Mathematics

Set Theory and Relations

355 Questions

Set theory involves the study of collections of objects and includes operations like union, intersection, and finding complements. Questions cover power sets, Cartesian products, and properties of empty sets. This foundational mathematical topic frequently appears in various competitive exams and university entrance tests.

set operationspower setscartesian productsset complementsproperties of empty setsset partitions

Set Theory and Relations Questions

Multiple choice general knowledge math & puzzles

Number set, that combines the integers, rational numbers, and irrational numbers into one big set of numbers. This set is called the

  1. Big fat set o' numbers

  2. Real numbers

  3. Collective number set

  4. Quaternary numbers
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The set of Real numbers (R) is the union of rational numbers (which include integers) and irrational numbers. 'Collective number set' is not a standard mathematical term.

Multiple choice general knowledge science & technology

The intersection of two fuzzy sets is calculated by the ____ of corresponding membership values

  1. minimum

  2. maximum

  3. average

  4. difference
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In fuzzy set theory, the intersection operation uses the minimum (MIN) of the membership values from each set. This means an element belongs to the intersection only to the degree that it belongs to BOTH sets - the more restrictive condition. The union uses maximum, so (B) is incorrect for intersection. Average (C) and difference (D) are not standard fuzzy set operations.

Multiple choice general knowledge

Which of these is true about a set?

  1. The elements are kept in order

  2. They can only hold numbers

  3. There are no duplicates

  4. All of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A set is defined as a collection that cannot contain duplicate elements - this is the fundamental property that distinguishes sets from lists. Sets can store any type of data, not just numbers, and they do not maintain insertion order.

Multiple choice technology mainframe

A mapset can contain only one map.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A mapset can contain multiple maps, not just one. A single mapset is a collection of one or more maps that are logically grouped together and loaded together into CICS. This allows related screens to be organized and maintained as a unit.

Multiple choice technology mainframe

Set linkage for a set can be a combination of

  1. NEXT

  2. FIRST

  3. PRIOR

  4. LAST

  5. All of the above.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Set linkage in IDMS determines how records in a set are connected: NEXT links each member to the following member (forward chain), and PRIOR links each member to the previous member (backward chain). FIRST and LAST are not valid linkage options; the linkage is about connection direction between members, not position.

Multiple choice

Consider the set {a, b, c} with binary operators + and × defined as follows

+ a b c
a b a c
b a b c
c a c b
* a b c
a a b c
b b c a
c c c b

For example, a + c = c, c + a = a, c ×b = c and b × c = a. Given the following set of equations: (a × x) + (a × y) = c (b × x) + (c × y) = c

The number of solution(s) (i.e., pair(s) (x, y) that satisfy the equations) is

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
+ a b c
a b a c
b a b c
c a c b
* a b c
a a b c
b b c a
c c c b

We can write from given expression a + a = b,         a + b = a,         a + c = c b + a = a,         b + b = b,         b + c = c c + a = a           c + b = c           c + c = b

a x a = a,          a x b = b,          a x c = c b x a = b,          b x b = c,          b x c = a c x a = c,          c x b = c,          c x c = b a x x + a x y = c                        --- (1) b x x + c x y = c             --- (2) First we see, these are the equation which give result c. a + c = c b + c = c c + b = c For equation (A), we find a x x = b  x = b We find a x x = a then x = a, a x y = c, y = c, x = a,b,c  a x y = c then y = c, a x x = c, x = c, x = c, y = a,b a x y = b, y = b for equation (B), we find b x x = b, c x y = c, b x x = a then x = c x = a, y = a, b    c x y = c then y = a, b b x x = c then    x = b    we find c x y = b then    y = c    x = a,b,c y = a,b,c so, total number of solution pairs (x,y) is 3.

Multiple choice

The set (1, 2, 3, 5, 7, 8, 9) under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?

  1. It is not closed

  2. 2 does not have an inverse

  3. 3 does not have an inverse

  4. 8 does not have an inverse
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The definition of binary operation multiplication modulo 10 is a follows aob = (a x b)% 10 And the identity element for the given set = 1 For closure property, 2 or 7 = 4 $\notin$ set So, the set is not closed Here, 2 and 8 do not have an inverse, so 307 = 1 Hence, 7 is the inverse of 3.

