- #1
mateomy
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A passage from "Excursions in Number Theory":
"A transcendental number is not a solution of any algebraic equation. Pi is a familiar example of such a number and there are infinitely many others. A circle, centered at the origin, with radius pi (or any other transcendental number) has on it no points both of whose coordinates are rational. For all points of such a circle must satisfy the equation
[tex]
x^2 + y^2 = \pi^2
[/tex]
and
[tex]
\pi = \sqrt{x^2 + y^2}
[/tex]
...for rational x and y would make pi merely irrational and not transcendental."
Maybe this is a simple minded question, but how can a circle have a radius pi?
"A transcendental number is not a solution of any algebraic equation. Pi is a familiar example of such a number and there are infinitely many others. A circle, centered at the origin, with radius pi (or any other transcendental number) has on it no points both of whose coordinates are rational. For all points of such a circle must satisfy the equation
[tex]
x^2 + y^2 = \pi^2
[/tex]
and
[tex]
\pi = \sqrt{x^2 + y^2}
[/tex]
...for rational x and y would make pi merely irrational and not transcendental."
Maybe this is a simple minded question, but how can a circle have a radius pi?