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### Square Root of 2 as a 'Vulgar Fraction'

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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
reasoning?)

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

Integers, Rational and Irrational Numbers - Ask Dr. Math FAQ
http://mathforum.org/dr.math/faq/faq.integers.html

Irrational Numbers - Jim Loy
http://www.jimloy.com/algebra/irration.htm

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

Continued Fractions - Ask Dr. Math archives
http://mathforum.org/dr.math/problems/gold10.7.97.html

It turns out that any fraction of the form

1
1 + ---------------
1
2 + -----------
1
2 + -------
1
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
http://mathforum.org/dr.math/
```

```
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.

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
http://mathforum.org/dr.math/
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Associated Topics:
High School Number Theory

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