Consider the following statements:

- $S_1:$ There exists infinite sets $A$, $B$, $C$ such that $A \cap (B \cup C)$ is finite.
- $S_2:$ There exists two irrational numbers $x$ and y such that $(x+y)$ is rational.

Which of the following is true about $S_1$ and $S_2$?

- Only $S_1$ is correct
- Only $S_2$ is correct
- Both $S_1$ and $S_2$ are correct
- None of $S_1$ and $S_2$ is correct

### 2 Comments

Hello Esha

Yes! it can be.

although in question there is 'There exist' which means if we find a single such combination of A,B,C that result finite outcome then statement will be true.

even if he says A,B,C are not disjoint sets then still we can find such arrangement that will make that statement true like

A={a^{*}} B={ab^{*}} and C={aab^{*}}

A∩(B∪C)=a^{* }∩ (a+aa)b^{* }= {a,aa}

## 4 Answers

### 5 Comments

Answer is C

S1 : take set universal set = U = {set of natural numbers} = { 1,2,3,4,5,6,7,8...... infinite } ,

set (A) = { set of even numbers } = { 2, 4, 6, 8, 10, 12 ......infinite } ,

set (B) = { set of prime numbers } = { 2, 3, 5, 7, 11, 13........infinite } ,

and set (C) = { set of odd numbers } = { 1, 3, 5, 7, 9, 11, 13........infinite }.

now A ∩ ( B ∪ C ) = { set of even numbers } ∩ ({ set of prime numbers } ∪ { set of odd numbers } )

= { set of even numbers } ∩ { { 2, 3, 5, 7, 11, 13........infinite } ∪ { 1, 3, 5, 7, 9, 11, 13........infinite } }

= { set of even numbers } ∩ { 1 , 2, 3, 5, 7, 9, 11, 13.......infinite }

= { set of even numbers } ∩ { { 2 } ∪ {1 , 3, 5, 7, 9, 11, 13.......infinite }}

= { set of even numbers } ∩ { { 2 } ∪ { set of odd numbers } }

= { set of even numbers } ∩ { { 2 } ∪ { set of odd numbers } } = { 2 } = only one element i.e. 2 = finite set

NOTE :- all, prime numbers are odd number except 2 .

S2: True becouse two irrational no. are -sqrt(2) and +sqrt(2) , when we add = -sqrt(2) +sqrt(2) = 0 ( 0 is a rational number)

for S1 both @Akash Kanase & @Mithlesh Upadhyay are correct

for S2:- Sum and product of two irrational number's may be either Rational or Irrational number . actually it depends upon what values you take for irrational numbers

For **Sum**

Case1:- a + b = c // prove that sum is rational

a= π , b= 1-π // here both a,b are irrational

c= π + (1-π ) = 1 c=1 which is rational

Case2:- a=π , b = π // // prove that sum is irrational

c= π +π = 2π c=2π which is irrational

for **product**

Case1:- a*b=c //prove product is rational

c= $\sqrt{2}$ * $\sqrt{2}$ = 2 c=2 which is rational

Case2:- a= π , b = π

c= π*π = π^2 c=π^2 which is irrational