# BdS misunderstands Fermat's Last Theorem

Discussion in 'Pseudoscience' started by BdS, Mar 24, 2016.

Messages:
422
0n + 0n = 0n

3. ### BdSRegistered Senior Member

Messages:
422
0n+1n =1n
0n+2n=2n
0n+3n=3n
0n+4n=4n
...

If x > 0 then x_binary = off else x_binary = on
If y > 0 then y_binary = off else y_binary = on
If x or y = on then z_boolean = true else z_boolean = false

0n+0n=0n
on+on=true

0n+1n=1n
on+off=true

0n+2n=2n
on+off=true
...

If x and y > 0 then z always = false
xn+yn=zn
off+off=false

// Moderator Edit -- Your post is off topic. Do you want your additions moved to a new thread or deleted? What should the new topic title be?

Last edited by a moderator: Mar 26, 2016

5. ### BdSRegistered Senior Member

Messages:
422
Cute.
How is my post off topic when
0n+0n=0n
0n+1n =1n
0n+2n=2n
0n+3n=3n
0n+4n=4n
...
are whole number solutions to xn+yn=zn when n is greater than 2?

"The mod who moved this is an idiot"

7. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Fermat's Last Theorem is: There are no positive integer solutions to $a^n + b^n = c^n$ when n is an integer larger than 2.
The Guardian quote in the OP is ambiguous in that 0 sometimes is and sometimes isn't considered a "whole number" or "natural number". In the context of a 300-year-old math problem, 0 is not a "whole number".

See: https://en.wikipedia.org/wiki/Fermat's_Last_Theorem#Equivalent_statements_of_the_theorem

Since 0 is not a positive integer then your family of trivial equations: $0^n + m^n = m^n$ is not at issue. Nor did you write them in standard notation, for this forum does not support superscripts except by using unicode (0ⁿ + mⁿ = mⁿ) or tex markup. Nor did you connect them to the other content of your posts or to any discussion in the thread. You have engaged in trivialities and pseudo-mathematics not even challenging yourself with the question of how the world could have missed such a trivial answer for 300 years or praised Wiles for proving what you think you have disproven.

danshawen likes this.
8. ### BdSRegistered Senior Member

Messages:
422
Well I included 0, because I dont want to do a half job.
Numbers are suppose to speak for themselves. Not my fault you dont understand them, as is proven in the fact that you thought my posts were off topic.
He's got a special place in my heart, congratulations sir! Thats why I gave him a special place in my post...
If x and y > 0 then z always = false
xn+yn=zn
off+off=false
Then I upgraded it to include 0

Was I the only one to receive a anonymous love note from you on your first day on the job?
Im engaging with pseudo-moderation now.

9. ### DaeconKiwi fruitValued Senior Member

Messages:
3,133
How were your posts relevant to the topic of Fermat's last theorem, considering it deals with numbers larger than 2 while your posts involve 0?

10. ### BdSRegistered Senior Member

Messages:
422
"There are no whole number solutions to the equation xn+yn=zn when n is greater than 2," - http://www.sciforums.com/threads/ox...rize-for-solving-fermats-last-theorem.155684/
Is 0 a whole number?
https://en.wikipedia.org/wiki/List_of_types_of_numbers
http://www.dictionary.com/browse/whole-number

Regardless, it still needs to include 0. I know you dont know why, because you couldn't even see that
0n+0n=0n
0n+1n =1n
0n+2n=2n
0n+3n=3n
0n+4n=4n
...
are whole number solutions to xn+yn=zn when n is greater than 2.

it's quite disappointing that you couldn't even get past "0" is not a positive integer. You haven't even scratched the surface yet.

There are no whole number solutions to the equation xn+yn=zn when n is greater than 2,

Last edited: Mar 28, 2016

Messages:
3,133

12. ### BdSRegistered Senior Member

Messages:
422
when a = 0 the calculation = true, it computes and is correct.
0^n + 1^n = 1^n = true, the number computes and is correct. 0 and 1 are whole numbers.
1^n + 1^n = 2 = false, because there is no z^n solution to 2 in whole numbers when n >= 3

13. ### James RJust this guy, you know?Staff Member

Messages:
31,999
BdS:

Consider the integer solutions of the equation $x^n+y^n=z^n$ for $n\gt 2$.

