Symmetric Difference Proof for Sets A, B, and C

bedi
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Homework Statement



Let % be the symmetric difference.
Prove that for any sets A, B, C;

A%B=C iff B%C=A iff A%C=B

Homework Equations



(I will use forwardslash as I cannot find the backslash on my keyboard.)

The Attempt at a Solution



Take x in A%B. Then x is either in A/B or in B/A and in C. Choose the former. Then x is in A and x is in C which implies x is in AnC. This is what I know.
Now, using these I will prove that B%C=A.
Take x in (B/C)U(C/B). Then x is either in the first or in the second. Actually it cannot be in the first because that would imply that x is not in C, which contradicts that x is in C. So x is in C/B. But I already know that x is in A, so nothing to prove in that direction. Take x in A... I don't know how to proceed.
 
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hi bedi! :smile:
bedi said:
A%B=C iff B%C=A …

(to prove from left to right …)

you need to start "B%C = B%(A%B) = … " :wink:
 

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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