Uncovering the Flaw: A Critique of the Proof of 1 > 0 Using Trichotomy Axiom

[sponsoredwalk]

There is a proof in this document:

http://docs.google.com/viewer?a=v&q...pPBSj7&sig=AHIEtbSAu9399TLl5Ysmu2o_LwCOymEFxA

trying to prove that 1 > 0 by just using the trichotomy axiom
and the basic other axioms.

The proof is to test whether 1 > 0, 1 < 0 or 1 = 0.
If we assume 1 = 0 then if we invoke the
additive identity axiom
a + 0 = a
a + 0 = a + 0
a + 0 ≠ a + 1

So 1 ≠ 0

By the document, if we assume 1 < 0
1 + (-1) < 0 + (-1)
0 < - 1
0 • (-1) < (-1)•(-1)
0 < 1
But 0 < 1 contradicts the assumption that 1 < 0.

The document then says it must be that 1 > 0 by
the trichotomy axiom.

BUT!

If we had assumed 1 > 0,
1 + (-1) > 0 + (-1)
0 > - 1
(0)•(-1) > (-1)•(-1)
0 > 1
BUT THIS CONTRADICTS OUR ASSUMPTION
THAT 1 > 0

Tell me this is a joke?

I know about the sign change is supposed to occur
when we multiply by (-1) but the PDF
doesn't seem to know this, in fact if we
do the sign change then by the method
I've used we can show:

By the document, if we assume 1 < 0
1 + (-1) < 0 + (-1)
0 < - 1
0 • (-1) < (-1)•(-1)
0 > 1

Which was our assumption to begin with, it satisfies everything
(which is totally bogus).

I must be missing something, I mean a random PDF off the net
couldn't have bad information in it, could it?

I guess I don't know how to "prove" this, and it's probably not something
you can prove but seeing as the PDF raised the question some pointers
would really help!


(A gift)

https://www.youtube.com/watch?v=FZgKheUt_SU

 

[statdad]

Things fall apart right from the git-go. If you assume that 0 = 1 [/tex]<br /> then ina + 0 = a <br /> a + 0 = a + 0<br /> a + 0 ≠ a + 1<br /> <br /> the final line doesn&#039;t make sense - you can&#039;t say the two sides are different after you assume 0 = 1.

 

[sponsoredwalk]

If zero equaled 1 we'd have:

2 + 0 = 2

But 2 + 1 = 3

How can 2 + 1 = 3 and 2 + 1 = 2?

0 is axiomatically defined as the additive inverse of every element in the set so I
was simply showing that we can't assume that 1 = 0, notice the trichotomy axiom
forces us to consider either 1 > 0, 1 < 0 or 1 = 0 & by simply showing that assuming
1 = 0 at all breaks the additive identity axiom. Therefore we can discount that aspect
of the proof and try 1 > 0 and 1 < 0.

 

[statdad]

You missed my point. At the start the assumption was made that 1 = 0 ("If we assume 1 = 0..." so that from that point on those two symbols represent the same object. But later
there is

a+0 = a+0
a+0 /= a+1

where, apparently, the claim of non-equality is made simply because 1 was substituted for 0 and 1 itself is not equal to 0. But, because of the original assumption, there is no valid reason to do that shown in the ``work'' .

I'm not sure what you mean that 0 is the additive inverse of the elements.

 

[sponsoredwalk]

Hmm, I know what you mean, I guess I should approach this differently then.

Well, first off what I mean by additive inverse was the field axiom

a + 0 = a, basically I'm working off the fundamental axioms:

Closure:
If a,b ∈ P, (a + b) ∈ P (for some set P)
If a,b ∈ P, a•b ∈ P

Commutativity:
a + b = b + a
a•b = b•a

Associativity:
a + (b + c) = (a + b) + c
a•(b•c) = (a•b)•c

Identity:
a + 0 = 0 + a = a
a•1 = 1•a = a

Invertability:
a + (-a) = (-a) + a = 0
a•a-¹ = a-¹•a = 1

Distributivity:
a•(b + c) = a•b + a•c

Trichotomy:
Only one of the following holds:
a = 0, a < 0, a > 0

Basically I was trying to show that if we assume 0 is equal to 1 we reach a contradiction
because if we invoke the additive identity axiom above, that a + 0 = a, we reach a
contradiction because a + 0 is not the same as a + 1, i.e. 2 + 0 is not the same as
2 + 1.

Honestly I initially thought it was just a consequence of the trichotomy axiom but
seeing as this proof was proving something so fundamental I thought maybe this
was a valid way to show why 0 couldn't be 1, obviously not

As for the rest of the proof, does what I've written above hold or have I missed something?