Question

14.Let A = X belongs W/X<2 , B= X belongs N/1<x less than or equal to 4 and C= 3,5 verify that (AUB) x C = (AxC) U (BxC) ​

Answer 1

Step-by-step explanation:

Given :-

A = { X:X€W,X<2 }

B = { X:X€N,1<X≤4 }

C = { 3,5 }

To find :-

Verify that (AUB) x C = (AxC) U (BxC) ?

Solution :-

Given that

A = { X:X€W,X<2 }

=> A = { 0,1 }

B = { X:X€N,1<X≤4 }

=> B = { 2,3,4 }

C = {3,5}

Now

I) (AUB)× C :-

AUB = { 0,1 } U { 2,3,4 } = { 0,1,2,3,4 }

(AUB) × C

=> {0,1,2,3,4} × {3,5}

=> { (0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),

(4,3),(4,5)} ---------------(1)

II) (A×C) U (B×C) :-

A×C = { 0,1 } × { 3,5 }

=> { (0,3),(0,5),(1,3),(1,5) }

B×C = { 2,3,4 } × { 3,5 }

=> { (2,3),(2,5),(3,3),(3,5),(4,3),(4,5) }

(A×C) U (B×C)

=> { (0,3),(0,5),(1,3),(1,5) } U { (2,3),(2,5),(3,3),(3,5),(4,3),(4,5) }

=> { (0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),

(4,3),(4,5)} ---------------(2)

From (1) & (2)

(AUB) x C = (AxC) U (BxC)

Answer:-

Verified (AUB) x C = (AxC) U (BxC)

Used formulae:-

→ AUB is the set of elements in either A or in B or in both A and B.

→ AUB = { X: X€A or X€ B }

→ A×B is the set of all order pairs in which First element belongs to first set and second element belongs to second set.

→ A×B = { (a,b) : a € A and b € B }

→ Where, € represents " belongs to "

→ N is the set of natural numbers.

→ N = { 1,2,3,...}

→ W is the set of whole numbers

→ W = { 0,1,2,3,...}