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Irrational circles about the orgin

  1. Dec 8, 2012 #1
    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?
     
  2. jcsd
  3. Dec 8, 2012 #2
    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.
     
  4. Dec 8, 2012 #3
    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.
     
  5. Dec 8, 2012 #4
    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.

    would that make more sense?
     
  6. Dec 8, 2012 #5
    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 existance 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: Dec 8, 2012
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