Pi as a transcendental

[mateomy]

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

x^2 + y^2 = \pi^2

and

\pi = \sqrt{x^2 + y^2}

...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?

[CompuChip]

In the same way as it has radius 2 or 43/19?
You can even construct one as follows: You can make a circle with a piece of rope by pinning it to some paper at one end, pulling it tight and drawing the position of the other end as you move it around a circle (for example by tying a pencil to it).
Just start with a piece of rope of length 1/2 (centimeter, meter, foot, yard, whatever), this will give you a circle with circumference 2 pi * 1/2 = pi. Take a new piece of string, lay it around the circle, cut it where it goes around once and use that to draw a new circle. It will have radius pi :)

[HallsofIvy]

A passage from "Excursions in Number Theory":

"A transcendental number is not a solution of any algebraic equation.
The standard definition is that a transcendental number is not a solution to any polynomial equation with integer coefficients. It can, then, be shown that any polynomial equation with rational coefficients can be changed to an equation with integer coefficients having the same roots (multiply through by the least common denominator of all coefficients). Any rational equation can be written as a fraction (one polynomial divided by another) equal to 0. Multiplying both sides by the denominator then gives a polynomial equation having the same roots. Finally, an equation involving roots can be converted to a polynomial equation having the same roots by taking powers. That is the sense in which "a transcendental number is not a solution to 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

x^2 + y^2 = \pi^2

and

\pi = \sqrt{x^2 + y^2}

...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 line segment can have any number as length. You cannot construct a segment of length \pi with compasses and straight edge but that has nothing to do with some line segment having that length.

(Compuchips construction uses more than compasses and straight edge.)

[mateomy]

That makes it easier to comprehend (Both HallsofIvy and CompuChip), I guess I get stuck on primarily thinking of pi as an infinitely long decimal, rather than a geometric construct.

Thanks for the explanations.

[CompuChip]

If you don't think about it as "an infinitely long decimal" but simply as "a number" just like 2 or 3/4 then you will be fine as well, in this case :)

[CRGreathouse]

Remember, numbers like pi that have infinitely many decimal places are the usual case. Numbers like 12.456 are the exception -- the unusual numbers that can be represented as an integer divided by a power of 10. There's nothing 'wrong' or 'strange' about pi.