What does this mean,(A Δ B)^c?

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What does this mean,

(A Δ B)^c? And what does it equal to?

I know that (A Δ B)= (A-B) U (B-A).
 
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matrix_204 said:
What does this mean,

(A Δ B)^c? And what does it equal to?

I know that (A Δ B)= (A-B) U (B-A).

Might mean the complement of (A Δ B)
 
Yes, I know, but how do I find the complement. Thats where I'm stuck. What is the complement equal to?
 
matrix_204 said:
Yes, I know, but how do I find the complement. Thats where I'm stuck. What is the complement equal to?

I don't really understand what you are asking, the complement is going to be all members of the universe that are not in the given set so from what you have told me it would be the set of all x such that x is not a member of the symmetric difference between A and B.
 
So for example, all x such that x is not in (A Δ B)= (A-B) U (B-A)!
What I was asking is what's (A Δ B)^c= ??
Because in the textbook it's not given and I got stuck for one of the problems before I could proceed.
 
Note that your triangle thingy set does not contain points common to A and B, nor does it contain points in the complement of the union of A and B.

Does that help?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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