Relations Test 18

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Relations Test 18

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Weekly Quiz Competition

Question 1
1 / -0

from the given statement $$N$$ denotes the natural number and $$W$$ denotes the whole number, so which statement in the following is correct

A
N=W

B
N $$\subset$$ W

C
W $$\subset$$ N

D
N $$\cong$$ W

Solution

Natural numbers are naturally occurring counts of an object $$1, 2, 3, 4,… $$forming an infinite list in which the first number is $$1$$ and every next number is equal to the preceding number $$+1.$$
Whole number include the set of all Natural numbers and also $$0.$$
Let $$N$$ and $$W$$ elements of natural and whole numbers.
$$N=\{1,2,3,4,....\infty\}$$
$$W=\{0,1,2,3,4,...\infty\}$$
So, here we can say $$N$$ is subset of $$W.$$
$$\therefore$$ $$N\subset W$$ is correct statement.
Question 2
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Which one of the following relations on $$R$$ is equivalence relation

A
$$x \space R_1 \space y \Leftrightarrow |x| = |y|$$

B
$$x \space R_2 \space y \Leftrightarrow x \ge y$$

C
$$x \space R_3 \space y \Leftrightarrow x | y$$

D
$$x \space R_4 \space y \Leftrightarrow x < y$$

Solution

Let's take option A,
$$x \space R_1 \space y \Leftrightarrow |x| = |y|$$
Now, $$|x|=|x|$$
$$\Rightarrow xR_{1}x$$
Hence, $$R_1$$ is reflexive.

Now, let $$xR_1 y$$
$$\Rightarrow |x|=|y|$$
or $$|y|=|x|$$
$$\Rightarrow yR_{1}x$$
Hence, $$R_1$$ is symmetric

Now, let $$xR_1 y, yR_1 z$$
$$\Rightarrow |x|=|y|, |y|=|z|$$
$$\Rightarrow |x|=|z|$$
$$\Rightarrow xR_{z}x$$
Hence, $$R_1$$ is transitive.
Hence, $$R_1$$ is an equivalence relation.

Clearly, $$R_2$$ is not an equivalence relation as $$2 \ge 3$$ does not implies $$3\ge 2$$
Also, $$R_3$$ is not an equivalence relation as $$x$$ divides $$y $$, does not implies $$y$$ divides $$x$$
Also $$R_4$$ is not an equivalence relation because $$2 <3$$ but $$3 >2$$
Question 3
1 / -0

Given the relation $$R = \left\{(1,2), (2,3)\right\}$$ on the set $$A = \left\{1,2,3\right\}$$, the minimum number of ordered pairs which when added to $$R$$ make it an equivalence relation is

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

$$R$$ is reflexive if it contains $$(1,1) (2,2) (3,3)$$

$$\because (1,2) \in R,\space (2,3) \in R$$

$$\therefore R$$ is symmetric if $$(2,1), (3,2) \in R$$

Now, $$R = \left\{(1,1), (2,2), (3,3), (2,1), (3,2), (2,3), (1,2)\right\}$$

$$R$$ will be transitive if $$(3,1); (1,3)\in R$$. Thus, $$R$$ becomes and

equivalence relation by adding $$(1,1) (2,2) (3,3) (2,1) (3,2) (1,3) (1,2)$$. Hence,

the total number of ordered pairs is $$7$$.
Question 4
1 / -0

If $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$, then $$A \times B$$ is equal to

A
$$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$

B
$$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$$

C
$$\left\{(2,1), (3,2)\right\}$$

D
$$\left\{(1,2), (2,3)\right\}$$

Solution

If $$A$$ and $$B$$ are any two non-empty sets.
then $$A\times B$$$$=\left\{(x,y):x\in A and y\in B\right\}$$
As $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$
$$A \times B$$$$=\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$
Hence, option A.
Question 5
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$$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is

A
$$2^5$$

B
$$2^{10} - 1$$

C
$$2^{12} - 1$$

D
$$none\ of\ these$$

Solution

`if $$A$$ and $$B$$ are two sets which contains 3 and 4 elements respectively.
Then no of elements in relation from $$A$$ to $$B$$ is $$2^(3\times 4)$$
means $$2^{12}$$
Question 6
1 / -0

Let $$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$. Which of the following is relations from $$X$$ to $$Y$$

A
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$

B
$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$

C
$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$

D
$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$

Solution

We have
$$X = \left\{1,2,3,4\right\}$$ and $$Y = \left\{1,3,5,7,9\right\}$$.
$$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$$
$$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$$

Let's take option A,
$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$
$$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$$
Since, $$R_1 \subseteq X\times Y$$
So, $$R_1$$ is a relation from $$X$$ to $$Y$$

