complement of c
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
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.
The complement of a subset B within a set A consists of all elements of A which are not in B.
Draw your Venn Diagram as three overlapping circles. Each circle is a set. The union of the sets is what's contained within all 3 circles, making sure not to count the overlapping portion twice. An easier problem is when you have 2 sets, lets say A and B. In a Venn Diagram that looks like 2 overlapping circles. A union B = A + B - (A intersect B) A intersect B is the region that both circles have in common. You subtract that because it has already been included when you added circle A, so you don't want to add that Again with circle B, thus you subtract after adding B. With three sets, A, B, C A union B union C = A + B - (A intersect B) + C - (A intersect C) - (B intersect C) + (A intersect B intersect C) You have to add the middle region (A intersect B intersect C) back because when you subtract A intersect C and B intersect C you are actually subtracting the very middle region Twice, and that's not accurate. This would be easier to explain if we could actually draw circles.
complement of c
The answer depends on whether you mean A intersecting (B union C) or (A intersecting B) union C.
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
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.
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.
yes. a + b = 90 degrees: complements: C + D = 180: supplements: a = 5, b = 85, c = 95, d = 85. b & c are supplements. b = d
Venn diagram is represented with the help of circles. Union of a, b and c is shown by the three fully shaded somewhat overlapped circles. Result will be the elements that is in all three sets(a,b,c).
(A' ∩ B') = (A È B)'
The Demorgans Law includes the union, intersection, and complement in mathematics. Examples are A intersection B and B union A. Those are the basic examples.
b
The complement of an angle C is (90 - C) So the complement of an angle of 17.7° is (90 - 17.7) = 72.3°
The complement of a subset B within a set A consists of all elements of A which are not in B.