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Square Root of 2 as a 'Vulgar Fraction'

Date: 05/04/2001 at 00:44:37
From: Dave
Subject: Square root of  2 as a 'vulgar fraction'

As I am semi-retired, and have always been interested in numbers 
from a recreational angle, I have recently been trying to work out 
the square root of 2 as a 'vulgar fraction', and so far have found 
that 99/70 (1+29/70) comes very close to 2. (In fact, it is out by 
only 1 in 9,800, and when multiplied in decimal notation, comes to 
2.00020408163+).  I can only assume that when I work my way up through 
the higher numbers this result will get even closer to 2.   

The question that I am asking is: will I eventually find two whole 
numbers, which when dividing the larger by the smaller, and then 
multiplying the result by itself, will actually give me the exact 
value of 2, or will they keep on 'just missing' by 1, as they have 
done so far?   

The reason I ask this is that when I draw a right-angled triangle, 
with two equal sides of, say, 1 foot, the length of the hypotenuse is 
(by Pythagoras) the exact square root of 2, and because it is a 
1.414+-foot-long line (as far as I can accurately measure it), it must 
have a definite value, as it is of finite length. Therefore, I would 
logically think that there is also a definite pair of numbers which, 
as a vulgar fraction, would also be finite, and in the same way also 
the exact square root of two; otherwise there would seem to be a 
conflict of logical thought. (Or am I overlooking some perceptual 

Perhaps you may wish to enlighten me on this little matter.

Date: 05/04/2001 at 09:05:31
From: Doctor Peterson
Subject: Re: Square root of  2 as a 'vulgar fraction'

Hi, Dave.

One of the crucial discoveries of the ancient Greeks was exactly this: 
the square root of two can't be expressed as a fraction. We call it an 
irrational number. This was very disturbing to the Greeks, who had 
assumed that all lengths they could construct geometrically could be 
"measured" using whole numbers of the same (tiny) units. They made the 
same "logical" assumption you made, and when they realized they were 
wrong, it marked the beginning of math as we know it, with its demand 
for proof.

You can find the proof of this fact everywhere; a search of our 
archives for the key words "irrational square root" shows at least a 
dozen answers to this question or a closely related one. You can start 
with our FAQ on irrational and other types of numbers, and one of the 
links listed there:

   Integers, Rational and Irrational Numbers - Ask Dr. Math FAQ   

   Irrational Numbers - Jim Loy   

You may be interested in a method for finding rational approximations 
to the square root of 2 using continued fractions. This answer will 

   Continued Fractions - Ask Dr. Math archives   

It turns out that any fraction of the form

    1 + ---------------
        2 + -----------
            2 + -------
                2 + ---

to any depth will be a good approximation, getting closer and closer 
the farther you go. We get 3/2, 7/5, 17/12, 41/29, 99/70, and so on. 
But this process never stops; you will never get a fraction that is 
exactly the square root of 2.

- Doctor Peterson, The Math Forum   

Date: 05/04/2001 at 09:09:54
From: Doctor Rob
Subject: Re: Square of  2 as a 'vulgar fraction'.

Thanks for writing to Ask Dr. Math, Dave.

Your task is doomed to failure.

Numbers that can be written as fractions are called rational numbers.
The discovery by Pythagoras and his pupils that sqrt(2) is NOT a
rational number was one of the most important in ancient mathematics.
There are several proofs of this, based on divisibility.  Here is one.

Suppose that sqrt(2) = a/b, where a and b are positive integers, and
a/b is reduced to lowest terms (that is, a and b have greatest common
divisor 1). Then

   sqrt(2) = a/b
         2 = a^2/b^2
     2*b^2 = a^2

Now a^2 is a divisor of 2*b^2. Recall that a and b had greatest common 
divisor 1. Then a^2 and b^2 also have greatest common divisor 1. That 
implies that a^2 must be a divisor of 2. The only perfect square 
divisor of 2 is 1, so a^2 = 1, and a = 1. Then you have

   2*b^2 = 1

so 2 is a divisor of 1, a contradiction.  The conclusion of this
argument is that sqrt(2) cannot be expressed as a/b for any positive
integers a and b.

Numbers like sqrt(2) that are not rational numbers are called
irrational numbers. Other examples are sqrt(n) for any positive 
integer n not a perfect square, pi, and e.

The best approximations you can get are these:

   c[1] = 1/1 = 1
   c[2] = 3/2 = 1 + 1/2
   c[3] = 7/5 = 1 + 1/(2+1/2)
   c[4] = 17/12 = 1 + 1/[2+1/(2+1/2)]
   c[5] = 41/29 = 1 + 1/(2+1/[2+1/(2+1/2)])
   c[6] = 99/70 = 1 + 1/[2+1/(2+1/[2+1/(2+1/2)])]
   c[7] = 239/169 = 1 + 1/(2+1/[2+1/(2+1/[2+1/(2+1/2)])])
   c[8] = 577/408 = 1 + 1/[2+1/(2+1/[2+1/(2+1/[2+1/(2+1/2)])])]

They satisfy

   c[n+1] = 1 + 1/(1+c[n])

You mention having found c[6] already.  This is related to the fact 
that 99^2 - 2*70^2 = 1.

I hope you derive enjoyment from these facts, as I do.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Number Theory

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