More examples of proof by contradiction

High School Geometry, High School Mathematics, Problem Solving and Proofs

We have had good discussions on mathematical proofs, so I am planning to create a mathematical proof series that will discuss the basics such as direct proof, indirect proof, and proof by mathematical induction.  But before I do that, let me continue with more examples of proof by contradiction.

Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Finding a contradiction means that your assumption is false and therefore the statement is true. Below are several more examples of this proof strategy.

Example 1:  \sqrt{2} irrational.

Example 2: \sqrt{6} is irrational. The proof of this is basically the same as example 1, so it is left as an exercise.

Example 3: Proof that there are infinitely many primes.

Example 4: Knights and Liars

Example 5: \sqrt{2} + \sqrt{3} is irrational.

Proof:

Suppose the statement is false. Then there is a rational number r such that \sqrt{2} + \sqrt{3} = r. Now, squaring both sides we have 2 + 2\sqrt{6} + 3 = r^2. This means that

\sqrt{6} = \displaystyle\frac{r^2-5}{2}

Now, from example 2, \sqrt{6} is irrational, and we know that (r^2-5)/2 is rational since r is rational. This makes the left hand side of the equation irrational and the right hand side rational —  a contradiction!

Therefore, our supposition is false, which implies that  \sqrt{2} + \sqrt{3} is irrational.

Example 6: If a line is tangent to a circle, then it is perpendicular to the radius at the point of tangency.

Proof:

Consider a circle with center O, radius OP, and with line m tangent to the circle at P. We have to show that m \perp \overline{OP}.

Suppose \overline{OP} is not \perp m. Then there exists a segment  OQ which is \perp m. If so, it follows that OQ < OP (Why?). But Q lies outside the circle since m is a tangent line. Therefore OQ > OP. But this contradicts the previous statement that OQ < OP. Hence, our supposition is false. Therefore, \overline{OP} \perp m.

Mathematical proofs is one of the most difficult topics in mathematics, but you can only see a few examples in books. In the Introduction to Mathematical Proofs series that I am planning to write, I am going to give as many and as varied examples as possible. This series is intended to give an intuitive introduction to mathematical proofs for Grades 9-12 students.

Credit: Example 5 was adapted from Numbers: Rational and Irrational by Ivan Niven

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