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Sunday, August 16, 2015

Approximating square root of three

Sometimes I figure things out just for the fun of it.

When I get bored I can play with some math of mine. And noticed this thing years ago, but don't think I've put it out there.

With my BQD Iterator let F = 1, u=x, v=y, and D = -3. Then:

1. x2 - 3y2 = 1

2. (x+3y)2 - 3(x+y)2 = -2

3. (4x+6y)2 - 3(2x + 4y)2 = 4

Divide off 4, and you have: (2x+3y)2 - 3(x + 2y)2 = 1

4. (5x + 9y)2 - 3(3x + 5y)2 = -2

5. (14x + 24y)2 - 3(8x + 14y)2 = 4

Divide off 4 again, and you get: (7x + 12y)2 - 3(4x + 7y)2 = 1

And x = 1, y = 0 works, so I have:

1. 12 - 3(0)2 = 1

2. (1)2 - 3(1)2 = -2

3. (2)2 - 3(1)2 = 1

4. (5)2 - 3(3)2 = -2

5. (7)2 - 3(4)2 = 1

and you can keep going out to infinity.

And you're getting approximations to square root of 3, with the ratio of x/y, like 7/4 = 1.75.

But wow that is SLOW.

Oh yeah, you can use the algebraic ones to move things along as well. So let's plug the last result of x = 7, y =4, into the last algebraic one: (7x + 12y)2 - 3(4x + 7y)2 = 1

(7*7 + 12*4)2 - 3(4*7 + 7*4)2 = 1

(97)2 - 3(56)2 = 1

And is approximately 1.732 to three significant figures. And yes, you can now plug x = 97 and y = 56 back into it, if you wish to continue. Or you could move further out with an algebraic solution. So yeah there are an infinity of such equations to generate an approximate square root of 3. Guess that's kind of cool.

I'm curious! So I'll plug them back into what I have. Not quite curious enough to extend to the next algebraic one:

(7*97 + 12*56)2 - 3(4*97 + 7*56)2 = 1

(1351)2 - 3(780)2 = 1

So I now have 1351/780 which is approximately 1.73205, which I think is more satisfying.

You can also look at 13512/7802 is approximately 3.000001.

That was fun! And it was something quick to do.

I LOVE to play with actual numbers. My discoveries mean I can always get my number theory fix whenever I need it. Often I'll figure something out just to feel better.

Works marvelously for some reason.


James Harris
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