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Squaring the Circle

Date: 12/22/97 at 12:19:20
From: deloach
Subject: Squaring the circle

I understand that the problem of squaring the circle implies that, 
using only a compass and straightedge, you can't construct a square 
with the same area as a circle with a given radius. My question /
unease is hard to express, but it's something like: can you determine 
such a square at all? How can you determine precisely the square root 
(the side of the square) of a number that has been arrived at from 
multiplication with a number such as pi, a number that never repeats 
the decimal? Aren't you just determining a number that's as accurate 
as you wish, but not THE answer?

Date: 12/22/97 at 14:11:27
From: Doctor Rob
Subject: Re: Squaring the circle

Equivalently, you cannot construct a line segment whose length is the
perimeter of a circle whose radius is given. If you could, you could 
then construct a line segment whose length was Sqrt[Pi] times the 
radius, and that would be the length of the side of the square in 

You can't determine this square root exactly, just as you cannot 
determine the square root of 2 exactly, since it, too, is an 
irrational number  You can, however, construct a line segment whose 
length is Sqrt[2] times the length of a given line segment. Sqrt[2] 
and Sqrt[Pi] are two valid real numbers.

In a sense, I think you are asking about irrational real numbers, or
perhaps even transcendental real numbers. How do we specify one of 
them if we cannot write its full decimal expansion?

For algebraic ones, they are specified by the polynomial equation they
solve. Sqrt[2] is the positive root of x^2 - 2 = 0, for example.  
Often we don't worry about the actual value of Sqrt[2], but leave it 
as a symbol, and whenever we see Sqrt[2]^2, replace it with 2. The 
same trick works with the imaginary unit i, where i^2 + 1 = 0.

For transcendental ones, we often give an infinite series, an infinite
product, or an infinite sequence which converges to the number, or a
definite integral whose value is the number. We also often just give 
it a symbol, and work with the symbol, until we *really* need to know 
the numerical value to some degree of accuracy.  For example,

   Pi = 4*[1 - 1/2 + 1/3 - 1/4 + 1/5 - ...],

   Pi = 4*(1-1/3^2)*(1-1/5^2)*(1-1/7^2)*...,

   Pi = 4*Integral  1/(1+x^2) dx,

    e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + ...,

    e =    lim      (1 + 1/n)^n.

Then, at least in theory, one can compute the number in question to as 
much accuracy as desired. For Sqrt[Pi], there is the following 
infinite product:

Sqrt[Pi] = Sqrt[2]*(2/3)*(4/5)*(6/7)*(8/9)*...

To actually compute Sqrt[Pi], one would undoubtedly compute Pi to many
digits, then take its square root by one of the standard methods.  
This is because the above expressions are too slowly convergent to be 
useful for computations of extreme accuracy, but there are very fast 
methods to compute Pi as the limit of very rapidly convergent infinite 

-Doctor Rob,  The Math Forum
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Associated Topics:
High School Constructions
High School Geometry

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