Proving "[a]_n ∩ [b]_n is Empty or Equals [b]_n

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Homework Statement


Prove that either [a]_{n}\cap<b>_{n}</b>=empty set or [a]_{n}=<b>_{n}</b>.

Homework Equations


The Attempt at a Solution


I want to assume there is an element x in [a]_{n}\cap<b>_{n}</b> and show this implies [a]_{n}=<b>_{n}</b>.
This tells me x is in [a]_{n} and <b>_{n}</b>.
That's where I get stuck.
 
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sorry, I meant to show this implies [a]=.
 
If x=a mod n then n|x-a. x=b mod n means n|x-b.

If x|x-a and n|x-b... what can you say about a-b?
 
x-a=x-b
x-x=a-b
0=a-b
b=a
 
x-a is probably not equal to x-b
 
Ok then I'm not really sure where to go with a-b then.
 

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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