9 12 15 18 21

Question

9. IfA-thirteen, 6, nine, 12, 15, xviii, 21), B 14, 8, 12, 16, 20 ),C= { 2, 4, 6, 8, 10, 12, 14, xvi }, D= {5, 10, 15, 20 }; observe(G) A - B(v) C-A(vi) D(ix)A-C-AD-B(ii) A D 4) B-A(vii) B- C 5) B D(ii)Vi1ViliC-B(ten)

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Solution

Given, A = { iii , vi , 9 , 12 , fifteen , xviii , 21 } , B = { 4 , 8 , 12 , xvi , 20 } , C = { 2 , 4 , 6 , 8 , ten , 12 , 14 , 16 } and D = { 5 , 10 , 15 , 20 } .

(i)

A − B means that the set contains elements of set A just not of set B .

So, set A − B can be expressed as

A − B = A − ( A ∩ B ) = { 3 , 6 , 9 , 12 , 15 , 18 , 21 } − ( { 3 , six , ix , 12 , 15 , 18 , 21 } ∩ { 4 , eight , 12 , 16 , twenty } ) = { three , six , ix , 12 , 15 , 18 , 21 } − { 12 } = { 3 , six , 9 , fifteen , 18 , 21 }

Hence, A − B = { 3 , 6 , nine , 15 , 18 , 21 } .

(ii)

A − C means that the set contains elements of set A but non of set C .

So, prepare A − C tin exist expressed as

A − C = A − ( A ∩ C ) = { 3 , 6 , ix , 12 , fifteen , 18 , 21 } − ( { 3 , vi , 9 , 12 , xv , 18 , 21 } ∩ { 2 , four , 6 , eight , ten , 12 , xiv , 16 } ) = { 3 , 6 , 9 , 12 , xv , 18 , 21 } − { 6 , 12 } = { 3 , 9 , 15 , 18 , 21 }

Hence, A − C = { 3 , ix , 15 , 18 , 21 } .

(iii)

A − D means that the set contains elements of fix A simply not of ready D .

So, set A − D can exist expressed as

A − D = A − ( A ∩ D ) = { 3 , 6 , 9 , 12 , 15 , eighteen , 21 } − ( { 3 , 6 , nine , 12 , fifteen , eighteen , 21 } ∩ { five , 10 , 15 , 20 } ) = { 3 , 6 , 9 , 12 , 15 , xviii , 21 } − { 15 } = { 3 , 6 , 9 , 12 , 18 , 21 }

Hence, A − D = { 3 , 6 , 9 , 12 , 18 , 21 } .

(iv)

B − A ways that the set contains elements of set B merely not of set A .

And so, set B − A can be expressed as

B − A = B − ( B ∩ A ) = { 4 , viii , 12 , xvi , 20 } − ( { 4 , 8 , 12 , sixteen , 20 } ∩ { 3 , 6 , 9 , 12 , 15 , eighteen , 21 } ) = { four , 8 , 12 , xvi , 20 } − { 12 } = { four , 8 , 16 , 20 }

Hence, B − A = { four , 8 , 16 , 20 } .

(5)

C − A means that the fix contains elements of set C but not of prepare A .

So, set C − A tin can be expressed every bit

C − A = C − ( C ∩ A ) = { ii , 4 , 6 , viii , ten , 12 , xiv , 16 } − ( { 2 , 4 , 6 , 8 , x , 12 , 14 , sixteen } ∩ { iii , 6 , 9 , 12 , 15 , xviii , 21 } ) = { 2 , iv , half-dozen , viii , 10 , 12 , fourteen , sixteen } − { half-dozen , 12 } = { two , 4 , 8 , 10 , 14 , 16 }

Hence, C − A = { 2 , 4 , 8 , x , 14 , 16 } .

(half dozen)

D − A means that the set contains elements of set D but non of set A .

So, set D − A tin can be expressed as

D − A = D − ( D ∩ A ) = { v , 10 , 15 , 20 } − ( { 5 , 10 , 15 , 20 } ∩ { iii , 6 , 9 , 12 , 15 , 18 , 21 } ) = { 5 , x , 15 , 20 } − { xv } = { five , 10 , 20 }

Hence, D − A = { 5 , x , 20 } .

(vii)

B − C ways that the set contains elements of set B but not of gear up C .

So, set B − C tin be expressed as

B − C = B − ( B ∩ C ) = { 4 , 8 , 12 , 16 , 20 } − ( { 4 , viii , 12 , 16 , twenty } ∩ { ii , 4 , 6 , 8 , 10 , 12 , 14 , 16 } ) = { 4 , 8 , 12 , 16 , 20 } − { 4 , 8 , 12 , sixteen } = { 20 }

Hence, B − C = { 20 } .

(viii)

B − D ways that the ready contains elements of set up B merely not of set D .

And so, set B − D can be expressed as

B − D = B − ( B ∩ D ) = { four , eight , 12 , 16 , twenty } − ( { iv , 8 , 12 , 16 , xx } ∩ { 5 , 10 , 15 , 20 } ) = { 4 , 8 , 12 , 16 , 20 } − { xx } = { 4 , viii , 12 , 16 }

Hence, B − D = { 4 , 8 , 12 , sixteen } .

(ix)

C − B means that the set contains elements of set C simply not of prepare B .

So, ready C − B can exist expressed as

C − B = C − ( B ∩ C ) = { two , 4 , 6 , 8 , ten , 12 , 14 , sixteen } − ( { 4 , 8 , 12 , 16 , 20 } ∩ { 2 , 4 , vi , 8 , ten , 12 , 14 , sixteen } ) = { 2 , 4 , 6 , 8 , 10 , 12 , xiv , 16 } − { 4 , 8 , 12 , 16 } = { 2 , six , x , fourteen }

Hence, C − B = { 2 , 6 , 10 , 14 } .

(10)

D − B ways that the prepare contains elements of set D only non of set B .

So, set D − B can be expressed as

D − B = D − ( B ∩ D ) = { 5 , 10 , fifteen , 20 } − ( { 4 , 8 , 12 , 16 , 20 } ∩ { 5 , 10 , 15 , 20 } ) = { 5 , ten , 15 , 20 } − { xx } = { five , ten , 15 }

Hence, D − B = { five , x , 15 } .

(xi)

C − D ways that the set contains elements of set C simply not of prepare D .

Then, set C − D tin can be expressed every bit

C − D = C − ( C ∩ D ) = { two , four , 6 , 8 , x , 12 , 14 , 16 } − ( { 2 , 4 , 6 , 8 , x , 12 , 14 , 16 } ∩ { v , 10 , fifteen , 20 } ) = { ii , 4 , 6 , eight , 10 , 12 , fourteen , 16 } − { 10 } = { 2 , iv , 6 , viii , 12 , 14 , xvi }

Hence, C − D = { ii , 4 , 6 , viii , 12 , 14 , 16 } .

(xii)

D − C means that the set contains elements of set up D but not of set C .

So, prepare D − C tin can be expressed every bit

D − C = D − ( C ∩ D ) = { 2 , 4 , 6 , viii , 10 , 12 , 14 , 16 } − ( { 2 , 4 , 6 , 8 , 10 , 12 , 14 , sixteen } ∩ { 5 , x , xv , twenty } ) = { 5 , x , 15 , twenty } − { x } = { five , 15 , 20 }

Hence, D − C = { 5 , fifteen , twenty } .

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