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

[@minalu8184]

Thank u this was very helpful 👍

[@WrathofMath]

Glad to hear it! Thanks for watching!

[@cholwemulilanduba8283]

How do you solve A U (G\M)

[@WrathofMath]

Thanks for watching and I am not sure what you mean. Assuming that G and M are sets, A U (G\M) is just a set, so what is the problem? What is being asked? Do we know what A, G, and M are, and want to find A U (G\M)?

[@nilushadilshan1520]

Thank you so much ❤

[@WrathofMath]

My pleasure!

[@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.

[@fikirdesalegn2092]

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

[@WrathofMath]

Thanks for watching! That would be the empty set - try to prove it! The intuition is that since A is a subset of B, it contains less than B, which means A' contains MORE than B', so taking A' away from B' would get rid of everything, leaving the empty set.

[@anaislugo3384]

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}

[@kroshasbs8539]

God bless you!

[@azizhani33]

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)?