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.
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Because there is no way to define the divisors, the equations cannot be evaluated.
In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
all major intersection have various other signs offering information to road users
If a=b and c=d then (a+c)=(b+d) ? This is proved very simply by the direct application of perhaps the most fundamental statement in all of Algebra: "If equals are added to equals, the sums are equal."
complement of c
Let x be in A intersect B. Then x is in A and x is in B. Then x is in A.
No. Suppose A = {1,2}, B = {1,2,3,4,5,6} and C = {1,2,3,5,7,11}. The intersection of A with B is {1,2}, the intersection of A with C is also {1,2}, but B is not equal to C.
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.
first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".
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.
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A - B is null.=> there are no elements in A - B.=> there are no elements such that they are in A but not in B.=> any element in A is in B.=> A is a subset of B.
Because there is no way to define the divisors, the equations cannot be evaluated.
In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
all major intersection have various other signs offering information to road users
If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.