Sets and Relations

A ={x

z : 2(x+2)(x2 $-$ 5x + 6) = 1} 2(x+2)(x2 $-$ 5x + 6) = 20 $\Rightarrow$ x = $-$ 2, 2, 3 A = {$-$2, 2, 3} B = {x

Z : $-$ < 2x $-$ 1 < 9} B = {0, 1, 2, 3, 4} A $\times$ B has is 15 elements so number of subsets of A $\times$ B is 215.

To evaluate the relation $R$ on the set $A = \{0, 1, 2, 3, 4, 5\}$, we first need to understand the conditions for an element $(x, y)$ to be in $R$.

Specifically, $(x, y) \in R$ if and only if $\max\{x, y\} \in \{3, 4\}$.

Considering this, let's list the pairs: For $\max\{x, y\} = 3$, the possible pairs are: $(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3)$ For $\max\{x, y\} = 4$, the possible pairs are: $(0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)$ Combining these, the set $R$ consists of the following elements: $R = \{(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)\}$ This gives us a total of 16 elements in $R$, not 18 as initially claimed in statement $S_1$.

Next, we analyze the properties of the relation $R$: Reflexivity: A relation is reflexive if $(x, x) \in R$ for all $x \in A$.

For example, $(0, 0), (1, 1), (2, 2)$ are not in $R$, so $R$ is not reflexive.

Symmetry: A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ as well.

For all pairs $(x, y)$ listed, both $(x, y)$ and $(y, x)$ are present.

Thus, $R$ is symmetric.

Transitivity: A relation is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

An example where transitivity fails is $(0, 3)$ and $(3, 1)$ are in $R$ but $(0, 1)$ is not in $R$.

Therefore, $R$ is not transitive.

In conclusion, statement $S_2$ is correct as $R$ is symmetric but neither reflexive nor transitive.

We have a set S = {1, 2, 3, ..., 60}, and a relation R defined on the set S.

An element (a, b) belongs to the relation R if and only if b can be expressed as the product of two prime numbers p and q, where both p and q are greater than or equal to 3.

In terms of number theory, prime numbers are integers greater than 1 that have no divisors other than 1 and themselves.

We are interested in prime numbers that are greater than or equal to 3, because p and q must both be greater than or equal to 3.

The primes greater than or equal to 3 and less than or equal to 60 are {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59}.

We need to find all possible values of b = p $\times$ q such that b belongs to the set S.

We start by choosing the smallest prime number (which is 3) and keep multiplying it with all the prime numbers until the product exceeds 60 : 1.

If we start with p = 3, we can choose q to be 3, 5, 7, 11, 13, 17, or 19.

This gives us 7 valid products that are less than or equal to 60.

2.

If we start with p = 5, we can choose q to be 5, 7, or 11.

This gives us 3 valid products that are less than or equal to 60.

3.

If we start with p = 7, we can choose q to be 7.

This gives us 1 valid product that is less than or equal to 60.

So, we have a total of 7 + 3 + 1 = 11 possible values for b = p $\times$ q that satisfy the conditions.

Since a can be any number in the set S, there are 60 possible values for a for each of the 11 values of b.

Therefore, the total number of elements in the relation R is 60 $\times$ 11 = 660.

Let's evaluate the relation $R$ for the properties of reflexivity, symmetry, and transitivity.

Relation R : $R={(x, y) \in W \times W \mid}$ the words $x$ and $y$ have at least one letter in common}.

Reflexivity : Each word in English obviously has at least one letter in common with itself, so the relation is reflexive.

Symmetry : If a word $x$ has at least one letter in common with a word $y$, then $y$ necessarily has that same letter in common with $x$.

So, the relation is symmetric.

Transitivity : This property is not always satisfied.

For instance, consider the three words 'cat', 'bat', and 'bee'.

'Cat' and 'bat' share a letter (the 'a'), and 'bat' and 'bee' share a letter (the 'b'), but 'cat' and 'bee' do not share any letters.

Therefore, even though 'cat' is related to 'bat' and 'bat' is related to 'bee', 'cat' is not related to 'bee', so the relation is not transitive.

In conclusion, the correct answer is Option A : $R$ is reflexive, symmetric and not transitive.