Solving Rings and Idempotent Problems - RK

  • Thread starter Thread starter cap.r
  • Start date Start date
  • Tags Tags
    Rings
cap.r
Messages
64
Reaction score
0
I have a tablet so I have made a PDF of all my work and the problem. the file is attached to this post. please let me know if i am on the right track or give me a hint. I am currently stuck in attempt 2 and don't like my solution in attempt 1.

attempt 1: at the very last step I am using multiplicative inverses and I haven't proved that they must exist. but since I have shown that a multiplicative identity is required, it shouldn't be hard to prove that inverses exist also but i don't know if it will be required..?

attempt 2: took a different approach at the problem, and while it's a bit more complicated in the end and is unfinished (this is where i am stuck), I think it's the better attempt.



thank you,
RK
 

Attachments

Physics news on Phys.org
a^3 = a implies that aaa = a.
Multiply by "a inverse" to obtain aa=1.
Multiply by "a inverse" again to obtain a=a^-1.
So each element is it's own inverse.
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top