution

The converse swaps only; the contrapositive negates both and swaps:

\(\text{original}\)\(:\)\(P \Rightarrow Q\)
\(\text{converse}\)\(:\)\(Q \Rightarrow P \ \ (\text{swap})\)
\(\text{contrapositive}\)\(:\)\(\neg Q \Rightarrow \neg P \ \ (\text{negate and swap})\)

Converse: "If \(n\) is a multiple of \(3\), then \(n\) is a multiple of \(9\)" — this is false (take \(n=6\)). Contrapositive: "If \(n\) is not a multiple of \(3\), then \(n\) is not a multiple of \(9\)" — true, like the original.

Converse (false): multiple of 3 ⇒ multiple of 9. Contrapositive (true): not a multiple of 3 ⇒ not a multiple of 9.

Example 3 — Prove by contrapositive
Prove: if \(3n+5\) is even, then \(n\) is odd, where \(n\) is an integer.
Solution

State the contrapositive (negate both and swap):

"If \(n\) is even, then \(3n+5\) is odd." Assume \(n=2k\):

\(3n+5\)\(=\)\(3(2k)+5\)
\(=\)\(6k+5\)
\(=\)\(2(3k+2)+1\)

This has the form \(2m+1\), so \(3n+5\) is odd. The contrapositive holds.

The contrapositive is proved, so the original statement is true.

Example 4 — Why not the converse
"If \(x>5\), then \(x^2>25\)" is true. Is its converse true? Give the contrapositive.
Solution

Test the converse "if \(x^2>25\) then \(x>5\)" with \(x=-6\):

\(x\)\(=\)\(-6\)
\(x^2\)\(=\)\(36\)
\(36\)\(>\)\(25 \ \ (\text{hypothesis holds})\)
\(x=-6\)\(\not>\)\(5 \ \ (\text{conclusion fails})\)

So the converse is false. The contrapositive "if \(x^2\le 25\), then \(x\le 5\)" is true, matching the original.

The converse is false; the contrapositive is true and equivalent to the original.

Common pitfalls

Confusing the contrapositive with the converse. The converse \(Q\Rightarrow P\) only swaps the parts; the contrapositive \(\neg Q\Rightarrow\neg P\) swaps and negates. Only the contrapositive is equivalent to the original.
Negating an inequality wrongly. The negation of \(x>3\) is \(x\le 3\), not \(x<3\): the boundary value \(x=3\) is included in the negation.
Assuming the wrong thing. To prove \(P\Rightarrow Q\) by contrapositive you assume \(\neg Q\) and aim for \(\neg P\) — not the other way round.
Thinking the converse follows. Proving a statement does not prove its converse; the converse can be false even when the original is true.

Frequently asked questions

What is the contrapositive of a statement?

The contrapositive of "if \(P\) then \(Q\)" is "if not \(Q\) then not \(P\)": you negate both parts and swap them. It is logically equivalent to the original statement.

What is the difference between the converse, inverse and contrapositive?

The converse \(Q\Rightarrow P\) swaps the parts; the inverse \(\neg P\Rightarrow\neg Q\) negates both parts; the contrapositive \(\neg Q\Rightarrow\neg P\) negates both parts and swaps. Only the contrapositive is equivalent to the original.

Why does proof by contrapositive work?

Because a conditional and its contrapositive always have the same truth value. If you prove \(\neg Q\Rightarrow\neg P\), you have automatically proved the equivalent statement \(P\Rightarrow Q\).

Is the contrapositive the same as the converse?

No. The converse only swaps the hypothesis and conclusion; the contrapositive swaps them and negates both. The converse is not equivalent to the original, but the contrapositive is.

How do you start a proof by contrapositive?

Assume the negation of the conclusion, \(\neg Q\), then reason directly with algebra or definitions until you reach the negation of the hypothesis, \(\neg P\).

How do you negate an inequality like \(x>3\)?

The negation of \(x>3\) is \(x\le 3\). The negation includes the boundary, so it is "less than or equal to", not just "less than".