Farmer Brown's conjecture - sci.math

Message from discussion Farmer Brown's conjecture

T.H. Ray

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More options Oct 7 2008, 5:32 am

Newsgroups: sci.math

From: "T.H. Ray" <thray...@aol.com>

Date: Tue, 07 Oct 2008 05:32:41 EDT

Local: Tues, Oct 7 2008 5:32 am

Subject: Re: Farmer Brown's conjecture

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> One multiplied by one is two.
> One instance of one is one.
> One times one is one.

> The square root of two is one.

If the square root of two were one, the square root
of four would also be one, by your reasoning. After
all, one instance of four is one--right? One times four
is one. One instance of four cannot be four instances;
otherwise, 1=4.

To a mathematician, it very much matters what one
means by "one" and "two." These are not "instances;"
they are values. If one cannot differentiate one value
from another, arithmetic does not exist.

The technical term you are missing, is "multiplicative
identity." It means that the integer 1 does not change
the identity of any other value by which it is
multiplied. Similarly, the number zero is called
"additive identity" in that it does not change the
identity of any value to which it is added.

Getting to square roots--we derive the value square root
of two by adding one side of the square to one side
of the square. There are two "instances" of one,
not one. The hypotenuse is therefore one "instance" of
two. The multiplicative identity holds.

That the square root of two is irrational pertains to
its relation to the sides of the right triangle of
which the hypotenuse is a member. Complete the square
on the other side of the hypotenuse, and you have your
rational increase, counting the sides. As easy as 2+2=4.
No need to make new rulers.

Tom

> In a proper real world counting system, a proper
> ruler would consist
> of inch increments whereby each subsequent inch, was
> longer than the
> one before it, and if you knew the proper amount to
> increase each
> increment, your math could always result in rational
> numbers.

> I have 4 chickens, and each chicken lays 4 eggs.

> Thats 16 eggs.

> I walk into the barn and I see 16 eggs and I say, how
> many chickens
> laid
> 4 eggs, and I say 4.

> If I say, I want to optimize chicken capacity, and
> find out how many
> equal chicken egg units, I need to arrive at 16 eggs,
> then I can
> square
> that amount and end up with 4 chickens 4 eggs.

> I walk into the barn, there are two chickens and two
> eggs.
> Both are walking funny.

> Is it not, prudent, for me to assume, that one
> chicken did not lay 1.4
> to infinity eggs 1.4 to infinity times?

> Is it not more rational then to assume that each
> chicken laid one
> egg?

> If I say, that mysteriously 2 apples appeared on the
> teachers desk two
> times out of thin air and then disappeared into thin
> air, and they
> were the same apples each time, because I saw they
> had numbered
> stickers on them, so two instances of two apples
> appeared on the
> teachers desk.
> How many apples were there?

> 2 apples.

> If I say, mysteriously apples began to multiply on
> the teachers desk,
> as twice 2 apples appeared out of thin air.

> 4 apples. I am using the term multiplied, and the
> term twice.

> Rephrased two apples mysteriously appeared two times
> on the teachers
> desk.
> 2 apples.

> The apples multiplied as two times two apples
> appeared on the teachers
> desk.

> So then the reason the square root of two is one, is
> that the function
> of multiplying real objects, means to increase their
> number. Whereas
> 'instances of', which is easily confused with
> 'multiplied by' refers
> to events where the number of actual apples is
> ambiguous,
> And when you talk about square root, the opposite of
> square root is
> multiply, not instance of.

> Therefore, one times one, equals one instance of one,
> and is not the
> same as one multiplied one time, as one multiplied
> one time, means
> that the number of apples increased once, which means
> that one times
> one equals two.

> If we use Pythagorean Theorem, to prove the square
> root of 2 we are
> not dealing in real world objects, we are now dealing
> in imaginary
> lines and ambiguous smeared out infinitely divisible
> imaginary
> quantities.
> However, Farmer Brown's conjecture states, that given
> the proper
> ruler, with the proper incremented counting system,
> your result would
> be a rational finite number.

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