Re: History of Existance in MathMatical Proof ~ 899

"Blake Brown" <[EMAIL PROTECTED]> wrote in message
> I wanted to add a comment about the mathematical
> reasoning shown below, which comes from a previous post:
>
> > I was hoping to find something interesting.  Such as
> >
> > let  a = b
> > then
> >      a^2             =  ab
> >      a^2 + a^2   =  a^2 + ab
> >      2a^2           =  a^2 + ab
> >      2a^2 - 2ab  =  a^2 + ab - 2ab
> >      2a^2 - 2ab  =  a^2 - ab
> >      2(a^2 - ab)  =  (a^2 - ab)
> >
> > Now divide each side by (a^2 - ab) and you have:
> >
> >      2  =  1
>
> This astonishing result is actually possible only due to a
fallacy
> of reasoning in the equations established above.  Since a has
> been defined as equal to b, the a^2-ab is equal to zero.
> Because division by zero is not permitted, the last equation
> listed above cannot be transformed into the equation "2 = 1"
> because the division cannot be performed.
>
> However, an illustration like the one above is a very good
> sample case that demonstrates WHY division by zero is not
> allowed.  The prohibition against division by zero is not
> intuitive to most people, but the necessity of this principle
> cannot be ignored.
>
> The previous post introduced another mathematical oddity that
> is a more subtle puzzle.  However this is another case where
> careful application of the rules of mathematics can save the
> day.
>
> > I tried to present a proof involving square roots but it
> > became too difficult.  Let's try it in words:
> >
> > -1 / 1  =  1 / -1
> >
> > now take the square root of both sides.
> >
> > The square root of -1 is i.  The square root of 1 is 1.
>
> I will interject at this point that if we maintain rigor, we
must
> consider that all square roots have a positive root and a
negative
> root.  This is a pestering perplexity in most problems, because
> it requires that we consider two different answers that may
> influence all of our future equations in the problem at hand.
> We should make a note that the square root we are dealing
> with in an equation has two possible answers, and then
> follow the chains of inference and the rules of algebra on
> both of those possible answers in our search for a solution.
> However, this step is really a small one in the problem as
> it was originally presented.  In the purest terms of strictly
> mathematical reality, we can say that a set of equations have
> formed after the taking of the square root of both sides. The
> equation above should read: "(-1/1)^(1/2)=(1/-1)^(1/2)"
> The next step is: "+i = +i, -i = -i".  The two possible
equations
> resulting from our use of the square root which can be
> discarded are: "-i = +i, +i = -i".  For you see, when a square
> root can result in two answers, one or both of those answers
> may be false ones that do not apply as true statements.
> This is like the division by zero problem.  If an equation
> resulting from a square root turns false, we can base no
> more reasoning upon it or upon any of the equations it may
> logically entail.
>
> > Therefore we have:
> >
> > i / 1  =  1 / i
>
> This statement above is false.  The reasons are partially
> explained by the paragraph I wrote above.  By taking the
> square root of both sides, negative signs can be lost or
> introduced.  The above statement that "i/1 = 1/i" is clearly
> problematical, as seen by the inferences directly below.
>
> > multiply both sides by i, and you have:
> >
> > i^2 / 1  = i / i
> >
> > i^2  is -1, and i / i is 1:
> >
> > -1 / 1  =  1.
> >
> > But  that is the same as:
> >
> >  -1  =  1
>
> This problem was introduced by taking the suare root of both
> sides of the equation.  If I had "a^(1/2)=b^(1/2)", then any of
> the following may be true: a=b, a=-b, -a=b, or -a=-b.  This
> problem illustrates the fallacy in previous statements by the
> time it gets '-1 = 1'.  When you see something like that, it is
> not a new discovery of some rare, arcane truth; western
> civilization has a pretty good hold on mathematical reasoning,
> and the real paradoxes in math are fairly more advanced than
> the ones we see here.

It is nice to see that there are several participants on these
newsgroups who are knowledgeable in mathematics.  Your analysis
of my "proofs" are correct.

There are three kinds of mathematicians:
those who can count and those who can't.

There are two groups of people in the world:
Those who can be categorized into one of two
groups of people, and those who can't.

---
Wax