

A 3700-year-old proof that the diagonal of a unit square has length √2 - sethg
http://www.math.ubc.ca/~cass/Euclid/ybc/ybc.html

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smanek
In the 6th century BC Pythagoras' followers knew that the square root of two
was irrational (couldn't be expressed as the ratio of two integers). The story
goes that this messed with a lot of their theories so it was ordered to be
kept a secret, and anyone who dared reveal the fact was killed.

It seems kind of funny to me that they actually had a cult/religion based on
math. Instead of just accepting the existence of irrationals and attempting to
update their theories - they tried to suppress the new evidence that
contradicted their teachings.

~~~
tome
But what would "accepting the existence of the irrationals" mean? As far as
the Greeks knew, a "number" _was_ a/b where a and b were integers.

"Accepting" that the length of a triangle was not a "number" could have all
sorts of consequences, including rejection of their belief in validity of
mathematical reasoning, so I'm not surprised they were scared of it.

~~~
smanek
The Greeks did know about irrational numbers. They could draw a right triangle
whose legs were 1 and 1. They knew that the length of the hypotenuse (a number
that we call the square root of two) couldn't be represented as the ratio
between any two integers, yet it had a length.

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gjm11
Not convinced that that's what it is. It's certainly a ~3700-year-old
_statement_ that the diagonal has length sqrt(2), but there's no evidence that
it is, or was intended to be, or demonstrates that anyone had, a _proof_.

The author of those pages says "It amounts to a dissection of the square on
the hypotenuse of an isosceles right triangle into pieces which can be
reassembled to make up the two squares on the sides, and I can't see why the
figure is exactly what it is if it weren't understood to demonstrate this."
but it seems to me that even if all you want to do is write down that the
diagonal is sqrt(2) then you'll need at least the square and one diagonal, and
adding the other could as well be motivated by love of symmetry as by having
noticed that with it there you can dissect-and-reassemble into a 2x1 rectangle
or whatever.

Remarkable, none the less.

~~~
chancho
If you know that 1/2 * base * height = the area, then the second diagonal can
be seen as the height of the triangle cut by the first diagonal. Since b = 2
h,

1/2 b h = 1/2 ==> b h = 1 ==> 1/2 b b = 1 ==> b b = 2 ==> b = sqrt(2).

I can't read cuneiform so I don't know if it makes this argument.

~~~
aneesh
You'd first have to prove that the diagonals bisect each other (so that you
can say b = 2h). I'm not sure whether that result was known at the time, but
it's not too hard to prove.

~~~
chancho
You're right. That's pushing it for those handful of scratches on the tablet.
Who knows, maybe they had some lemma tablets lying around.

~~~
sethg
From my own limited experience, ancient documents tended to be very
elliptical; when everything had to be copied out by hand the reader was
expected to do a little more work to understand what documents meant. So this
tablet may have been the equivalent of notes accompanying a lecture: "As you
can see from the triangles on this diagram..."

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Tichy
Could also be a failed attempt at drawing the "nikolaus hut" (not sure what it
is called in english: [http://www.kreudenstein-
online.de/AppletNikol/AppletNikol.ht...](http://www.kreudenstein-
online.de/AppletNikol/AppletNikol.html)).

That, or their alphabet is REALLY expressive.

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daveying99
Where is the proof?

~~~
Avshalom
There is no proof specifically this is really annoying about most of the
Babylonian Texts, they never go into method very much.

However from the numbers they got (30, 1;24,51,10, and 30*1;24,51,10=42;25,35)
it looks less like they used a proof as they did use an iterative
approximation method as follows:

let a be some number

let a1 be an approximation of sqrt(a) such that a1 > sqrt(a)

then B1 = a/a1 is also an approximation of sqrt(a) but deviates in the other
direction s B1 < sqrt(a).

we have now bracketed sqrt(a)

so a new better bracketing can now be made by

a2 = (a1 +B1)/2 and B2 = a/a2

and then

a3 = (a2 + B2)/2, B3 = a/a3

etc. the answer for sqrt(2) of 1;24,51,10 happens to equal a3 in this case.

note that the above is only speculation, but the method DOES produce the
Babylonian numbers so and it uses only arithmetic that we know they had, so it
seems pretty likely

(All this information and more can be found in Mathematical Cuneiform Texts by
Otto Neugebauer)

