1.8 Proving Subsets

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: 1.7 Subset

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

Proof for subsets

Let

\begin{aligned} A = {\frac{π}{4}, \frac{9 π}{4}} \\ B = {θ \in [0, 6 π] : 4 \cos (θ - \frac{π}{4}) - 4 = 0} \end{aligned}

Prove that $A \subseteq B$ .

To prove $A \subseteq B$ , we follow our nose.
\begin{proof}
Let

x \in A ⟹ x = \frac{π}{4} or x = \frac{9 π}{4}

Case 1. $x = \frac{π}{4}$

\begin{aligned} x = \frac{π}{4} & ⟹ \cos (θ - \frac{π}{4}) = \cos (0) \\ ⟹ 4 \cos (\frac{π}{4} - \frac{π}{4}) = 4 \cdot 1 \\ ⟹ 4 \cos (\frac{π}{4} - \frac{π}{4}) - 4 = 0 \\ ⟹ \frac{π}{4} \in B \end{aligned}

Case 2. $x = \frac{9 π}{4}$

\begin{aligned} x = \frac{9 π}{4} & ⟹ 4 \cos (\frac{9 π}{4} - \frac{π}{4}) - 4 & = 4 \cos (2 π) - 4 \\ = 4 \cdot 1 - 4 \\ = 0 \\ ⟹ \frac{9 π}{4} \in B \end{aligned}

Thus, we showed if $x \in A$ , then $x \in B$ , which implies that $A \subseteq B$ by definition.
\end{proof}