Maths 1 Marks

It is given that A = {1, 2, 3}.
An equivalence relation is reflexive, symmetric and transitive.
The smallest equivalence relations containing (1, 2) is equal to
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, only four pairs are left (2, 3), (3, 2), (1, 3) and (3, 1).
So, if we add the pair (2, 3) to R, then for symmetry we must add (3, 2).
Also, for transitivity we required to add (1, 3) and (3, 1).
Thus, the only equivalence relation is the universal relation.
Therefore, the total number of equivalence relations containing (1, 2) is 2.

The number of onto functions that can be defined from a finite set A containing n elements onto a finite set B containing n elements = 2ⁿ - n.

R is reflexive and transitive but not symmetric.

x₁ ,x₂  are two different elements of R
Let  f(x₁) = f (x_{​​​​​​2})
2x₁ = 2x₂
x₁ = x_2, hence f is one-one.

No two different students of the class can have the same roll number.
Therefore, f must be one-one.
We can assume without any loss of generality that roll numbers of students are from 1 to 50.
This implies that 51 and higher natural numbers in N cannot be the roll numbers of any student in the class. 
So, 51 and higher natural numbers can not be the images of any element of A under f. Hence, f is not onto.

Given A ={1, 2, 3, 4 , 5, 6, 7} and R = {(a, b): both a and b are either odd or even number}
Therefore,
R = {(1, 1), (1, 3), (1, 5), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 7), (7, 5), (7, 3), (5, 3), (5, 1), (3, 1)
(2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)}
From the relation R it is seen that R is symmetric, reflexive and transitive as well. Therefore, R is an equivalent relation.
From the relation R it is seen that {1, 3, 5, 7} are related with each other only and {2, 4, 6} are related with each other.

The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add anyone, say (2, 3) to R₁, then for symmetry, we must add (3, 2) this loses transitivity.
To maintain transitivity along with symmetry, we are forced to add (1, 3) and (3, 1).
Thus, the only equivalence relation bigger than R₁ is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two.

The smallest relation R₁ containing (1, 2) and (2, 3) which is reflexive and transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.
Now, if we add the pair (2, 1) to R₁ to get R₂, then the relation R₂ will be reflexive, transitive but not symmetric.
Similarly, we can obtain R₃ by adding (3, 2) to R₁ to get the desired relation.
However, we can not add the pairs (2, 1), (3, 2) and (3, 1) to R₁ at a time, as by doing so the relation will become symmetric, which is not required. At a time, we can only add two pairs out of the remaining three viz. (2,1), (3,2) and (3,1). This can be done in three ways
adding (2,1) and (3,2)
or  (2,1) and (3,1)
or  (3,2) and (3,1)
Thus, the total number of desired relations is three.

One-one function from {1, 2, 3} to itself is simply a permutation on three symbols 1, 2, 3.
Therefore, a total number of one-one maps from {1, 2, 3} to itself is the same as the total number of permutations on three symbols 1, 2, 3 which is 3! = 6

Since f is one – one three element of {1, 2, 3} must be taken to 3 different element of the co – domain {1, 2, 3} under f. Hence, f has to be onto.

{by definition ONTO FUNCTION- every element of co-domain have a pre-image in domain}

Suppose f is not one-one.
Then there exists two elements, say 1 and 2 in the domain whose image in the co-domain is the same.
Also, the image of 3 under f can be only one element.
Therefore, the range set can have at the most two elements of the co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one.