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+28 votes
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Consider the following statements:

  • $S1:$ There exists infinite sets $A$, $B$, $C$ such that $A \cap (B \cup C)$ is finite.
  • $S2:$ There exists two irrational numbers $x$ and y such that $(x+y)$ is rational.

Which of the following is true about $S1$ and $S2$?

  1. Only $S1$ is correct
  2. Only $S2$ is correct
  3. Both $S1$ and $S2$ are correct
  4. None of $S1$ and $S2$ is correct
asked in Set Theory & Algebra by Veteran (59.5k points)
edited by | 1.6k views
0
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.?
+1

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

+34 votes
Best answer

$S1$:

CounterExample : $a^{*} \cap ( b^{*} \cup c^{*})$ , Here $a^{*},b^{*},c^{*}$ all are infinite,$\text{ Result is finite language} = \{\in\}.$

So S1 is True

$S2$->

Counterexample : $x= 1+ \sqrt {2}\;, y = 1- \sqrt{2}$ ,
$x+y = 2$.

So S2 is True.

Answer -> $C$

answered by Boss (42.8k points)
edited by
0
Also for S1

A = Natural Numbers

B = Negative Integers

C = Negative Integers

B U C = Negative Integers

A ∩ ( B U C) = ∈
0
@Akash

{∈}. $\neq$ ∈

So shouldn't result set of S1 be ∈?
0
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)
+27 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)

answered by Loyal (6k points)
reshown by
+1

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?

+4
22/7 is no longer irrational.
+4
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.
0
Sir ,but 2π is also not rational number.

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

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

Option (C) is the correct answer.

answered by Loyal (6.7k points)
–5 votes
ans should be D.

intersection of infinite sets need not be finite.

addition of two irrational is also irrational.
answered by Loyal (8.2k points)
+2
Addition of two irrational no can be rational.
Answer:

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