[net.math] -1 = 1

stevesu@bronze.UUCP (Steve Summit) (09/10/83)

The proof that a = b was pretty obvious.  Here's one I discovered
by accident and still can't quite figure out:

1 = 1                   reflexive property of equality

2/2 = 2/2               another name for 1

  2/2    2/2
-1   = -1               raise -1 to both sides

  1              2       b/c      th         b
-1 = sqrt( ( -1 )  )    a    ==  c  root of a

                                     2
-1 = sqrt( 1 )          reduce ( -1 )	

-1 = 1                  reduce sqrt( 1 )

The mistake is probably in the fourth step, but no book I've seen
places restrictions on a, b, or c in that identity.

					Steve Summit

laura@utcsstat.UUCP (Laura Creighton) (09/10/83)

you cannot take the square root of something in that way. You have
to split your question instead.  

if k**2 is 16, thn you have to say that k is + or - the square root of 16.
thus k is +4 or k is -4. In the same way with your question, you prove
that 1 = 1 or 1 = -1. While the second statement is false, the union
of them is true.

Laura Creighton
utzoo!utcsstat!laura

laura@utcsstat.UUCP (Laura Creighton) (09/10/83)

Actually, you proved that -1 is equal to 1, or -1 is equal to -1,
but for the reasons I stated.

laura

asente@decwrl.UUCP (Paul Asente) (09/11/83)

It's really quite straightforward:  sqrt is a multiple valued relation
in mathematics, although you don't usually think of it that way:

sqrt(4) = {+2, -2}	(I really mean a set here!)

Transitivity does not hold in this situation, any more than it does in
the situation

"John is a boy; Tom is a boy; therefore John is Tom."

	-paul asente

debray@sbcs.UUCP (Saumya Debray) (09/12/83)

To say that

	  1              2       b/c      th         b
	-1 = sqrt( ( -1 )  )    a    ==  c  root of a

is only partially correct.  There are n values for the nth root of any
number (not all of which might be real, e.g. cube roots of 1). Thus, we
don't have a one-to-one mapping between (bth. root of a) and
((bth. root of a) to the bth. power).

Saumya Debray
SUNY at Stony Brook

ecn-ec:ecn-pc:ecn-ed:vu@pur-ee.UUCP (09/13/83)

O	The mistake: a **(b/c) = c th root of ( a ** b ) is right,
BUT it might be  - c th root of ( a ** b ), in case there are
two different real root, i.e. in case c is even.

	So  -1 = - sqrt(1) and that's more than legal.

	Hao-Nhien Vu (pur-ee!norris)