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### Investigation Involving Square Root of 2

```
Date: 07/09/2001 at 14:14:43
From: Loka
Subject: Further investigation involving Square root of 2

Please Dr. Math, can you help me on this problem?

I have had to work out: (1+sqrt(2))^2 = 1 + 2(sqrt(2) + 2).

I have gone on to work out (1+sqrt(2))^4, and then (1+sqrt(2))^8 in
similar fashion, so that I ended up with the answer 577 + 408sqrt(2).
(1+sqrt(2))^16. I found the answer to be 665857 + 470832sqrt(2).

Here's the bit I'm stuck on: how can you explain the fact that
(665857/470832)^2 = 2?

Please can you help me? I'll be very grateful.
```

```
Date: 07/09/2001 at 17:56:08
From: Doctor Rob
Subject: Re: Further investigation involving Square root of 2

Thanks for writing to Ask Dr. Math, Loka.

The fact is that (665857/470832)^2 is not equal to 2.  It is very
close, but that's not the same as equal. In fact, it is true that

(665857/470832)^2 = 2 + 1/470832^2.

The reason this is true is that

(1+sqrt[2])*(-1+sqrt[2]) = 1,
(1+sqrt[2])^16*(-1+sqrt[2])^16 = 1^16 = 1,
(665857+470832*sqrt[2])*(665857-470832*sqrt[2]) = 1,
665857^2 - 470832^2*2 = 1,
2 = (665857^2-1)/470832^2,
(665857/470832)^2 = 2 + 1/470832^2.

Since 1/470832^2 = 1/221682772224, the difference is probably too
small to show up on your calculator.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/10/2001 at 12:06:33
From: Doctor Peterson
Subject: Re: Further investigation involving Square root of 2

Hi, Loka.

Your problem is very interesting, especially if we look a little
deeper than the problem itself; I've been thinking about it, trying to
find a way to answer it at a reasonably simple level. I'll show you
what I've found.

Dr. Rob already pointed out to you that your calculation does not give
exactly 2; you probably knew that, since if it did, you would have a
rational expression for the square root of 2, and we know it is
irrational.

(In fact, my calculator gives 2.0000000000045109504449427720993 for
the square.)

The interesting question is, why should it be so close? He answered
that question as well, but you may have missed it, as I did when I
first read it. I'd like to elaborate a bit, and show you some of the
interesting sights along the way to the conclusion.

My first thought when I see a close rational approximation to an
irrational is "continued fractions." This is a way to generate a
sequence of increasingly accurate approximations to an irrational
number; you can read about it here and elsewhere in our archives:

Square Root of 2 as a 'Vulgar Fraction'
http://mathforum.org/dr.math/problems/dave.05.04.01.html

Now, you got a very accurate fraction faster than with continued
fractions; but I guessed that (1+sqrt(2))^n would be the nth
"convergent" (approximation to the root) generated by the continued
fraction, and I found I was right. Let me demonstrate. The continued
fraction for sqrt(2) is

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

The nth convergent is found by stopping after n steps; if you look
closely, you find that, if we call the nth convergent x[n], then

x[n+1] = 1 + 1/(1+x[n])
= (2+x[n])/(1+x[n])

Now, supposed we are finding (1+sqrt(2))^n by repeatedly multiplying,
rather than squaring as you did to find large powers faster. Then

(1+sqrt(2))^(n+1) = (1+sqrt(2))^n (1+sqrt(2))

If for the nth power we got a + b sqrt(2), then the (n+1)th will be

(a + b sqrt(2))(1 + sqrt(2)) = (a+2b) + (a+b) sqrt(2)

Defining our nth approximation to sqrt(2) as y[n] = a/b, we can write

y[n+1] = (a+2b)/(a+b)
= (a/b + 2)/(a/b + 1)
= (y[n]+2)/(y[n]+1)

As you can see, the approximations we are producing are just the
convergents obtained by continued fractions. By squaring repeatedly,
you are jumping ahead in this sequence with less work, skipping from 2
to 4 to 8 rather than plodding along one at a time.

I could show you that this will in fact approach sqrt(2); but that
would take us too far from the original problem. What Dr. Rob pointed
out does what we need, directly from 1+sqrt(2), rather than focusing
on convergents. Let's just suppose that

(sqrt(2)+1)^n = a + b sqrt(2)

Then a little thought will show that

(sqrt(2)-1)^n = +/-(a - b sqrt(2))

(where the sign will alternate for even and odd powers). Multiplying
these together,

(sqrt(2)+1)^n (sqrt(2)-1)^n = +/-(a + b sqrt(2))(a - b sqrt(2))
= +/-(a^2 - 2 b^2)

But since

(sqrt(2)+1)(sqrt(2)-1) = sqrt(2)^2 - 1^2 = 1

we've shown that

a^2 - 2 b^2 = +/-1

a^2 = 2 b^2 +/- 1

(a/b)^2 = 2 +/- 1/b^2

This is Dr. Rob's result: not just that a/b is not equal to the
square root of 2, but that the error is 1/b^2, which decreases as b
increases. (Note also that the direction of the error will alternate,
just as I showed above.) Since b increases very rapidly as you square
repeatedly, and the error is the square of that, you get a very close
approximation very quickly.

I hope this adds a little to your appreciation of the problem.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents
High School Number Theory

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