The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: The New Calculus: Never division by ZERO. No bullshit limit
theory. Sound analytic geometry.

Replies: 1   Last Post: Nov 11, 2017 7:45 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View

Posts: 1,176
Registered: 9/18/17
Re: The New Calculus: Never division by ZERO. No bullshit limit
theory. Sound analytic geometry.

Posted: Nov 11, 2017 7:45 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

>It is proved you idiot! Just because you do not understand proofs does not mean they are assertions.

I understand it, it is, like all you do, however fundamentally flawed.

What you do is
g(x) = (f(x+m)-f(x-n))/(m+n)=k

show then that f(x+m)-f(x-n)=k(m+n) and then proclaim that f(x+m)-f(x-n) has (m+n) as a factor which doesn't follow. Sure g(x)(m+n)=f(x+m)-f(x-n) but that doesn't mean that f(x+m)-f(x-n) has (m+n) as a factor in it. Prime example is f(x)=sin x, no matter how you deal with that, sin (x+m)-sin (x-n) does not have any factor in the form of (m+n). Ln x, a^x, tan x etc and countless others do not.

>Assertions are what YOU do. Proofs are what I do.

If you read basic mathematics books they provide countless proofs for anything.

>Dedekind Cuts and Cauchy sequences are NOT valid constructions as I proved but YOU cannot understand or don't want to.

I understood your proof and if it wasn't fundamentally flawed I would have listened.

Do this, prove that dedekinds cuts doesn't work using the EXACT definition, no modification of yours, the EXACT definition of it. Or better yet, use the definition of Dedekind?MacNeille completion on an ordered field. If you can do that, you will have taken the leg out from even more and I would be surprised and amazed.

But I recon you cannot even understand that definition.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.