# Algebraic Laws in Set Theory

For any two sets A and B*Commutative Laws*

A ∪ B = B ∪ A and A ∩ B = B ∩ AFor any three sets A, B and C,*Associative Laws*

A ∩ (B ∩ C) = (A ∩ B) ∩ CFor any three sets A, B and C,*Distributive laws*

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)*Identity laws*

*A ∪ φ = A, where φ is the null set*

*A ∩ U = A, where U is the universal set**De-Morgan’s law*

(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′- A ∪ A = A and A ∩ A = A

# Some Important Results

If A and B are any two sets, then- A – B = A ∩ B′
- B – A = B ∩ A′
- (A – B) ∪ B = A ∪ B
- (A – B) ∩ B = f
- (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

# Some Important Results on the Number of Elements in Sets

If A, B and C are three finite sets then*n*(A ∪ B) =*n*(A) +*n*(B) –*n*(A ∩ B)*n*(A ∪ B) =*n*(A) +*n*(B) if set A and set B are two disjoint sets.*n*(A – B) =*n*(A) –*n*(A ∩ B)*n*(A ∪ B ∪ C) =*n*(A) +*n*(B) +*n*(C) –*n*(A ∩ B) –*n*(B ∩ C) –*n*(A ∩ C) +*n*(A ∩ B ∩ C)*n*(A′ ∪ B′) =*n*((A ∩ B)′) =*n*(U) –*n*(A ∩ B)*n*(A′ ∩ B′) =*n*((A ∪ B)′) =*n*(U) –*n*(A ∪ B)

Example-1

A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options—air conditioning, radio and power windows—were already installed. The survey found:

15 had air conditioning

2 had air conditioning and power windows but no radios

12 had radio

6 had air conditioning and radio but no power windows

11 had power windows

4 had radio and power windows

3 had all three options

- 4
- 3
- 1
- 2

Solution

From the given conditions, we have

When we add all the values, we get a total of 23 cars.

So, 2 cars don’t have the air conditioning or the radio, or the power windows.

Example-2

If A = { 1, 3, 5, 7}, B = { 1, 2, 3, 4}
What is the value of (A U B) and ( A ∩ B)?

Solution

(A U B) = {1, 2, 3, 4, 5, 7}

(A ∩ B) = {1, 3}

(A ∩ B) = {1, 3}

Example-3

How many numbers from 1 to 100 are not divisible by either 2 or 4 or 5?

Solution

Let us first understand the meaning of the statement given in the question—

It is given that the numbers are from 1 to 100, so while counting we will include both the limits, i.e., 1 and 100. Had this been “How many numbers in between 1 to 100 are …”, then we would not have included either 1 or 100.

Now to solve this question, we will first find out the number of numbers from 1 to 100 which are divisible by either 2 or 5 (since all the numbers which are not divisible by 2 will not be divisible by 4 also, so we do not need to find the numbers divisible by 4). And then we will subtract this from the total number of numbers i.e., 100. It can be seen below:

Total number of numbers = numbers which are divisible + numbers which are not divisible

So,

*n*(2 U 5) =*n*( 2 ) +*n*( 5 ) –*n*( 2 ∩ 5)Now,

*n*( 2 ) = 50*n*( 5 ) = 20

*n*( 2 ∩ 5) = 10

*n*(2 U 5) = 50 + 20 – 10 = 60

Numbers which are not divisible = total number of numbers-numbers which are divisible

= 100 – 60 = 40

Example-4

Sandeep, Mohit and Hardeep are three financial analysts working at Due North Consultants. The sum of the number of projects handled by Sandeep and Hardeep individually is equal to the number of projects in which Mohit is involved. All three consultants are involved together in 6 projects. Sandeep works with Mohit in 14 projects. Hardeep has 2 projects with Mohit but without Sandeep, and 3 projects with Sandeep but without Mohit. The total number of projects is one less than twice the number of projects in which more than one consultant is involved. In how many projects is Mohit involved alone?

Solution

Let us see the situation given above through the Venn-diagram:

*a + g = b + d + f + c*…(i)

And

*d*= 6*b + d*= 14, so,

*b*= 8.

*f*= 2 and

*e*= 3

And finally,

*a + b + c + d + e + f + g*= 2 (

*b + d + e + f*) – 1…(ii)Using (i) and (ii),

*c*= Number of projects handled by Mohit alone = 1