Proving the Truth of 3(b) in Basic Set Theory

isabelle york
Messages
2
Reaction score
0
how do I go about doing 3(a) and 3(b)?

I'm guessing that for 3(a), it is true, since we have for LHS:

P((A or B) and C)

we can consider the case P(A and C) by excluding B, and this is a subset of the RHS when we also exclude B: (P(A) and P(C)).

We can consider excluding B because it's in an OR function.

Thanks x
isabelle
 

Attachments

  • mathsprob2.png
    mathsprob2.png
    14.3 KB · Views: 505
Physics news on Phys.org
When you exclude B from the left hand side you make the left hand side smaller - So the question is asking you to show that X is a subset of Y, and you showed that W is a subset of Y where W is a smaller set than X is.
 
Office_Shredder said:
When you exclude B from the left hand side you make the left hand side smaller - So the question is asking you to show that X is a subset of Y, and you showed that W is a subset of Y where W is a smaller set than X is.

I've managed to do 3(a), it is false. I used the counter example: A= a, B= b, C = a, b.

How do I do 3(b)? 3(b) is 3(a) reversed.

I'm pretty sure 3(b) is true since the RHS will always end up being 'larger', but i don't know how to go about proving it.

EDIT: ignore my reasoning in the OP, I was confused then.
 
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...
Back
Top