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Jan 29, 2018 · I suspect you meant to write $$\frac {\left (\frac BC\right)}A=\frac {B} {CA}$$ which is correct and can be established by multiplying by $\frac CC$. The only clue in what you wrote is the size of the f...
Jan 17, 2017 · I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof
Mar 30, 2019 · The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs.
Jul 6, 2019 · 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 $$ (a...
Aug 23, 2016 · If I have the elements from sets A and B, and I want to find the set A ∪ (B ∩ C), I end up with just the elements of A. On the other hand, if I have the elements from A and B and want to find (A ∪ ...
Apr 20, 2018 · Since the given polynomial is homogeneous and symmetrical w.r.t. a,b,c there can be only one linear factor a+b+c. You can substitute - (b+c) for a and prove that the result is zero and verify that a fac...
So perhaps express the difference in terms of $a-b$, $b-c$, and $c-a$. Remark: One cannot expect to be able to prove that something holds for all $a,b,c$ by looking only at a special case.
Feb 17, 2015 · How to prove P (A∩B∩C) = P (A|B∩C)P (B|C)P (C)? I've tried to approach this a couple times, but always end up getting stuck : ( Any help is appreciated! :)
Sep 24, 2019 · A comment on line 3: this is not the distribution law. You're using the associative law for intersection, and the idempotent law $\bar C \cap \bar C = \bar C$.
$$ (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.
