Proof by Contradiction: Odd and Its Square

We have had several examples on proof by contradiction in this blog like irrationality of square root of 3 and the sum of root of 2 and root of 3. In this post, we will have another detailed example.

In proof by contradiction, we want to change the statement “if P then Q” to “if NOT Q then NOT P.” These two statements are equivalent, so if we can prove the latter, then we have proved the former. Consider the following theorem.

Theorem: For any integer $latex x$, if $latex x^2$ is odd, then $latex x$ is odd.

We can assign the statements above to P (the hypothesis) and Q (the conclusion) as follows.

P: $latex x^2$ is odd

Q: $latex x$ is odd.

If we are going to change this statement to “if NOT Q then NOT P,” then, we have to find the opposite of P and opposite of Q.  Continue reading

Tangent Theorem 2

In the previous post on tangent theorem, we have learned that if a line is perpendicular to a radius of a circle at a point on the circle, then the line is tangent to the circle. In this post, we prove the converse of this statement. We prove that if a line is tangent to a circle, then it is perpendicular to the radius at the point of tangency. Note that the point of tangency is the point where the circle and the line intersect.


Line m is tangent to circle O at point A


Line m is perpendicular to OA


We will use proof by contradiction to prove the statement above.  Continue reading

Proof by Contradiction: 0.999… = 1

In the previous post, we proved that

$latex \displaystyle a < \frac{a + b}{2}$

both by direct and indirect proof. The proof that we have used in the previous post was proof by contradiction. In proof by contradiction, we show that by assuming the proposition to false would imply a contradiction.  One of the most famous of this proof is the proof that there are infinitely many prime numbers.

In this post, we will prove that $latex 0.999 \cdots 1$. Recall that we already proved this theorem using direct proof.  In proving by contradiction, first, we assume that

$latex 0.999 \cdots < 1$

then find a contradiction somewhere. We then conclude that our assumption cannot be be false.  Continue reading