Re: Farmer Brown's conjecture

On Oct 7, 5:27 pm, "T.H. Ray" <thray...@xxxxxxx> wrote:
rick sobie wrote



On Oct 7, 10:32 am, "T.H. Ray" <thray...@xxxxxxx>
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?


Division is the opposite of multiplication and by good fortune, both
were observed.
The cells multiplied by dividing and that is what you would expect
given the uniqueness of the number one.


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.

One cannot be a prime number.



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

You just saw it.

Old rules are always overthrown.


.

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