Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
1st Edition
ISBN: 9780137548552
Author: Michael Akritas
Publisher: PEARSON+
Question
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Chapter 2.2, Problem 8E
To determine

Prove that (AB)(BA)=(AB)(BA) from Exercise 4.

Prove that (AB)c=AcBc from Exercise 6.

Prove that (AB)C=(AC)(BC) from Exercise 7.

Expert Solution & Answer
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Explanation of Solution

Calculation:

Define the events A and B. The events A and B are considered as equal when AB and BA. The event A would be subset of B for an element x, if xA then xB is true.

Pair of events (AB)(BA)=(AB)(BA):

Let x be an element.

x(AB)(BA)x(AB)orx(BA)(xAbutxB)or(xBbutxA)xAorxBx(AB)andx(AB)x(AB)(AB)

The event (AB)(BA)(AB)(BA) because x(AB)(BA) then x(AB)(AB) is also true.

x(AB)(AB)x(AB)andx(AB)xAorxB(xAbutxB)or(xBbutxA)x(AB)orx(BA)x(AB)(BA)

The event (AB)(BA)(AB)(BA) because x(AB)(AB) then x(AB)(BA) is also true.

Hence, it is proved that (AB)(BA)=(AB)(BA).

Pair of events (AB)c=AcBc:

x(AB)cxABx(AB)orx(BA)orx(AB)c[x(AB)orx(AB)c]or[x(BA)orx(AB)c]xBcorxAcx(AcBc)

The event (AB)cAcBc because x(AB)c then x(AcBc) is also true.

x(AcBc)xBcorxAc[x(AB)orx(AB)c]or[x(BA)orx(AB)c]x(AB)orx(BA)orx(AB)cxABx(AB)c

The event AcBc(AB)c because x(AcBc) then x(AB)c is also true.

Hence, it is proved that (AB)c=AcBc.

Pair of events (AB)C=(AC)(BC):

x(AB)Cx(AB)orxC[xAandxB]orxC[xAorxC]and[xBorxC]xACandxBCx(AC)(BC)

The event (AB)C(AC)(BC) because x(AB)C then x(AC)(BC) is also true.

x(AC)(BC)xACandxBC[xAorxC]and[xBorxC][xAandxB]orxCx(AB)orxCx(AB)C

The event (AC)(BC)(AB)C because x(AC)(BC) then x(AB)C is also true.

Hence, it is proved that (AB)C=(AC)(BC).

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