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, 12:27 pm

Newsgroups: sci.math

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

Date: Tue, 07 Oct 2008 12:27:15 EDT

Local: Tues, Oct 7 2008 12:27 pm

Subject: Re: Farmer Brown's conjecture

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rick sobie wrote

> On Oct 7, 10:32 am, "T.H. Ray" <thray...@aol.com>
> wrote:
> > > 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.

> No you are mistaken. I am saying multiplied by, not
> instance of.

That's not what you said. As long as I am laid up in
bed today, though, we might as well indulge in a little
silliness.

> Mr. Barnum peered with awe into his microscope and
> proclaimed "They
> are multiplying!"
> "The cells! They are multiplying by cell division.
> One cell multiplied
> one time into two cells I saw it with my own eyes"

Mr. Barnum saw one cell divide into two, not multiply.
Or don't you accept that 1+1=2?

> One cell multiplied into two cells, the cells went
> forth and
> multiplied.

> Once multiplied into two, and the reverse, one cell
> divided into two,
> and since one cell multiplied into two, one
> multiplied one time equals
> two, hence the square root of two is one.

Division is certainly the inverse (not reverse) of
multiplication; however, division of any integer,
=> 1, by itself is one. Did you not get my point about
multiplicative identity? Look it up.

> That is my proof and I would love to see your proof,
> that one
> multiplied one time equals one.

You just saw it.

Tom

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