Clearly, $$R_2$$ is not a relation from $$X$$ to $$Y$$ as $$(5,5) \notin X\times Y$$
Similarly, $$R_3$$ is not a relation from $$X$$ to $$Y$$ as $$(5,7) \notin X\times Y$$
In the same manner, $$R_4$$ is also not a relation from $$X$$ to $$Y$$ as $$(7,9) \notin X\times Y$$
Question 7
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If $$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements, then the number of relations from $$A$$ to $$B$$ is:

A
$$2^{mn}$$

B
$$2^{mn} - 1$$

C
$$2mn$$

D
$$m^n$$

Solution

$$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements.
Number of elements in $$A\times B$$ $$=mn$$
No of relations $$=$$ number of subsets of $$A\times B$$ $$=2^{mn}$$
Hence, option A is correct.
Question 8
1 / -0

The relation $$R$$ is defined in $$A = \left\{1,2,3\right\}$$ by $$a\ R$$ $$b$$ if $$|a^2 - b^2| \le 5$$. Which of the following is false?

A
$$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$$

B
$$R^{-1} = R$$

C
Domain of $$R = \left\{1,2,3\right\}$$

D
Range of $$R = \left\{5\right\}$$

Solution

$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$
Clearly $$R^{-1} = R$$
Domain $$=\{1,2,3\}$$ Range $$=\{1,2,3\}$$
Hence option 'D' is correct choice.
Question 9
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If $$R = \{(x, y):3x + 2y = 15 \text{ and }x\,, y\,\in \, N\}$$, the range of the relation R is .........

A
$$\{1, 2\}$$

B
$$\{1, 2,..., 5\}$$

C
$$\{1, 2,..., 7\}$$

D
$$\{3, 6\}$$

Solution

Since x and y belong to natural numbers we need to find positive integral solutions for $$3x+2y=15$$

$$2y=15-3x$$

$$\Rightarrow y =\dfrac{15-3x}{2}$$

For $$x=0,2,4$$ $$y$$ will become fractional which is not possible

For $$x\geq 5$$ y becomes 0 or negative which again is not possible.

Hence for $$x=1,3$$ y= 6,3$$ respectively.

Hence the range is $$\{3,6\}$$
Question 10
1 / -0

Let $$n(A) = n$$. Then the number of all relations on $$A$$ is

A
$$\displaystyle 2^{n}$$

B
$$\displaystyle 2^{\left ( n \right )!}$$

C
$$\displaystyle 2^{n^{2}}$$

D
none

Solution

For any set $$A$$ such that $$n(A)=n$$
then number of all relations on $$A$$ is $${ 2 }^{ n^{2} }$$
As the total number of Relations that can be defined from a set $$A$$ to $$B$$ is the number of possible subsets of $$ A \times B$$. If $$n(A) = p$$ and $$n(B) = q$$ then $$n(A \times B) = pq$$ and the number of subsets of $$A \times B$$ = $$2^{pq}$$.

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Re-Attempt Weekly Quiz Competition

Relations Test 18

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Relations Test 18

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Question 1

N=W

N $$\subset$$ W

W $$\subset$$ N

N $$\cong$$ W

Solution

Question 2

$$x \space R_1 \space y \Leftrightarrow |x| = |y|$$

$$x \space R_2 \space y \Leftrightarrow x \ge y$$

$$x \space R_3 \space y \Leftrightarrow x | y$$

$$x \space R_4 \space y \Leftrightarrow x < y$$

Solution

Question 3

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 4

$$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$

$$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$$

$$\left\{(2,1), (3,2)\right\}$$

$$\left\{(1,2), (2,3)\right\}$$

Solution

Question 5

$$2^5$$

$$2^{10} - 1$$

$$2^{12} - 1$$

$$none\ of\ these$$

Solution

Question 6

$$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$$

$$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$$

$$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$$

$$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$$

Solution

Question 7

$$2^{mn}$$

$$2^{mn} - 1$$

$$2mn$$

$$m^n$$

Solution

Question 8

$$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$$

$$R^{-1} = R$$

Domain of $$R = \left\{1,2,3\right\}$$

Range of $$R = \left\{5\right\}$$

Solution

Question 9

$$\{1, 2\}$$

$$\{1, 2,..., 5\}$$

$$\{1, 2,..., 7\}$$

$$\{3, 6\}$$

Solution

Question 10

$$\displaystyle 2^{n}$$

$$\displaystyle 2^{\left ( n \right )!}$$

$$\displaystyle 2^{n^{2}}$$

none

Solution

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