Mea culpa. Like I said, Jurassic math major.
I was thinking the [(X+Y)(2)] and [(X-Y)(2)] constructs.
Alan Steward
At 21:25 07/30/2001 -0400, you wrote:
>Sorry!
>multiplying (x - y)(x+y) is x(2) - xy + xy - y(2) or x(2) - y(2)
>Read the post from ForenSys for the correct answer
>----- Original Message -----
>From: "Alan Steward" <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Monday, July 30, 2001 8:39 PM
>Subject: Re: RBDos 6.5++ LASTKEY function seems faulty
>
>
> > 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
> > > >
> > > >
> >