If $x=0$ then trivially $y=z$ is a solution (provided $y,z$ are integers).
If $y=0$ then trivially $x=z$ is a solution (provided $x,z$ are integers).

These trivial solutions hold for any $n$ at all, including $n \gt 2$.

This seems to sum up the point you were trying to make. Do you agree?

14. ### BdSRegistered Senior Member

Messages:
422
Its not trivial at all...

Hasn't even started...

15. ### James RJust this guy, you know?Staff Member

Messages:
31,999
What hasn't been covered that you want to say, then, BdS?

16. ### BdSRegistered Senior Member

Messages:
422
1 of the numerous minor points, yes I agree.

0n+0n=0n
0n+1n =1n
0n+2n=2n
0n+3n=3n
0n+4n=4n
...

If x > 0 then x_binary = off else x_binary = on
If y > 0 then y_binary = off else y_binary = on
If x or y = on then z_boolean = true else z_boolean = false

0n+0n=0n
on+on=true

0n+1n=1n
on+off=true

0n+2n=2n
on+off=true
...

If x and y > 0 then z always = false
xn+yn=zn
off+off=false

17. ### James RJust this guy, you know?Staff Member

Messages:
31,999
You're repeating yourself. We already covered that.

What is the relevance of this?

18. ### YazataValued Senior Member

Messages:
5,021

BdS admitted in his subject line that he doesn't understand Fermat's Last Theorem.

I think that he was asking for somebody to explain it to him in language that a non-mathematician can understand.

19. ### James RJust this guy, you know?Staff Member

Messages:
31,999
Yazata:

This thread was split from another thread, I think. The title is therefore not BdS's chosen title.

It seems to me like BdS understands Fermat's theorem. If not, why would he be posting computer-program-like pseudocode apparently dealing with some (trivial) solutions to it?

Hasn't that been done?

Let's ask him directly, just in case.

BdS:

Are you struggling to understand what Fermat's theorem is about? Do you need some help?

BdS likes this.
20. ### rpennerFully WiredValued Senior Member

Messages:
4,833
I disagree as BdS was led astray by an OP with a Guardian quote that used the term of art "whole numbers" when positive integers was meant. BdS shows that if "non-negative integers" was meant there would be families of trivial solutions, which nonetheless are solutions and this would render the original Guardian article praising Wiles as unintelligible.

Unfortunately, both "whole numbers" and "natural numbers" sometimes include zero and sometimes do not, depending on the author.

Also unfortunately, the WYSIWYG editor supports HTML/RTF superscripts upon cut-and-paste from preformatted material, but such superscript disappear upon posting. Thus Unicode superscripts, TeX superscripts or alternate notation are required to show exponentiation.

Last edited: Mar 28, 2016
BdS likes this.
21. ### YazataValued Senior Member

Messages:
5,021
Ok. It looked to me like he was admitting that he didn't understand it, and was being slapped around without much being done to help him. If I misread it, sorry.

Apparently because he thinks that they are solutions to it. If solutions seem to him to exist to a major mathematical problem and they are that trivial, then presumably he isn't understanding the problem the way that mathematicians do. That might need some more explanation.

I'll admit that I know next to nothing about mathematics. I'm not even all that interested in it, except where it is relevant to formal logic, which I am interested in. What drew me into this was my sense that a board participant was being abused.

BdS likes this.
22. ### BdSRegistered Senior Member

Messages:
422
I'm in no way in opposition, disagreement or trying to disprove Mr Wiles work and this is very trivial to his proof.

It didn't make sense when I read "There are no whole number solutions to the equation xn+yn=zn when n is greater than 2", sorry for the confusion. It does spice it up a bit when you include o.

23. ### James RJust this guy, you know?Staff Member

Messages:
31,999
Ok. Well, I'm glad we're cleared that up.