Question

is the set {-1,0,1} closed under addition? Explain your anwser.

Answer 1

someone else may need to check this, but I believe it is because adding any one of the set members to to any other set member yields a value that is in the set. e.g.
-1+0 = -1
-1+1 = 0
0+1 = 1
all values in the right hand side are within the set - therefore the set is closed.
I haven't worked with sets for a looong time - so I may have missed something.

Answer 2

so you just add the number on the right to the first number and if that number is in the set itsclosed?

Answer 3

asnaseer has it right

Answer 4

thankyouuuuuuuuuuu

Answer 5

thx for confirming @cazil

Answer 6

yw

Answer 7

basically I believe you need to apply the given operation (addition in this case) between every member of the set and show that the result is just another member of the same set.

Answer 8

e.g. {1, 2, 3} is not closed under addition because:
1+3=4
2+3=5

and 4 and 5 are not members of the set

Answer 9

I dont think it is closed under addition, because we forgot to check adding an element with itself.

Since 1+1 = 2, and 2 isnt in the set, it is not closed under addition. It is closed with respect to multiplication though.

Answer 10

I believe you are right joemath. I was under the wrong impression it had to be a unique member of the set. But the postulate states that a+b can be any number of a set. So a and b can both be the same number.

Here are two links with conflicting answers to this exact question. LOL

http://www.youtube.com/watch?v=-rNEAHCqsQo

http://www.algebra.com/algebra/homework/Number-Line/Number-Line.faq.question.160073.html

Closure axiom for addition: For any number a and b, a + b is a number.

Answer 11

thanks for clarifying @joemath - much appreciated
@desperate - sorry for giving you the wrong information :-(