Simple. Problem. I think.

Paradoxical Numbers

Discuss the language of the universe.

Moderators: Calilasseia, ADParker

Re: Simple. Problem. I think.

#41  Postby logical bob » Feb 12, 2016 11:08 am

Thommo wrote:Is that right? Isn't the algebraic completion of the reals unique up to isomorphism?

Sorry Thommo, who was that at?

Edited to include the quote when I realised I was at the top of a new page.
Last edited by logical bob on Feb 12, 2016 11:33 am, edited 1 time in total.
User avatar
logical bob
 
Posts: 4482
Male

Scotland (ss)
Print view this post

Ads by Google


Re: Simple. Problem. I think.

#42  Postby Sendraks » Feb 12, 2016 11:25 am

Gareth wrote:
Firstly, he was prepared to admit that he was wrong. That's huge. Show me a man who never admits when he's wrong and I'll show you a fool.
Secondly, he had the humility to ask for help, even if others might think it was a stupid question. That is a fast track to wisdom. I recommend it.


You'll find most of the skeptics here are fully prepared to admit where they are wrong. Its usually the theist types who dig themselves into entrenched positions and refuse to admit any error.

We also have a couple of folk over in the politics forums who are unable to deal with the possibility of being wrong.

Accepting you're wrong is just BAU stuff.
"One of the great tragedies of mankind is that morality has been hijacked by religion." - Arthur C Clarke

"'Science doesn't know everything' - Well science knows it doesn't know everything, otherwise it'd stop" - Dara O'Brian
User avatar
Sendraks
 
Name: D-Money Jr
Posts: 15239
Age: 104
Male

Country: England
Print view this post

Re: Simple. Problem. I think.

#43  Postby Thommo » Feb 12, 2016 11:34 am

logical bob wrote:Sorry Thommo, who was that at?


Ughaibu. He said it was a matter of definition rather than algebra, but the algebra follows from the definition - the answer would be the same whatever we assigned to be the square root of -1, up to isomorphism.

He obviously realised at about the time I posted though, since he edited saying he got it just moments after I hit submit, so it became a bit redundant as no explanation was actually needed.
User avatar
Thommo
 
Posts: 27172

Print view this post

Re: Simple. Problem. I think.

#44  Postby VazScep » Feb 12, 2016 11:35 am

scott1328 wrote:The Pythagoreans killed the man who first demonstrated that √2 is irrational. It overthrew their mathematics until Eudoxus theory of proportion salvaged the mess.
I think there might have been some mix-ups in the story there. There's this story about Hippasus being killed for revealing the secret for constructing the Dodecahedron, though.
Here we go again. First, we discover recursion.
VazScep
 
Posts: 4590

United Kingdom (uk)
Print view this post

Re: Simple. Problem. I think.

#45  Postby VazScep » Feb 12, 2016 11:44 am

Thommo wrote:
logical bob wrote:Sorry Thommo, who was that at?


Ughaibu. He said it was a matter of definition rather than algebra, but the algebra follows from the definition - the answer would be the same whatever we assigned to be the square root of -1, up to isomorphism.

He obviously realised at about the time I posted though, since he edited saying he got it just moments after I hit submit, so it became a bit redundant as no explanation was actually needed.
Not sure what the point of confusion is (though it obviously isn't yours).

The roots of a polynomial function are those that are mapped to 0. The n-th roots of w are the roots of the function p(z) = z^n - w, which are just the solutions to the equation z^n = w.

So for any n, there's a definite answer to what the set of nth-roots are. In the reals, you can prove that there are only two fourth and eighth roots of 1. In the complex numbers, you can prove that there are 4 fourth roots and 8 eighth roots of 1, and, indeed, n nth-roots of 1.
Here we go again. First, we discover recursion.
VazScep
 
Posts: 4590

United Kingdom (uk)
Print view this post

Re: Simple. Problem. I think.

#46  Postby Thommo » Feb 12, 2016 11:52 am

Yes, it's just a special case of the number of solutions of a polynomial of degree n being n, if you count repeated roots separately.

I forget how the proof goes though. It's all algebraic either way.
User avatar
Thommo
 
Posts: 27172

Print view this post

Re: Simple. Problem. I think.

#47  Postby DavidMcC » Feb 12, 2016 12:45 pm

Gareth wrote:DavidMcC.

Thanks for that. I don't get it though. If x to the fourth is 1, what else could x be but 1?
I didn't crib it by the way. I thought of squaring both sides all be myself, so as not to get stuck with impossible square roots.

...

