Irrational circles about the orgin

  • Thread starter Thread starter ramsey2879
  • Start date Start date
  • Tags Tags
    Circles Irrational
ramsey2879
Messages
841
Reaction score
3
I recall a post previously where the Op was wondering if any circle about the orgin having an irrational radius could pass through a rational point. The answer then was if the irrational radius was the square root of the sum of two rational squares then of course.

Now I am wondering what if the radius was a non-algebraic number? Could that circle pass through a rational point?
 
Physics news on Phys.org
Hint: Suppose that r is transcendental and is the radius of a circle centered at the origin. Show that the existence of a rational point on this circle implies that r is algebraic.
 
Petek said:
Hint: Suppose that r is transcendental and is the radius of a circle centered at the origin. Show that the existence of a rational point on this circle implies that r is algebraic.
I was thinking that way but then I was wondering if transcendental numbers actually exist since it seems strange to me that one could travel a full circle and not come upon a rational point. Sorry but I have other things that I am working on and don't have time to look up the theory of transcendental numbers for now.
 
ramsey2879 said:
I was thinking that way but then I was wondering if transcendental numbers actually exist since it seems strange to me that one could travel a full circle and not come upon a rational point. Sorry but I have other things that I am working on and don't have time to look up the theory of transcendental numbers for now.

I'm honestly confused at this point. You used the term "non-algebraic number" in your original post. Non-algebraic numbers are exactly the same as transcendental numbers (if we're confining this discussion to real numbers). If I changed my hint to

Hint: Suppose that r is non-algebraic and is the radius of a circle centered at the origin. Show that the existence of a rational point on this circle implies that r is algebraic.[/color]

would that make more sense?
 
Yes I can see that x^2 + y^2 = r^2 would make r algebraic, but my real question was how can one move completely around a circle without passing over a rational point. Didn't that call into question the existence of non-algebraic values as real numbers?
PS, I took a look at some of the information of non-algebraic (transcendental) numbers and guess that they do exist. Moreover I noted that there are far more algebraic irrational numbers and transcendental numbers, so I guess that it must be true that such a circle has no rational point lying thereon.
 
Last edited:
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top