1.10 Proving Set Equality

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: 1.9 Set Equality, 1.5 Set Builder Notation

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Proving Set Equality

Let

\begin{aligned} A & = {x \in R : \frac{d}{d x} [x^{3} - 12 x] = 0} \\ B & = {- 2, 2} \end{aligned}

Prove that $A = B$ .

Step 1. Show $A \subseteq B$ . Let $x \in A$
\begin{proof}

\begin{aligned} x \in A & ⟹ x \in R and \frac{d}{d x} [x^{3} - 12 x] = 0 & Let x = A \\ ⟹ 3 x^{2} - 12 = 0 \\ ⟹ 3 x^{2} = 12 & Add 12 to both sides \\ ⟹ x^{2} = \frac{12}{3} = 4 & Divide by 3 \\ ⟹ \sqrt{x^{2}} = \sqrt{4} & Take square root \\ ⟹ | x | = 2 & Solve absolute value \\ ⟹ x = \pm 2 \\ ⟹ x \in B & x = A, therefore \\ ⟹ A \in B \end{aligned}

\end{proof}
Step 2. Show $B \subseteq A$ . Let $x \in B$
\begin{proof}

\begin{aligned} x \in B & ⟹ x = \pm 2 \\ ⟹ \frac{d}{d x} [x^{3} - 12 x] |_{x = \pm_{2}} \overset{?}{=} 0 \\ ⟹ 3 x^{2} - 12 |_{x = \pm 2} {\begin{cases} 3 \cdot 2^{2} - 12 = 3 \cdot 4 - 12 = 0 \\ 3 \cdot (- 2)^{2} - 12 = 3 \cdot 4 - 12 = 0 \end{cases} \\ ⟹ B \subseteq A \end{aligned}

Thus, by definition, we know $A = B$ .
\end{proof}