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Home »  Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: (i) A x C ⊂ B x D (ii) A x (B ∩ C) = (A x B) ∩ (A x C)
 Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: (i) A x C ⊂ B x D (ii) A x (B ∩ C) = (A x B) ∩ (A x C)
CBSE | Class 11 | Exercise 2.2 | Maths | RD Sharma | Relations
 Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: (i) A x C ⊂ B x D (ii) A x (B ∩ C) = (A x B) ∩ (A x C)

Solution:

We are given that A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) A x C ⊂ B x D

Let us first consider LHS A x C

A\text{ }\times \text{ }C\text{ }=\text{ }\left\{ 1,\text{ }2 \right\}\text{ }\times \text{ }\left\{ 5,\text{ }6 \right\}

A\text{ }\times \text{ }C=\text{ }\left\{ \left( 1,\text{ }5 \right),\text{ }\left( 1,\text{ }6 \right),\text{ }\left( 2,\text{ }5 \right),\text{ }\left( 2,\text{ }6 \right) \right\}

Now, taking RHS

B\text{ }\times \text{ }D\text{ }=\text{ }\left\{ 1,\text{ }2,\text{ }3,\text{ }4 \right\}\text{ }\times \text{ }\left\{ 5,\text{ }6,\text{ }7,\text{ }8 \right\}

B\text{ }\times \text{ }D\text{ }=\text{ }\left\{ \left( 1,\text{ }5 \right),\text{ }\left( 1,\text{ }6 \right),\text{ }\left( 1,\text{ }7 \right),\text{ }\left( 1,\text{ }8 \right),\text{ }\left( 2,\text{ }5 \right),\text{ }\left( 2,\text{ }6 \right),\text{ }\left( 2,\text{ }7 \right),\text{ }\left( 2,\text{ }8 \right),\text{ }\left( 3,\text{ }5 \right),\text{ }\left( 3,\text{ }6 \right),\text{ }\left( 3,\text{ }7 \right),\text{ }\left( 3,\text{ }8 \right),\text{ }\left( 4,\text{ }5 \right),\text{ }\left( 4,\text{ }6 \right),\text{ }\left( 4,\text{ }7 \right),\text{ }\left( 4,\text{ }8 \right) \right\}

Since, all elements of A × C is in B × D.

∴We can say A × C ⊂ B × D

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let us first consider the LHS of above equation

\left( B~\cap ~C \right)=~\varnothing

A\text{ }\times \text{ }\left( B~\cap ~C \right)=\left\{ 1,\text{ }2 \right\}\text{ }\times \varnothing

A\text{ }\times \text{ }\left( B~\cap ~C \right)=\varnothing

Now, taking RHS

\left( A\text{ }\times \text{ }B \right)\text{ }=\text{ }\left\{ 1,\text{ }2 \right\}\text{ }\times \text{ }\left\{ 1,\text{ }2,\text{ }3,\text{ }4 \right\}

\left( A\text{ }\times \text{ }B \right)=\text{ }\left\{ \left( 1,\text{ }1 \right),\text{ }\left( 1,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left( 1,\text{ }4 \right),\text{ }\left( 2,\text{ }1 \right),\text{ }\left( 2,\text{ }2 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 2,\text{ }4 \right) \right\}

\left( A\text{ }\times \text{ }C \right)\text{ }=\text{ }\left\{ 1,\text{ }2 \right\}\text{ }\times \text{ }\left\{ 5,\text{ }6 \right\}

\left( A\text{ }\times \text{ }C \right)=\text{ }\left\{ \left( 1,\text{ }5 \right),\text{ }\left( 1,\text{ }6 \right),\text{ }\left( 2,\text{ }5 \right),\text{ }\left( 2,\text{ }6 \right) \right\}

As it is evident that there is no common element between A × B and A × C, we can write:

\left( A\text{ }\times \text{ }B \right)~\cap ~\left( A~\times ~C \right)\text{ }=~\varnothing

\therefore A\text{ }\times \text{ }\left( B~\cap ~C \right)\text{ }=\text{ }\left( A~\times ~B \right)~\cap ~\left( A~\times ~C \right)

i

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