Multiple choice

Let $X,Y,Z$ be sets of sizes x, y and z respectively. Let $W = X \times Y$ and $E$ be the set of all subsets of $W$. The number of functions from $Z$ to $E$ is
  1. $z^{2^{xy}}$
  2. $z \times 2^{xy}$
  3. $z^{2^{x+y}}$
  4. $2^{xyz}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The number of functions from a set A to a set B Where, |A | = m and, |B| = n is nm Also W = X x Y $\Rightarrow$| W | = xy $\because$| E | = 2xy |Z| = z So, the number of functions from Z to E is (2xy)z = 2xyz

Multiple choice

Let $E, F$ and $G$ be finite sets. Let

$X = (E ∩ F) - (F ∩ G)$ and $Y = (E - (E ∩ G)) - (E - F)$.

Which one of the following is true?
  1. $X ⊂ Y$
  2. $X ⊃ Y$
  3. $X = Y$
  4. $X - Y ≠ \phi$ and $Y - X ≠ \phi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

X = (E ∩ F) - (F ∩ G) = elements in E and F but not in G. Y = (E - (E ∩ G)) - (E - F) = elements in E that are not exclusively in E (not in G) and are in F. Simplifying Y: elements in E that are in F but not in G. Therefore X = Y. Both represent elements that are in E and F but not in G.

Multiple choice

Let S = {1, 2, 3,........, m}, m >3. Let X1.........Xn be subsets of S each of size 3. Define a function f from S to the set of natural numbers as, f(i) is the number of sets Xj that contain the element i. That is $f(i)=\left | \left\{j \mid i\in X_j \right\} \right|$ then $ \sum_{i=1}^{m} f(i)$ is:

  1. 3m

  2. 3n

  3. 2m + 1

  4. 2n + 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Suppose, S = {1, 2, 3, 4} m = 4 X1 = { 1, 2, 3, } X2 = { 1, 3, 4} n = 2 $\therefore f(i)= f(1) + f(2) + f(3) + f(4) $ = 2 + 1 + 2 + 1 = 6 = 3 x 2 = 3n

Multiple choice

Given a set of elements N = {1, 2, ..., n} and two arbitrary subsets A⊆N and B⊆N, how many of the n! permutations $\pi$ from N to N satisfy min($\pi$(A)) = min($\pi$(B)), where min(S) is the smallest integer in the set of integers S, and $\pi$(S) is the set of integers obtained by applying permutation $\pi$ to each element of S?

  1. (n - |A ∪ B|) |A| |B|
  2. $(|A|^2 + |B|^2)n^2$
  3. $n! \frac{|A ∩ B|}{|A ∪ B|}$
  4. $\frac{|A ∩ B|^2}{^n \mathrm{C}_{|A ∪ B|}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We need min(π(A)) = min(π(B)). The minimum of permuted A equals minimum of permuted B only when both sets contain their smallest elements, and these smallest elements are the same. This happens when the smallest element of A∪B appears in both A and B (i.e., is in A∩B). The number of favorable permutations depends on |A ∩ B| and |A ∪ B|. The formula counts permutations where the minimum element from the intersection is placed first in both A and B's images.

Multiple choice

A set X can be represented by an array x [n] as follows:

x$\left [ i \right ]=\begin {cases} 1 & \text{if } i \in X \\ 0 & otherwise \end{cases}$ Consider the following algorithm in which x, y and z are boolean arrays of size n. algorithm zzz (x [ ], y[ ], z [ ] ) { int i; for (i = 0 ; i< n; ++i) z[i] = (x[i] Ù ~y[i]) Ú (~x[i] Ù y[i]) }

The set Z computed by the algorithm is

  1. (X U Y)
  2. (X $\cap$Y)
  3. (X − Y) $\cap$ (Y − X)

  4. (X − Y) U (Y − X)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{Here}\ z = (x \land y') \lor (x' \land y)$