Skip to main content

Section 3.2 Using Other Logical Equivalencies

There are many logical equivalencies, but fortunately, only a small number are frequently used when trying to construct and write proofs. Most of these are listed in TheoremĀ 1.1. We will illustrate the use of one of these logical equivalencies with the following proposition:

For all real numbers \(a\) and \(b\text{,}\) if \(a \ne 0\) and \(b \ne 0\text{,}\) then \(ab \ne 0\text{.}\)
First, notice that the hypothesis and the conclusion of the conditional statement are stated in the form of negations. This suggests that we consider the contrapositive. Care must be taken when we negate the hypothesis since it is a conjunction. We use one of De Morgan's Laws as follows

\begin{equation*} \neg (a \ne 0 \wedge b \ne 0) \equiv (a = 0) \vee (b = 0)\text{.} \end{equation*}

So the contrapositive is:

For all real numbers \(a\) and \(b\text{,}\) if \(ab = 0\text{,}\) then \(a = 0\) or \(b = 0\text{.}\)

The contrapositive is a conditional statement in the form \(X \to (Y \vee Z)\text{.}\) The difficulty is that there is not much we can do with the hypothesis \((ab = 0)\) since we know nothing else about the real numbers \(a\) and \(b\text{.}\) However, if we knew that \(a\) was not equal to zero, then we could multiply both sides of the equation \(ab = 0\) by \(\frac{1}{a}\text{.}\) This suggests that we consider using the following logical equivalency based on a result in TheoremĀ 1.1:

\begin{equation*} X \to (Y \vee Z) \equiv (X \wedge \neg Y) \to Z\text{.} \end{equation*}

Proof.

We will prove the contrapositive of this proposition, which is

For all real numbers \(a\) and \(b\text{,}\) if \(ab = 0\text{,}\) then \(a = 0\) or \(b = 0\text{.}\)
This contrapositive, however, is logically equivalent to the following:
For all real numbers \(a\) and \(b\text{,}\) if \(ab = 0\) and \(a \ne 0\text{,}\) then \(b = 0\text{.}\)

To prove this, we let \(a\) and \(b\) be real numbers and assume that \(ab = 0\) and \(a \ne 0\text{.}\) We can then multiply both sides of the equation \(ab = 0\) by \(\frac{1}{a}\text{.}\) This gives

\begin{equation*} \frac{1}{a}(ab) = \frac{1}{a} \cdot 0\text{.} \end{equation*}

We now use the associative property on the left side of this equation and simplify both sides of the equation to obtain

\begin{align*} \left( \frac{1}{a} \cdot a \right) b \amp = 0\\ 1 \cdot b \amp = 0\\ b \amp = 0 \end{align*}

Therefore, \(b = 0\) and this proves that for all real numbers \(a\) and \(b\text{,}\) if \(ab = 0\) and \(a \ne 0\text{,}\) then \(b = 0\text{.}\) Since this statement is logically equivalent to the contrapositive of the proposition, we have proved the proposition.