I have already said what the 4 4th roots of 1 are. Two of them are "imaginary", two real. Your answer (1) was limited to positive real solutions, and that is why you missed three of them out.
May The Voice be with you!
DavidMcC
 
Name: David McCulloch
Posts: 14913
Age: 67
Male

Country: United Kigdom
United Kingdom (uk)
Print view this post

Ads by Google


Re: Simple. Problem. I think.

#48  Postby scott1328 » Feb 12, 2016 2:37 pm

VazScep wrote:
scott1328 wrote:The Pythagoreans killed the man who first demonstrated that √2 is irrational. It overthrew their mathematics until Eudoxus theory of proportion salvaged the mess.
I think there might have been some mix-ups in the story there. There's this story about Hippasus being killed for revealing the secret for constructing the Dodecahedron, though.

Hippasus must have been a real dick. He is credited for the proof that √2 that irrational

Greek mathematicians termed this ratio of incommensurable magnitudes alogos, or inexpressible. Hippasus, however, was not lauded for his efforts: according to one legend, he made his discovery while out at sea, and was subsequently thrown overboard by his fellow Pythagoreans “…for having produced an element in the universe which denied the…doctrine that all phenomena in the universe can be reduced to whole numbers and their ratios.”[8] Another legend states that Hippasus was merely exiled for this revelation. Whatever the consequence to Hippasus himself, his discovery posed a very serious problem to Pythagorean mathematics, since it shattered the assumption that number and geometry were inseparable–a foundation of their theory.


Apparantly this is the real paradox that Gareth should explore: How did Hippasus die?

He should also take it as object lesson in not pissing off Mathematicians by discovering an actual flaw in their mathematics.
User avatar
scott1328
 
Name: Some call me... Tim
Posts: 8695
Male

United States (us)
Print view this post

Re: Simple. Problem. I think.

#49  Postby VazScep » Feb 12, 2016 3:51 pm

Thommo wrote:Yes, it's just a special case of the number of solutions of a polynomial of degree n being n, if you count repeated roots separately.

I forget how the proof goes though. It's all algebraic either way.
The Fundamental Theorem of Algebra? It's mostly analysis.
Here we go again. First, we discover recursion.
VazScep
 
Posts: 4590

United Kingdom (uk)
Print view this post

Re: Simple. Problem. I think.

#50  Postby LjSpike » Apr 21, 2016 3:32 pm

I haven't read all the way through, but I spot a problem:
b). then x squared is minus 1

You can't have a square being a negative number (or any even power for that fact) as an NEGATIVE x NEGATIVE = POSITIVE. You also can't fix this statement by saying it means to be the root of x as the root of x can be 2 values, negative or position

So from point B I've been put off, because secondary school education tells me its not possible.
Or at least, its not real.

So if anything your number is a dream.
LjSpike
 
Posts: 89
Age: 20
Male

United Kingdom (uk)
Print view this post

Re: Simple. Problem. I think.

#51  Postby VazScep » Apr 21, 2016 5:17 pm

LjSpike wrote:I haven't read all the way through, but I spot a problem:
b). then x squared is minus 1

You can't have a square being a negative number (or any even power for that fact) as an NEGATIVE x NEGATIVE = POSITIVE. You also can't fix this statement by saying it means to be the root of x as the root of x can be 2 values, negative or position

So from point B I've been put off, because secondary school education tells me its not possible.
Take A-level maths, and you should be introduced to complex numbers, a universe in which there are numbers that square to any negative number you like.
Here we go again. First, we discover recursion.
VazScep
 
Posts: 4590

United Kingdom (uk)
Print view this post

Re: Simple. Problem. I think.

#52  Postby campermon » Apr 21, 2016 6:29 pm

LjSpike wrote:I haven't read all the way through, but I spot a problem:
b). then x squared is minus 1

You can't have a square being a negative number (or any even power for that fact) as an NEGATIVE x NEGATIVE = POSITIVE. You also can't fix this statement by saying it means to be the root of x as the root of x can be 2 values, negative or position

So from point B I've been put off, because secondary school education tells me its not possible.
Or at least, its not real.

So if anything your number is a dream.


https://en.wikipedia.org/wiki/Imaginary_unit

:thumbup:
Scarlett and Ironclad wrote:Campermon,...a middle aged, middle class, Guardian reading, dad of four, knackered hippy, woolly jumper wearing wino and science teacher.
User avatar
campermon
RS Donator
 
Posts: 17434
Age: 50
Male

United Kingdom (uk)
Print view this post

Previous

Return to Mathematics

Who is online

Users viewing this topic: No registered users and 1 guest