On 15 Oct 2008 13:52:56 -0400, Warp wrote:
> There are 11 types of programmers: Those who understand Gray code and
> those who don't.

There are 10.010000010010000000010000100001000100010000000010000100000001...
types of mathematicians: Those who understand base-(squareroot of 2)
and those who don't.

-- 
Joel Yliluoma - http://iki.fi/bisqwit/

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Joel Yliluoma wrote:

> There are 10.010000010010000000010000100001000100010000000010000100000001...
> types of mathematicians: Those who understand base-(squareroot of 2)
> and those who don't.

I think 10.01 (base-phi) is more fun.

And I *still* think the cute thing about binary is that there are *2* 
types of people, and we have base *2* numbers. ;-)

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Invisible wrote:

> I think 10.01 (base-phi) is more fun.

Or perhaps 011 (Fibonacci code).

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On Thu, 16 Oct 2008 13:09:40 +0100, Invisible wrote:
> Joel Yliluoma wrote:
>
>> There are 10.010000010010000000010000100001000100010000000010000100000001...
>> types of mathematicians: Those who understand base-(squareroot of 2)
>> and those who don't.
>
> I think 10.01 (base-phi) is more fun.

That would be 10.0011, wouldn't it?
Still, rather neat.

Hmm, actually 10.01 gives the same result.

bisqwit@chii:~$ bc -l
x=(1+sqrt(5))/2;
x^1+x^-3+x^-4
1.99999999999999999999
x^1+x^-2
1.99999999999999999999

-- 
Joel Yliluoma - http://iki.fi/bisqwit/

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>> I think 10.01 (base-phi) is more fun.
> 
> That would be 10.0011, wouldn't it?
> Still, rather neat.
> 
> Hmm, actually 10.01 gives the same result.

http://en.wikipedia.org/wiki/Phinary

"A base-φ numeral that includes the digit sequence "11" can always be 
rewritten in standard form, using the algebraic properties of the base φ 
— most notably that φ+1 = φ2."

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> "A base-φ numeral that includes the digit sequence "11" can always be 
> rewritten in standard form, using the algebraic properties of the base φ 
> — most notably that φ+1 = φ2."

That should of course be "φ+1 = φ^2".

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> That would be 10.0011, wouldn't it?
> Still, rather neat.
>
> Hmm, actually 10.01 gives the same result.
>
> bisqwit@chii:~$ bc -l
> x=(1+sqrt(5))/2;
> x^1+x^-3+x^-4
> 1.99999999999999999999
> x^1+x^-2
> 1.99999999999999999999

Tut tut, youth of today, taking short cuts with numerical methods ... just 
do the algebra and you'll see they are both *exactly* 2, not this 
1.9999999999 mess ;-)

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scott wrote:

> Tut tut, youth of today, taking short cuts with numerical methods ... 
> just do the algebra and you'll see they are both *exactly* 2, not this 
> 1.9999999999 mess ;-)

...says the guy who thinks that "1/0" is a definite value. :-P



Hmm, let's see now...

φ = (1 + sqrt(5)) / 2

φ + 1 = (1 + sqrt(5)) / 2 + 1 = (3 + sqrt(5)) / 2

φ^2 = (1 + sqrt(5))^2 / 4 = (1 + 2 sqrt(5) + 5) / 4 = 1/4 + sqrt(5)/2 + 5/4

Nope, sorry, I'm not seeing it.

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> Hmm, let's see now...
>
> φ = (1 + sqrt(5)) / 2
>
> φ + 1 = (1 + sqrt(5)) / 2 + 1 = (3 + sqrt(5)) / 2
>
> φ^2 = (1 + sqrt(5))^2 / 4 = (1 + 2 sqrt(5) + 5) / 4 = 1/4 + sqrt(5)/2 + 
> 5/4
>
> Nope, sorry, I'm not seeing it.

You just need to do a couple more steps:

1/4 + sqrt(5)/2 + 5/4
=
6/4 + sqrt(5)/2
=
3/2 + sqrt(5)/2
=
(3+sqrt(5))/2

:-)

Or you can just solve x+1=x^2 your favourite way.

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>> Nope, sorry, I'm not seeing it.
> 
> You just need to do a couple more steps:
> 
> 1/4 + sqrt(5)/2 + 5/4
> =
> 6/4 + sqrt(5)/2
> =
> 3/2 + sqrt(5)/2
> =
> (3+sqrt(5))/2
> 
> :-)

I failed to notice the two constant terms at opposite ends of the 
equation. Clearly I need to invent that CAS real soon now... o_O

> Or you can just solve x+1=x^2 your favourite way.

Yeah, I guess.

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