Finding a Set from its Intersection and Relative Complement | Set Theory

If we are given A intersect B, A-B, and B-A, can we determine what A and B are? We'll be going over this problem in today's set theory lesson with an example that will demonstrate the reasoning necessary to solve this problem!

In general, for two sets A and B, A is equal to (A intersect B) U (A - B) because (A intersect B) gives us the elements of A that are also in B, and (A - B) gives us the elements of A that are not in B. Similarly, B is equal to (A intersect B) U (B - A).

Note the use of "relative complement" in the title of this video; because A - B is the complement of B relative to A (as in, the set of all elements that are not in B, but are in A).

I hope you find this video helpful, and be sure to ask any questions down in the comments!

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[psinghcpr]

I have a fun problem for you : Let G be a simple graph with 3n² vertices (n≥2). It is known that the degree of each vertex of G is not greater than 4n,  there exists at least a vertex of degree one, and between any two vertices, there is a path of length ≤3. Prove that the minimum number of edges that G might have is equal to( 7n² - 3n)/2.
Hope you find it interesting.

[anaislugo]

Given sets​ A, B, ​ C, and​ U, find the elements in A​B'.
A​{​0, 1, 3 ​}
B​{​1, 3, 6​}
C​{​ 5, ​7}
U​{0, ​1, 2, ​ 3, 4, ​ 5, 6, ​ 7}

[azizhani]

Let us use inductive reasoning and start backward from the end result. Let us assume that A = {1, 2, 3} and B = {1, 2, 4}. Therefore A - B = {s | s belongs to A and s does not belong to B} = {3}. Also B - A = {s | s belongs to B and s does not belong to A} = {4}. Additionally, A intersect B = {s | s belongs to A and s also belongs to B} = {1, 2}.
Can we then conjecture that A = (A Ո B) U (A - B) and B = (B Ո A) U (B - A)?

[fikirdesalegn]

If A is subset of B then what is B'-A'?