4,999 views

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

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.?

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}

### Subscribe to GO Classes for GATE CSE 2022

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

Also for S1

A = Natural Numbers

B = Negative Integers

C = Negative Integers

B U C = Negative Integers

A ∩ ( B U C) = ∈
@Akash

{∈}. $\neq$ ∈

So shouldn't result set of S1 be ∈?
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)
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
Will s1 always give the cardinality of 1?

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)

edited by

if we take pI as 22/7 then PI+PI means (22/7)+(22/7) is irrational..correct me if i wrong?

sir,i read it from here..can you give another example?

22/7 is no longer irrational.
And we cannot take $\pi$ as 22/7, we can only approximate $\pi$ as 22/7, and the moment we do so, it is no longer irrational.
Sir ,but 2π is also not rational number.

If S2 is true,then 2π should be rational
"Adding 2 integers can give 0"

is this true or false?
Yes, adding 2 integers can give zero.
@Manasi  $2 + 4 = 6$ and $6 \neq 0$. Guess you got it.
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

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

ans should be D.

intersection of infinite sets need not be finite.

addition of two irrational is also irrational.
by

### 1 comment

Addition of two irrational no can be rational.