Question

IFA=(x:x is a letter of word ALGEBRA)
B=(x:x is a letter of word BANGLORE)
Find a∩b

Answer 1

Required answer:-

A∩B = { A L G E B R}

Given :-

A = {x :x is a letter of word ALGEBRA}

B = {x :x is a letter of word BANGLORE}

To find :-

A∩B

Solution:-

They given the sets in set builder form we shall convert into roaster form So,

A = { A L G E B R A}

B = {B A N G L O R E}

Now we need to find the A∩B That means common letters in the both A and B

A∩B = {A L G E B R A} ∩ { B A N G L O R E }

Here the common letters in both sets are

A, L , G , E , B , R  

So,

A∩B = { A L G E B R}

Here in ALGEBRA "A" is repeated for 2 times we have to take only for 1 time

So,

A∩B = { A L G E B R}

Know  more :-

How to find the intersection and union of a given set ?

Intersection means :- If the two sets were given the common elements that present in both are called intersection (∩) symbol.

Eg :-

A = {V , I , B , G , Y , O , R }

B = {L, S , U , D , P , N, O, R   }

A∩ B = { O, R} i.e these elements are present in both sets i.e called as common elements.

____________________

Union of a set :- If two sets were given  A, B The elements should belongs to A or B or both A nd B (U) symbol.

Eg :-

A = {A, E, I ,O , U }

B = {B, C , D , F , G }

A U B = { A,B , C , D, E, F, G , I , O , U} i.e both elements of A, B are present.

Answer 2

Answer:

${ \large{ \underline{ \sf{Given}}}}$

A = {x:x is a letter of word ALGEBRA}

B= {x:x is a letter of word BANGLORE}

${ \large{ \sf { \underline{To \: Find}}}}$

Union of A and B

${ \large{ \sf{ \underline{Solution}}}}$

• Firstly they given in set builder form, so we have to write the elements in roaster form of given sets. Then, we can find the intersection of set of A and B. Lets start our solution.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{★ ━━━━━━━━━━★}$

Roaster form :

${ \dashrightarrow{ \sf{A = [A, L, G, E, B, R]}}}$

${ \dashrightarrow{ \sf{B =[ B, A, N, G, L, O, R, E]}}}$

Intersection:

Here intersection of two sets means common elements. The common elements of two sets will be the intersection.

${ \dashrightarrow{ \sf{A∩B}}} \\ \\ {\dashrightarrow{\sf{A∩B = [A, L, G, E, B, R]∩ [B, A, N, G, L, O, R, E]}}} \\ \\ \dashrightarrow{\sf{A∩B = [A, L, G, E, B, R]}}$

$\therefore {\sf{A∩B = [A, L, G, E, B, R]}}$