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 reasoning?) 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 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. 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 http://mathforum.org/dr.math/ |
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