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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 $S_2$?

  1. Only $S_1$ is correct
  2. Only $S_2$ is correct
  3. Both $S_1$ and $S_2$ are correct
  4. None of $S_1$ and $S_2$ is correct
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Best answer
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$S_1:$ $a^{*} \cap ( b^{*} \cup c^{*}) = \{\epsilon\}$ which is finite but $a^{*},b^{*}$ and $c^{*}$ all are infinite.

So $S_1$ is True.

$S_2:$ Let, $x= 1+ \sqrt {2}\;, y = 1- \sqrt{2},$ both of which are irrational.
$x+y = 2,$ which is rational.

So $S_2$ is True.

Answer: $C$
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41 votes

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)

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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

 

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Answer: (C)

Explanation: S1: A ∩ (B ∪ C)
Here S1 is finite where A, B, C are infinite
We’ll prove this by taking an example.
Let A = {Set of all even numbers} = {2, 4, 6, 8, 10…}
Let B = {Set of all odd numbers} = {1, 3, 5, 7………..}
Let C = {Set of all prime numbers} = {2, 3, 5, 7, 11, 13……}

 

B U C = {1, 2, 3, 5, 7, 9, 11, 13……}
A ∩ (B ∪ C)
Will
be equals to: {2} which is finite.
I.e. using A, B, C as infinite sets the statement S1 is finite.

 

So, statement S1 is correct.
S2: There exists two irrational numbers x, y such that (x+y) is rational

 

To prove this statement as correct, we take an example.
Let X = 2- √3, Y = 2+√3 => X, Y are irrational


X+Y = 2+√3 + 2- √3 = 2+2 = 4
So, statement S2 is also correct.
Answer is Option C

Both Statements S1, S2 are correct.

Answer:

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