Let A be the set of primes less than 6 and B be the set of positive odd numbers less than 6. How many different sums of the form a + b are possible, if a is in A and b is in B?
A = {2,3,5} and B = {1, 3, 5}
2+1 =
2+3 =
2 + 5 =
3+1 =
3+3 =
3+5 =
5+1 =
5+3 = already have that
5+5 =
make sure you don't count equal sums twice,
e.g.
3+3 = 6
5+1 = 6
You are just looking at the different sums
=(1,2,3,4.....,18) A=(prime numbers)and B=(odd numbers greater than 3)IfA and B are subsets of the universal set .(a)Find the number of elements in And B. (b)Find the set (i)AnB (ii)n(AnB) (iii)AuB) (iv)n(AuB)
To find the possible sums of the form a + b, where a is in set A and b is in set B, we need to determine all the possible combinations of elements from A and B and then calculate their sums.
Set A consists of primes less than 6, which are {2, 3, 5}.
Set B consists of positive odd numbers less than 6, which are {1, 3, 5}.
Now, let's calculate the sums:
- a = 2, b = 1: a + b = 2 + 1 = 3
- a = 2, b = 3: a + b = 2 + 3 = 5
- a = 2, b = 5: a + b = 2 + 5 = 7
- a = 3, b = 1: a + b = 3 + 1 = 4
- a = 3, b = 3: a + b = 3 + 3 = 6
- a = 3, b = 5: a + b = 3 + 5 = 8
- a = 5, b = 1: a + b = 5 + 1 = 6
- a = 5, b = 3: a + b = 5 + 3 = 8
- a = 5, b = 5: a + b = 5 + 5 = 10
From these calculations, we can see that the possible sums of the form a + b are: 3, 4, 5, 6, 7, 8, and 10.
Therefore, there are a total of 7 different sums of the form a + b possible when a is in set A and b is in set B.
To find out the number of different sums of the form a + b, where a is in set A and b is in set B, we need to consider all possible combinations of elements from each set and count the unique sums.
First, let's list the elements of set A and set B:
Set A: {2, 3, 5}
Set B: {1, 3, 5}
To get the sums, we need to take one element from set A and one element from set B. We can represent this as a Cartesian product of the two sets:
A x B = {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
From this product, we can calculate the sums:
2 + 1 = 3
2 + 3 = 5
2 + 5 = 7
3 + 1 = 4
3 + 3 = 6
3 + 5 = 8
5 + 1 = 6
5 + 3 = 8
5 + 5 = 10
Now, let's remove the duplicate sums:
Unique sums: {3, 4, 5, 6, 7, 8, 10}
Therefore, there are 7 different sums of the form a + b, where a is in set A and b is in set B.