## Venn Diagrams Representing sets by using Venn diagrams The diagrams are oval shaped. They were named after John Venn, an English mathematician who introduced them. For example A = {1,2,3} in Venn diagram can be represented as follows:

$\mu$ is a universal set which can be a set of counting numbers and A is a subset of it. If we have two sets, say set A and B and these sets have some elements in common and we are supposed to represent them in Venn diagrams, their ovals will overlap. For example if A = {a,b,c,d,e,} and B = {a,e,i,o,u} in Venn diagrams they will look like this:

If the two sets have no elements in common, then the ovals will be separate. For example; if A = {1,2,3} and B = {5,6}. In Venn diagram they will appear like here below:

If we have two sets, A and B and set A is a subset of set B then the oval for set A will be inside the oval of set B. For example; if A = {b,c} and B = {a,b,c,d} then in Venn diagram it will look like this:

If we have to represent the union or intersection of two or more sets using Venn diagrams, the appearance of the Venn diagrams will depend on whether the sets under consideration have some elements in common or not. ## Information from Venn Diagrams ### Case 1: Sets with elements in common #### Example 1 If A = {5,6,7,9,10} and B = {3,4,7,9,11} represent A union B and A intersection B in Venn diagrams. **Solution** **Case 1:** A union B

**Case 2:** A intersection B

#### Example 2 A = {a,b,c,d,e,f}, B = {a,e} and C = {b,c,e,d}. Represent in Venn diagrams A∪B∪C and A∩B∩C. **Solution:** Case 1. A∪B∪C **Case 2:** A∩B∩C ### Case 2: Sets with no elements in common

For example; A = {a,b,c,}, B = {d,f}, C = {h,g} on Venn diagram will appear like this:

## Number of elements in two sets Number of elements in two sets say set A and B i.e. $n(A \cup B)$ is given by: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ **Proof:** consider the Venn diagram below:

From our Venn diagram: n(A) = x + y, n(B) = y + z, n(A∩B) = y and n(A∪B) = x + y + z thus; n(A) + n(B) = (x + y) + (y + z) = (x + y + z) + y but x + y + z = n(A∪B) and n(A∩B) = y so, n(A) + n(B) = n(A∪B) + n(A∩B) make n(A∪B) be the subject of the formula **For example**; if n(A) = 15, n(A∩B) = 3 and n(A∪B) = 24. Find n(B) **Solution:** **Recall that:** $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ n(B) = n(A∪B) + n(A∩B) – n(A) = 24 + 3 – 15 n(B) = 12 **Therefore,** n(B) = 12 ## Word problems For example; at Mtakuja primary school there are 180 pupils. If 120 pupils like one of the sports, either netball or football and 50 pupils likes netball while 30 pupils likes both netball and football. How many pupils 1. likes football. 2. Likes neither of the sport **Solutions.** Let $\mu$ be the universal set N be the set of pupils who likes netball F be the set of pupils who likes football **Thus,** n(F) = ? n(N∩F) = 30 n(N∪F) = 120 n($\mu$) = 180 **But we know that** $n(N \cup F) = n(N) + n(F) - n(N \cap F)$ **Thus,** n(F) = n(N∪F) + n(N∩F) – n(N) = 120 + 30 – 50 n(F) = 100 **Therefore** there are 100 pupils who likes football. We have a total of 180 pupils at Mtakuja primary school But only 120 pupils likes one of either the sport. so, those who likes neither of the sport will be 180 – 120 = 50 **Therefore** 50 pupils likes neither of the sport. **Alternatively:** by using Venn diagram

n(F) only = 120 – 30 -20 = 70 n(F) = those who likes both netball and football + those who likes football only n(F) = 30 + 70 = 100 **Therefore,** there 100 pupils who likes football.