If my algebraic memory is correct from several decades ago, the flaw is:
[X(2) - Y(2)] is not the same as [(X-Y)(X+Y)]
because
[(X-Y)(X+Y)] is equal to [ X(2) - 2XY + Y(2) ]
The "factoring" of [X(2) - Y(2)] is erroneous.
Jurassic math major,
Alan Steward
At 18:15 07/30/2001 -0400, you wrote:
>Congratulations! you found the flaw.
>I've always taught that "cancelling" is a dirty word and should not be used.
>Too many teachers use that word and what they should be saying is that we
>"divide" both sides by x - y (or whatever they are dividing by)
>----- Original Message -----
>From: "ForenSys" <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Monday, July 30, 2001 6:06 PM
>Subject: Re: RBDos 6.5++ LASTKEY function seems faulty
>
>
> > > (x-y) (x+y) = y(x-y)
> >
> > Of course, you can't divide both sides by (x-y) because x-y=0 and you
>can't
> > divide by zero.
> >
> > BUT, if TOLERANCE is set greater than or equal to 1, then 1=2 in RBase.
> >
> > Regards,
> >
> > Stephen Markson
> > ForenSys The Forensic Systems Group
> > www.ForensicSystemsGroup.com
> > 416 482 2140
> >
> > ----- Original Message -----
> > From: "Bernard Lis" <[EMAIL PROTECTED]>
> > To: <[EMAIL PROTECTED]>
> > Sent: Sunday, July 29, 2001 3:15 PM
> > Subject: Re: RBDos 6.5++ LASTKEY function seems faulty
> >
> >
> > > I can't resist this -- your while 2 > 1 reminded me that 2 is really
> > > equal to 1. Proof follows:
> > >
> > > x=1 y=1 therefore x = y
> > > multiplying both sides by x gives
> > > x(2) = xy that's read as x squared
> > > subtracting y(2) from both sides we get
> > > x(2) - y(2) = xy - y(2)
> > > factoring: difference of 2 squares and common factor
> > > (x-y) (x+y) = y(x-y)
> > > canceling x-y on both sides we get:
> > > x + y = y or
> > > 2 = 1
> > >
> > > Yours truly,
> > > Bernie Lis
> >
> >
