Reputation
349
Next privilege 500 Rep.
Access review queues
Badges
2 11
Impact
~11k people reached

  • 0 posts edited
  • 0 helpful flags
  • 31 votes cast
Jan
5
accepted Folland 2.36 portion of proof
Jan
5
comment Folland 2.36 portion of proof
This is the conclusion I arrived at as well. Thank you!
Jan
5
asked Folland 2.36 portion of proof
Dec
22
accepted Question about simple functions as described in Folland's Real Analysis
Dec
22
comment Question about simple functions as described in Folland's Real Analysis
I'm a fool. Thank you so much.
Dec
22
comment Question about simple functions as described in Folland's Real Analysis
Yes, I understand that part of the proof. My question is why does this proof work for the more general case. I'm not seeing why $\phi$ being simple and 0 almost everywhere imply that it's integral is 0 since this is something we are trying to prove in the first place.
Dec
21
asked Question about simple functions as described in Folland's Real Analysis
Dec
17
awarded  Yearling
Dec
17
asked Improper Riemann and Lebesgue Integrals
Oct
21
accepted Different ternary representations
Oct
19
comment Different ternary representations
Oh, no I have not. I will look this up.
Oct
19
asked Different ternary representations
Oct
15
accepted Kernel of cononical ring homomorphism
Oct
15
comment Kernel of cononical ring homomorphism
This helps clear things up. Thank you!
Oct
15
asked Kernel of cononical ring homomorphism
Oct
6
comment Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring
Sorry, this isn't quite right what I meant was that I get $ax+ay\rho+bx\rho+by\rho^2=ax+by(\rho+\frac{D-1}{4})+(ay+bx)\rho$, which give $(ax+by(\frac{D-1}{4}))+(ay+bx+by)\rho$
Oct
6
accepted Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring
Oct
6
comment Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring
So, $(a+b\rho)(x+y\rho)=(ax+\frac{1}{4}by(\rho+\frac{D-1}{4})+(ay+bx)\rho^2$. Then since $D\equiv 1$ $\frac{D-1}{4}$ must be an integer. And so the $by$ goes with the $\rho$ (I for some reason assumed it was already there?). Thank you!
Oct
6
revised Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring
added 45 characters in body
Oct
6
asked Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring