This is really a version of deMorgans that states the complement of the intersection of any number of sets equals the union of their complements.
We prove it in general and this is a specific case.
Take x contained in the complement of the intersection of all sets Aj that is to say x is not in the intersection of all Aj . Now there must be at least one set that does not contain x since if all sets contain x then x would be in their intersection as well. Call this set A. Since x is not in A, x must be in the complement of A. But then x is also in the union of all complements of Aj , because A is one of those sets. This proves that the right hand set is contained in the left one.
We now prove it the other way, that is to say, the left hand side is contained in the right. Remember that if A and B are sets and A is contained in B and B is contained in A we can say that as sets A=B.
Consider x contained in the union of all complements of Aj . That means there is at least one complement that contains x, or in other words, at least one of the Aj does not contain x. But then x is not in the intersection of all Aj , and hence it must be in the complement of that intersection. That proves the other inequality, so both sets must be equal.
complement of c
If a equals 3 and b equals minus 5 then a minus b equals what
a - b = c -(a - b) = -c b - a = -c
A - B = 11A = 11 + whatever ' B ' is.
C minus B equals AB
complement of c
If a equals 3 and b equals minus 5 then a minus b equals what
a - b = c -(a - b) = -c b - a = -c
A - B = 11A = 11 + whatever ' B ' is.
52500-a = b
x = b/(a + c)
All the time
C minus B equals AB
That depends what the values of k and b are.
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
Commutative, not communtative The mathematical property of being able to change the order of the numbers and not change the answer. A plus B equals B plus A (1 plus 3 equals 4, 3 plus 1 equals 4) A times B equals B times A (2 times 5 equals 10, 5 times 2 equals 10) Addition and multiplication are commutative operators A minus B is not necessarily equal to B minus A (6 minus 4 equals 2, 4 minus 6 equals minus 2) A divided B is not necessarily equal to B divided A (9 divided by 3 equals 3, 3 divided by 9 equals one-third) Subtraction and division are not commutative operators
b - 4.75 = 5.2 b = 5.2 + 4.75 b = 9.95