# 2.18: Indirect Proof in Algebra and Geometry

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Proof by contradiction, beginning with the assumption that the conclusion is false.

## Indirect Proofs

Most likely, the first type of formal proof you learned was a direct proof using direct reasoning. Most of the proofs done in geometry are done in the two-column format, which is a direct proof format. Another common type of reasoning is indirect reasoning, which you have likely done outside of math class. Below we will formally learn what an indirect proof is and see some examples in both algebra and geometry.

Indirect Proof or Proof by Contradiction: 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.

In other words, if you are trying to show that something is true, show that if it was not true there would be a contradiction (something else would not make sense).

The steps to follow when proving indirectly are:

• 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.

The easiest way to understand indirect proofs is by example.

What if you wanted to prove a statement was true without a two-column proof? How might you go about doing so?

Example $$\PageIndex{1}$$

If $$x=2$$, then $$3x−5\neq 10$$. Prove this statement is true by contradiction.

Solution

Remember that 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 "If $$x=2$$, then $$3x−5\neq 10$$":

If $$x=2$$, then $$3x−5=10$$.

Take this statement as true and solve for x.

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

But $$x=5$$ contradicts the given statement that $$x=2$$. Hence, our assumption is incorrect and $$3x−5\neq 10$$ is true.

Example $$\PageIndex{2}$$

If \Delta ABC is isosceles, then the measure of the base angles cannot be $$92^{\circ}$$. Prove this indirectly.

Solution

Remember, to start assume the opposite of the conclusion.

The measure of the base angles are $$92^{\circ}$$.

If the base angles are $$92^{\circ}$$, then they add up to $$184^{\circ}$$. This contradicts the Triangle Sum Theorem that says the three angle measures of all triangles add up to $$180^{\circ}$$. Therefore, the base angles cannot be $$92^{\circ}$$.

Example $$\PageIndex{3}$$

If $$\angle A$$ and $$\angle B$$ are complementary then $$\angle A\leq 90^{\circ}$$. Prove this by contradiction.

Solution

Assume the opposite of the conclusion.

$$\angle A>90^{\circ}$$.

Consider first that the measure of \angle B cannot be negative. So if $$\angle A>90^{\circ}$$ this contradicts the definition of complementary, which says that two angles are complementary if they add up to $$90^{\circ}$$. Therefore, $$\angle A\leq 90^{\circ}$$.

Example $$\PageIndex{4}$$

If n is an integer and $$n^{2}$$ is odd, then n is odd. Prove this is true indirectly.

Solution

First, assume the opposite of “$$n$$ is odd.”

$$n$$ is even.

Now, square $$n$$ and see what happens.

If $$n$$ is even, then $$n=2a$$, where a is any integer.

$$n^{2}=(2a)^{2}=4a^{2}$$

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

Example $$\PageIndex{5}$$

Prove the SSS Inequality Theorem is true by contradiction. (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's two congruent sides is greater in measure than the included angle of the second triangle's two congruent sides.”)

Solution

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.

## Review

Prove the following statements true indirectly.

1. If $$n$$ is an integer and $$n^{2}$$ is even, then n is even.
2. If $$m\angle A\neq m\angle B$$ in $$\Delta ABC$$, then $$\Delta ABC$$ is not equilateral.
3. If $$x>3$$, then $$x^{2}>9$$.
4. The base angles of an isosceles triangle are congruent.
5. If $$x$$ is even and $$y$$ is odd, then $$x+y$$ is odd.
6. In $$\Delta ABE$$, if $$\angle A$$ is a right angle, then $$\angle B$$ cannot be obtuse.
7. If $$A$$, $$B$$, and $$C$$ are collinear, then $$AB+BC=AC$$ (Segment Addition Postulate).
8. If $$\Delta ABC$$ is equilateral, then the measure of the base angles cannot be $$72^{\circ}$$.
9. If $$x=11$$ then $$2x−3\neq 21$$.
10. If $$\Delta ABC$$ is a right triangle, then it cannot have side lengths 3, 4, and 6.

Video: Indirect Proof in Algebra and Geometry Examples - Basic

Activities: Indirect Proof in Algebra and Geometry Discussion Questions

Study Aids: Types of Reasoning Study Guide

Practice: Indirect Proof in Algebra and Geometry