Is the set V={3,5,7} closed under addition or not closed under addition? Explain why or why not or show a counterexample.
not
adding together any of the elements will give a result that is not part of the set
Well, let's see if V={3,5,7} is closed under addition. Adding any two numbers from the set, let's say 3 and 5, we get 8. Uh oh! 8 is not in the set V, is it? So, V={3,5,7} is not closed under addition because the sum of two numbers in the set is not always in the set itself. Looks like V needs to find some new numbers to join its addition party!
To determine if the set V = {3, 5, 7} is closed under addition, we need to check if the sum of any two elements in the set remains within the set.
First, let's check the sum of 3 and 5: 3 + 5 = 8. Since 8 is not an element of the set V, we can conclude that the set V is not closed under addition.
Therefore, the set V = {3, 5, 7} is not closed under addition.
To determine if the set V={3,5,7} is closed under addition, we need to check whether adding any two elements from the set always results in another element in the set.
Let's consider all possible pairs of elements from V and check if their sum is also in V:
1. 3 + 3 = 6 (not in V)
2. 3 + 5 = 8 (not in V)
3. 3 + 7 = 10 (not in V)
4. 5 + 5 = 10 (not in V)
5. 5 + 7 = 12 (not in V)
6. 7 + 7 = 14 (not in V)
As we can see, none of the possible sums are in V. Therefore, the set V={3,5,7} is not closed under addition.
We have shown a counterexample where adding elements from V does not result in another element in V, which proves that the set is not closed under addition.