Fraction rules A/B/C vs B/C/A - Mathematics Stack Exchange
网页2018年1月29日 · Your first rule is. A (B C) = AC B A (B C) = A C B. which you can establish by multiplying the fraction by C C C C. You have written the second the same way on the left, B (C A) = B CA B (C A) = B C A. which is incorrect. I suspect you meant to write. (B C) A = B CA (B C) A = B C A. which is correct and can be established by multiplying by C C C C.
Let A, B, and C be sets. Prove that (A-B) - C = (A-C) - (B-C)
网页2019年9月24日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
(A⊕B)⊕C = A⊕ (B⊕C) how to prove - Mathematics Stack Exchange
网页2018年10月22日 · 2. An element x x is in (A ⊕ B) ⊕ C (A ⊕ B) ⊕ C iff it is in A ⊕ B A ⊕ B or in C C but not in both. Iff it is either in exactly one of A A or B B, or otherwise it is in C C. Iff exactly one or all three of A, B, C A, B, C contains it. Iff the number of summands is odd in which it is contained. The same conclusion applies to A ⊕ (B ...
elementary set theory - Why does A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C ...
网页2016年8月23日 · Not generally, and more importantly: not relevant. ∪ means union: A ∪ B is set of elements in either set A or set B. ∩ means intersection: B ∩ C is set of elements in both set B and set C. A ∪ (B ∩ C) ⊆ (A ∩ B) ∪ (A ∩ C) If you have an element either from set A or from both sets B and C, then you have elements which are from ...
Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$
网页$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$ It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all. algebra-precalculus
Proving the identity: (A\\B) ∪ (B\\C) = (A∪B) \\ (B∩C)
网页2019年7月6日 · Then the indicator function of $(A\setminus B)\cup(B\setminus C)$ is $$ a(1-b)+b(1-c)-a(1-b)b(1-c)=a-ab+b-bc$$ because $$(1-b)b=b-b^2=b-b=0$$ The indicator function of the right-hand side is $$(a+b-ab)(1-bc)=a-abc+b-b^2c-ab+abc=a+b-bc-ab$$ and we see that the two functions are identical.
How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2 ...
网页2019年3月30日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Simplification of: AB + A'C + BC in boolean algebra
网页2016年9月6日 · I am trying to understand the simplification of the boolean expression: AB + A'C + BC I know it simplifies to A'C + BC And I understand why, but I cannot figure out how to perform the simplific...
elementary set theory - A\(B∩C) = (A\B) ∪ (A\C) how to prove ...
网页2017年1月17日 · The latter condition means that either x ∉ B or x ∉ C (since it does not belong to both B and C). Thus either x ∈ A ∖ B or x ∈ A ∖ C. That is, x ∈ (A ∖ B) ∪ (A ∖ C). Look at the implication that was just proved: x ∈ A ∖ (B ∩ C) x ∈ (A ∖ B) ∪ (A ∖ C). This is precisely the meaning of A ∖ (B ∩ C) ⊆ (A ∖ B ...
Showing that (A⊕B)⊕B is equal to A - Mathematics Stack Exchange
网页2019年1月18日 · I found this solution somewhat incomplete in a video I was watching. Here's the image and video (solution starts at ~2:11). Here's what he wrote: (A⊕B)⊕B=A. ⇒ x∈ (A⊕B)⊕B. = x∈ (A⊕B) xor x∈B. = (x∈A xor x∈B) xor x∈B.