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

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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.

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i've got solution of Statement 1 is true but i was not clear about statement 2 ,even though it's easier one .so need to check all edge cases for statement 2.

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

Also for S1

A = Natural Numbers

B = Negative Integers

C = Negative Integers

B U C = Negative Integers

A ∩ ( B U C) = ∈

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@Akash

{∈}. $\neq$ ∈

So shouldn't result set of S1 be ∈?

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a*= { ∈, a, aa, aaa, aaaa,...........}

b*={ ∈, b,bb,bbb, bbbb,..............}     c*={ ∈, c,cc,ccc,cccc, .............}

hence, a* ∩ ( b* U c*) = { ∈ }  which is a finite set (cardinality =1)

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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 👍

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nice question and nice explaination

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

1.) Let us define:

A = { 3n | n ∈ ℕ } = { 3, 6, 9, 12, … }

B = { 3n + 1 | n ∈ ℕ } = { 4, 7, 10, … }

C = { 3n + 2 | n ∈ ℕ } = { 5, 8, 11, … }

A∩(BUC) = A∩(elements not in A) = ∅. So the intersection is finite (in fact, empty).

 ∴ S1 is True.

2.) take x = $\sqrt{2}$ and y = -$\sqrt{2}$

( x + y ) = $\sqrt{2}$ + ( -$\sqrt{2}$ ) = 0 . since 0 is rational number.

∴ S2 is True.

Finally, Option ( C ) is correct.

Answer 5

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 6

S1 : TRUE

eg: A= set of all prime numbers : { 2, 3, 5, 7, 11,13.....}

B= set of multiples of 2: {2,4,6,8,10...}

C= set of multiples of 3 : {3,6,9,12,15...}

A ∩ (B U C)= {2,3}

 

S2: TRUE

x= 2+ √3  y= 2-√3

x+y = 4 which is rational