Re: History of Existance in MathMatical Proof ~ 899

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.