<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

5.5: Extension: Indirect Proof

Difficulty Level: At Grade Created by: CK-12

The indirect proof or proof by contradiction is part of 41 out of 50 states’ mathematics standards. Depending on the state, the teacher may choose to use none, part or all of this section.

Learning Objectives

  • Reason indirectly to develop proofs.

Until now, we have proved theorems true by direct reasoning, where conclusions are drawn from a series of facts and previously proven theorems. Indirect proof is another option.

Indirect Proof: When the conclusion from a hypothesis is assumed false (or opposite of what it states) and then a contradiction is reached from the given or deduced statements.

The easiest way to understand indirect proofs is by example.

Indirect Proofs in Algebra

Example 1: If \begin{align*}x=2\end{align*}x=2, then \begin{align*}3x - 5 \neq 10\end{align*}3x510. Prove this statement is true by contradiction.

Solution: In an indirect proof the first thing you do is assume the conclusion of the statement is false. In this case, we will assume the opposite of \begin{align*}3x - 5 \neq 10\end{align*}3x510

If \begin{align*}x=2\end{align*}x=2, then \begin{align*}3x - 5 = 10\end{align*}3x5=10

Take this statement as true and solve for \begin{align*}x\end{align*}x.

\begin{align*}3x - 5 &= 10\\ 3x &= 15\\ x &= 5\end{align*}3x53xx=10=15=5

\begin{align*}x = 5\end{align*}x=5 contradicts the given statement that \begin{align*}x = 2\end{align*}x=2. Hence, our assumption is incorrect and \begin{align*}3x - 5 \neq 10\end{align*}3x510 is true.

Example 2: If \begin{align*}n\end{align*}n is an integer and \begin{align*}n^2\end{align*}n2 is odd, then \begin{align*}n\end{align*}n is odd. Prove this is true indirectly.

Solution: First, assume the opposite of “\begin{align*}n\end{align*}n is odd.”

\begin{align*}n\end{align*}n is even.

Now, square \begin{align*}n\end{align*}n and see what happens.

If \begin{align*}n\end{align*}n is even, then \begin{align*}n = 2a\end{align*}n=2a, where \begin{align*}a\end{align*}a is any integer.

\begin{align*}n^2 = (2a)^2 = 4a^2\end{align*}n2=(2a)2=4a2

This means that \begin{align*}n^2\end{align*}n2 is a multiple of 4. No odd number can be divided evenly by an even number, so this contradicts our assumption that \begin{align*}n\end{align*}n is even. Therefore, \begin{align*}n\end{align*}n must be odd if \begin{align*}n^2\end{align*}n2 is odd.

Indirect Proofs in Geometry

Example 3: If \begin{align*}\triangle ABC\end{align*}ABC is isosceles, then the measure of the base angles cannot be \begin{align*}92^\circ\end{align*}92. Prove this indirectly.

Solution: Assume the opposite of the conclusion.

The measure of the base angles are \begin{align*}92^\circ\end{align*}92.

If the base angles are \begin{align*}92^\circ\end{align*}92, then they add up to \begin{align*}184^\circ\end{align*}184. This contradicts the Triangle Sum Theorem that says all triangles add up to \begin{align*}180^\circ\end{align*}180. Therefore, the base angles cannot be \begin{align*}92^\circ\end{align*}92.

Example 4: Prove the SSS Inequality Theorem is true by contradiction.

Solution: The SSS Inequality Theorem says: “If two sides of a triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is greater in measure than the included angle of the second triangle.” First, assume the opposite of the conclusion.

The included angle of the first triangle is less than or equal to the included angle of the second triangle.

If the included angles are equal then the two triangles would be congruent by SAS and the third sides would be congruent by CPCTC. This contradicts the hypothesis of the original statement “the third side of the first triangle is longer than the third side of the second.” Therefore, the included angle of the first triangle must be larger than the included angle of the second.

To summarize:

  • Assume the opposite of the conclusion (second half) of the statement.
  • Proceed as if this assumption is true to find the contradiction.
  • Once there is a contradiction, the original statement is true.
  • DO NOT use specific examples. Use variables so that the contradiction can be generalized.

Review Questions

Prove the following statements true indirectly.

  1. If \begin{align*}n\end{align*}n is an integer and \begin{align*}n^2\end{align*}n2 is even, then \begin{align*}n\end{align*}n is even.
  2. If \begin{align*}m \angle A \neq m \angle B\end{align*}mAmB in \begin{align*}\triangle ABC\end{align*}ABC, then \begin{align*}\triangle ABC\end{align*}ABC is not equilateral.
  3. If \begin{align*}x > 3\end{align*}x>3, then \begin{align*}x^2 > 9\end{align*}x2>9.
  4. The base angles of an isosceles triangle are congruent.
  5. If \begin{align*}x\end{align*}x is even and \begin{align*}y\end{align*}y is odd, then \begin{align*}x + y\end{align*}x+y is odd.
  6. In \begin{align*}\triangle ABE\end{align*}ABE, if \begin{align*}\angle A\end{align*}A is a right angle, then \begin{align*}\angle B\end{align*}B cannot be obtuse.
  7. If \begin{align*}A, \ B\end{align*}A, B, and \begin{align*}C\end{align*}C are collinear, then \begin{align*}AB + BC = AC\end{align*}AB+BC=AC (Segment Addition Postulate).
  8. Challenge Prove the SAS Inequality Theorem is true using indirect proofs.

Image Attributions

Show Hide Details
Files can only be attached to the latest version of section
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Geometry-Basic.5.5