Question

a.      Let A and B and C be sets, prove that A∩(BC) = (A∩B) ( A∩C). 

b. a.      The number x is multiplied by itself and the result y is less than the number x.

For example, x*x = x² = y and y < x. 

i.   State a set name in which numbers behave in such an assertion above.

ii.     State any number in the stated set in (a) above and verify the assertion. 

Answer 1

a. Distributive law $A\cap(B\cup C)=(A\cap B)\cup (A\cap C)$

Suppose that $x\in A\cap (B\cup C).$ Then $x\in A$ and $x\in B\cup C.$ By the definition of union, it follows that $x\in A$ and $x\in B$ or $x\in C$ (or both).

In other words, we know that the compound proposition $(x\in A) \land((x\in B)\lor(x\in C))$ is true.

By the distributive law for conjunction over disjunction, it follows that $((x\in A) \land(x\in B))\lor((x\in A) \land(x\in C)).$

We conclude that either $x\in A$ and $x\in B$ or $x\in A$ and $x\in C$ .

By the definition of intersection, it follows that $x\in A\cap B$ or $x\in A\cap C.$

Using the definition of union, we conclude that $x\in (A\cap B)\cup (A\cap C).$ We conclude that $A\cap(B\cup C)\subseteq(A\cap B)\cup (A\cap C).$

Now suppose that $x\in A\cap(B\cup C).$ Then, by the definition of union, $x\in A\cap B$ or $x\in A\cap C.$

By the definition of intersection, it follows that $x\in A$ and $x\in B$ or that $x\in A$ and $x\in C.$

From this we see that $x\in A,$ and $x\in B$ or $x\in C.$ Consequently, by the definition of union we see that $x\in A$ and $x\in B\cup C.$ Furthermore, by the definition of intersection, it follows that $x\in A\cap (B\cup C).$

We conclude that $(A\cap B)\cup (A\cap C)\subseteq A\cap(B\cup C).$

This completes the proof of the identity.

The membership table for these combinations of sets is shown in Table.

$$\begin{matrix} A & B & C & B\cup C & A\cap (B\cup C) \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{matrix}$$

$$\begin{matrix} A & B & C & A\cap B & A\cap C & (A\cap B)\cup (A\cap C) \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{matrix}$$

Because the columns for $A\cap(B\cup C)$ and $(A\cap B)\cup (A\cap C)$ are the same, the identity is valid.

b. $y=x^2<x$

i.

$$x^2<x=>x^2-x<0=>x(x-1)<0$$

$$=>x\in(0, 1)$$

The opened interval $(0, 1).$

ii. Let $x=\dfrac{1}{2}.$ Then

$$y(\dfrac{1}{2})=(\dfrac{1}{2})^2=\dfrac{1}{4}<\dfrac{1}{2}=x$$