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Division and Multiplication, right?
@zzr0ck3r @mathmate @King.Void. @SolomonZelman
what does it mean set closed?
A set of elements is closed under an operation if, when you apply the operation to elements of the set, you always get another element of the set.
It's either A and D or all of them. I'm not sure
ok let us go through them A) Is there a way to add two elements from the set and not get an element in the set?
Oh wait it can't be division because you can't divide by 0.....
not addition because 1+1 = 2 is not in the set
yes -1 + 1 = 0
while `-1 + 1 = 0` is true, the fact that `1+1=2` means that you can take the elements `1` and `1` (repeat selections are possible), add them up, and get a number that is NOT in the set. So this set is NOT closed under addition
0 x -1 = 0 0 x 1 = 0 1 - 1 = 0 So, is it multiplication and subtraction?
make a 3x3 table 3 rows for the elements -1,0,1 3 columns for the elements -1,0,1 |dw:1441669222547:dw|
Now let's say we will use the subtraction operation to fill out the table, we subtract the value on the left from the value up top eg: -1 minus -1 = -1+1 = 0 and this goes in the first row, first column the rest of the table is filled out the same way |dw:1441669277363:dw|
Does the table make sense?
So, it's only multiplication that will work?
let's find out by making a multiplication table |dw:1441669432403:dw|
so yes, the set is closed under multiplication take any two numbers (repeats allowed), multiply them, and you'll get some value that's in the set
here's what the addition table would look like |dw:1441669534485:dw|
And division won't work either, right. Because you can't divide by 0
division table |dw:1441669624730:dw|
Thanks for the help!!