Skip to main content

Section 3.3 Proofs of Biconditional Statements

One of the logical equivalencies in TheoremĀ 1.1 is the following one for biconditional statements.

\begin{equation*} \left( P \leftrightarrow Q \right) \equiv \left( P \to Q \right) \wedge \left( Q \to P \right) \end{equation*}

This logical equivalency suggests one method for proving a biconditional statement written in the form ā€œ\(P\) if and only if \(Q\text{.}\)ā€ This method is to construct separate proofs of the two conditional statements \(P \to Q\) and \(Q \to P\text{.}\)

We will illustrate this with a proposition about right triangles.

Recall that the Pythagorean Theorem for right triangles states that if \(a\) and \(b\) are the lengths of the legs of a right triangle and \(c\) is the length of the hypotenuse, then \(a^2 + b^2 = c^2\) . We also know that the area of any triangle is one-half the base times the altitude. So for the right triangle we have described, the area is \(A = \frac{1}{2} ab\text{.}\)

Proof.

We assume that we have a right triangle where \(a\) and \(b\) are the lengths of the legs of a right triangle and \(c\) is the length of the hypotenuse. We will prove that this right triangle is an isosceles triangle if and only if the area of the right triangle is \(\frac{1}{4} c^2\) by proving the two conditional statements associated with this biconditional statement.

We first prove that if this right triangle is an isosceles triangle, then the area of the right triangle is \(\frac{1}{4} c^2\text{.}\) So we assume the right triangle is an isosceles triangle. This means that \(a = b\text{,}\) and consequently, \(A = \frac{1}{2} a^2\text{.}\) Using the Pythagorean Theorem, we see that

\begin{equation*} c^2 = a^2 + a^2 = 2a^2\text{.} \end{equation*}

Hence, \(a^2 = \frac{1}{2} c^2\text{,}\) and we obtain \(A = \frac{1}{2} a^2 = \frac{1}{4} c^2\text{.}\) This proves that if this right triangle is an isosceles triangle, then the area of the right triangle is \(\frac{1}{4} c^2\text{.}\)

We now prove the converse of the first conditional statement. So we assume the area of this isosceles triangle is \(A = \frac{1}{4} c^2\) , and will prove that \(a = b\text{.}\) Since the area is also \(\frac{1}{2} ab\text{,}\) we see that

\begin{align*} \frac{1}{4} c^2 \amp = \frac{1}{2}ab\\ c^2 \amp = 2ab \end{align*}

We now use the Pythagorean Theorem to conclude that \(a^2 + b^2 = 2ab\text{.}\) So the last equation can be rewritten as follows:

\begin{align*} a^2 - 2ab + b^2 \amp = 0\\ (a - b)^2 \amp = 0\text{.} \end{align*}

The last equation implies that \(a = b\) and hence the right triangle is an isosceles triangle. This proves that if the area of this right triangle is \(A = \frac{1}{4} c^2\text{,}\) then the right triangle is an isosceles triangle.

Since we have proven both conditional statements, we have proven that this right triangle is an isosceles triangle if and only if the area of the right triangle is \(\frac{1}{4} c^2\text{.}\)