GATE CSE 2001 | Question: 2.2

Question

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 $S_2$?

• A. Only $S_1$ is correct
• B. Only $S_2$ is correct
• C. Both $S_1$ and $S_2$ are correct
• D. None of $S_1$ and $S_2$ is correct

Comments

CAN THERE BE A  POSSIBILITY IN S1 THAT A,B,C ARE NOT MUTUALLY EXCLUSIVE SETS(BECAUSE ONLY INFINITE IS MENTIONED AND NOTHING HAS BEEN SAID ABOUT THEIR RELATION WITH EACH OTHER)-THEN S1 CAN BE EITHER TRUE AND EITHER FALSE.?

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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}

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Video Solution

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for statement $S_{1}$ , let set $A$ be set of odd integers, set $B$ be set of multiple 2 and set $C$ be set of multiple 4.

$A\cup (B\cap C$) is empty set.

Answer 1

$S_1:$ $a^{*} \cap ( b^{*} \cup c^{*}) = \{\epsilon\}$ 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: $C$

Comments

S1 is true for the term 'There exist', right???

otherwise I can show B U C = a^n.b^m | n,m >0  &    A = a^n.b^m | n=m

then S1 is infinite

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Will s1 always give the cardinality of 1?

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Nice explanation 👍

Answer 2

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)

Comments

Yes, adding 2 integers can give zero.

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@Manasi  $2 + 4 = 6$ and $6 \neq 0$. Guess you got it.

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Ohh. Yes. I was thinking wrong.actually statement can be true for at least one case also.not necessary all cases.

Assumed : 2+(-2)=0

Answer 3

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

 

Answer 4

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.