Square Root Of 2 (from Hardy Course Of Pure Mathematics )

[moe darklight]

Square Root Of 2 (from Hardy "Course Of Pure Mathematics")

I was surprised to find it in my local bookstore amidst math "cheat" books in the one-shelf math section (and the fact that it was the last copy left... ??).

Section 5, Example 3:
Show that if m/n is a close approximation to \sqrt{2}, then (m+2n)/(m+n) is a better one, and that errors in both cases are in opposite directions.

Is he asking to simply carry out the calculation? Or is he asking to answer why this method works, because I have no clue for the latter... how did he get that formula?

Also, I've noticed that if the formula is carried out on m=\sqrt{2}, n=1; then the formula does not apply. Is this due to the irrationality of the number? (as in: the approximation would never actually reach \sqrt{2}, so it is always applicable)

 

[Borek]

Perhaps m & n should be integers?

Try to compare squares of both approximations to 2 - which one is closer?

 

[moe darklight]

Perhaps m & n should be integers?

Try to compare squares of both approximations to 2 - which one is closer?

yes I've followed through the calculations and it works out. I was just wondering if he also expects the reader to be able to come up with (m+2n)/(m+n) on his own, because I don't understand where the formula comes from.

 

[HallsofIvy]

yes I've followed through the calculations and it works out. I was just wondering if he also expects the reader to be able to come up with (m+2n)/(m+n) on his own, because I don't understand where the formula comes from.

That is certainly NOT implied by the question.

 

[epenguin]

So just prove it.

But after you have proved it, think about it and the textbook context and why it should have been obvious or suggested itself in the first place or what it is connected with.

The Polya booklet "How to Solve It" has about 5 rules and recommendations - one of them is "when you have solved it the job is not finished" (do essentially what I said +...).

Oh and if successful come back and tell us about it as we are too busy to do it ourselves.

Whaddayamean lazy?!

 

[robert Ihnot]

Borek: Perhaps m & n should be integers?

The case does work for m=2, n=1, the next approximations being 4/3, 10/7, 24/17.

However, the better way to do it is to choose 1,1: 1/1, 3/2, 7/5, 17/12...BECAUSE

in the Pellian 1^1-2(1^1) = -1, 3^2-2(2^2) = 1, 7^2-2(5^2) = -1, 17^2-2(12)^2 =1.

Thus the key to the problem is the equation: x^2-2(y^2) = +/- 1. This choice of integers x,y is as close as we can get to the square root of 2, considering the size of the numbers.

 

[Gib Z]

Is he asking to simply carry out the calculation? Or is he asking to answer why this method works, because I have no clue for the latter... how did he get that formula?

To investigate this problem I wanted to be able to write this as a sequence. As m/n would be the first term, we can rewrite

(m+2n)/(m+n) as 1 + n/(m+n)

by splitting the numerator, and then divide the fraction by n to get

1+ \frac{1}{1+ (m/n)}.

So now if the first term in our sequence be a_0, then

a_1 = 1+\frac{1}{1+a_0},

and in general a_n = 1 + \frac{1}{1+ a_{n-1}}.

So by direct substitution, what is the next term, and the term after that? If you are still hazy, you may want to look up the continued fraction representation of sqrt 2.

 

[robert Ihnot]

(I can't vouch for the rigor here, but...) We can split up the factors and be ready to multiply by the complement. (By the complement I mean x+\sqrt2y, has the complement of x-\sqrt2y.

Take the Pellian, split the factors: x-\sqrt2y=\pm(\sqrt2-1) Then multiply by 1-\sqrt2 resulting in x+2y-\sqrt2(x+y) =\pm(\sqrt2-1)(1-\sqrt2). And multiplying by the complement, we arrive at:

(x+2y)^2-2(x+y)^2 = \mp1.

Because we have \pm1 on the right hand side, we can continue indefinitively to obtain new solutions to the equation, each one resulting in a change of sign. Thus the equation \frac{x^2}{y^2}-2=\frac{\pm1}{y^2} alternates being over or under \sqrt2 and becomes increasingly accurate as y increases.

 

[moe darklight]

thanks. but I did not know what a pellian or continued fractions are, and I haven't seen examples of either in my pre-calc textbook. Am I supposed to have seen this by now? I just finished 1st semester year 1 Calculus and linear algebra... but this is really the first time I study math at school, and my study of pre-calc was pretty rushed.

 

[robert Ihnot]

moe darklight: Am I supposed to have seen this by now?

A caution about Hardy. His stuff is more of a survey work than a text.

A member of the Royal Society. He also wrote on Number Theory, a subject in which he had great interest along with Ramanujan, the great mathematician from India. Hardy must have added some of that in the book you have. Continued fractions and the Pellian equation are from Number Theory.