Re: Distance-dependent time contraction

From: mich (mich_at_efni.com)
Date: 06/27/04 

Date: Sun, 27 Jun 2004 09:35:13 -0400

 I'm replying to this thread because it puzzles me that what I thought a
very basic
mathematical rule may be more complicated. My knowledge of math is very
basic...

.... I always thought that just as minus was the opposite function of
positive, and
multiplication was the opposite of division, the squareroot was the opposite
function of the square..., and knowing that 2^2 and -2^2 are both equal to
4, I thought that the squareroot of 4 was indeed equal to + or - 2?

Andre

"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:BKFCc.165144$MG6.8457273@phobos.telenet-ops.be...
>
> "Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
news:TGFCc.1840$NL3.17890372@news-text.cableinet.net...
> >
> > "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:SoDCc.165007$8z6.8454274@phobos.telenet-ops.be...
> > |
> > | "Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
> > news:wkDCc.1673$Uo2.16328827@news-text.cableinet.net...
> > | >
> > | > "YBM" <ybmess@nooos.fr> wrote in message
> > | > news:40d9c10a$0$19025$636a15ce@news.free.fr...
> > | > | Androcles a écrit :
> > | > | > x = -3 or 3
> > | > | > y = -4 or 4
> > | > | > z = -5, given.
> > | > | > z = sqrt(x^2+y^2) - Pythagoras.
> > | > |
> > | > | Pythagoras ? Where did you see a triangle with sides
> > | > | measuring -3, -4 and -5 recently ?
> > | > Sure. I put a rule on the hypotenuse and read it from 5 to zero.
> > | > That's a negative distance, -5. It would be a positive distance if I
> > read it
> > | > from 0 to 5.
> > | >
> > | > | Do you know what
> > | > | properties any kind of *distance* has ?
> > | > Sure do.
> > | >
> > | > |
> > | > | > "This way sqrt is ALWAYS positive."-Dinky the Deranged.
> > | > | > So, according to Dinky, z = 5.
> > | > | > Egads, Pythagoras was wrong.
> > | > |
> > | > | z = 5 = sqrt( (-3)^2 + (-4)^2 )
> > | > | z = 5 = sqrt ( 3^2 + 4^2 )
> > | > | z = 5 = sqrt( (-3)^2 + 4^2 )
> > | > | z = 5 = sqrt ( 3^2 + (-4)^2 )
> > | > |
> > | > | Is any of these equations wrong ?
> > |
> > | > Yep. I said z = -5, GIVEN.
> > | > Since (-5)^2 = x^2 +y^2 for the values of x I gave, ALL of your
> > equations
> > | > are wrong. The correct answer is
> > | > z = -5 = sqrt( (-3)^2 + (-4)^2 )
> > | > z = -5 = sqrt ( 3^2 + 4^2 )
> > | > z = -5 = sqrt( (-3)^2 + 4^2 )
> > | > z = -5 = sqrt ( 3^2 + (-4)^2 )
> > |
> > | No, donkey ***, GIVEN
> > | z = -5
> > | the WHOLE WORLD writes
> > | - z = 5 = sqrt( (-3)^2 + (-4)^2 )
> > | - z = 5 = sqrt ( 3^2 + 4^2 )
> > | - z = 5 = sqrt( (-3)^2 + 4^2 )
> > | - z = 5 = sqrt ( 3^2 + (-4)^2 )
> > |
> > | Dirk Vdm
>
> > -z=5 <=> z = -5 has nothing to do with z = f(g(x),g(y)), shithead,
> > regardless of what the functions f and g or the variables x,y, or z may
be.
> > I can write that independently.
> > As it happens, I chose to write z = -5, you moron.
>
> Okay, if you are so afraid to write
> -z = 5
> because you go berzerk when you see
> -z = 5 = sqrt( (-3)^2 + (-4)^2 ) ,
> then you can always write
> z = - 5 = - sqrt( (-3)^2 + (-4)^2 )
> and then you have your
> z = -5
> Does *that* make you somehow feel better?
> Always glad to help :-)
>
> Dirk Vdm
>
>

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