Simple. Problem. I think.

Paradoxical Numbers

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Re: Simple. Problem. I think.

#21  Postby Gareth » Feb 11, 2016 8:17 pm

Still waiting for DMcC's answer to post #14
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Re: Simple. Problem. I think.

#22  Postby campermon » Feb 11, 2016 8:19 pm

scott1328 wrote:
Gareth wrote:That's okay. Look, you guys don't have to keep pointing out that I'm not as smart as I think I am... I have a wife to do that... but thanks anyway.

Listen to your wife. Presumably she loves you.


:lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:
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Re: Simple. Problem. I think.

#23  Postby campermon » Feb 11, 2016 8:21 pm

Gareth wrote:Still waiting for DMcC's answer to post #14


have a beer while you're waiting.

:thumbup:
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Re: Simple. Problem. I think.

#24  Postby DavidMcC » Feb 11, 2016 8:22 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?
...

Easy, the 4 4th roots of 1 are: +1, -1, +i -i, where i = square root of -1.
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Re: Simple. Problem. I think.

#25  Postby logical bob » Feb 11, 2016 8:26 pm

DavidMcC wrote:
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?
...

Easy, the 4 4th roots of 1 are: +1, -1, +i -i, where i = square root of -1.

And since you said at (a) that x^2 +1 = 0 it followed straight away that x = i or -i.
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Re: Simple. Problem. I think.

#26  Postby campermon » Feb 11, 2016 8:28 pm

Gareth wrote:...what else could x be but 1?


One of these?

Image

:ask:
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Re: Simple. Problem. I think.

#27  Postby laklak » Feb 11, 2016 8:56 pm

I don't always drink beer, but when I do I drink Dos Equis.

Image

Actually, the first bit is a bold faced lie, I do always drink beer. I think I'll have one now, because it's 5 o'clock somewhere.
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Re: Simple. Problem. I think.

#28  Postby campermon » Feb 11, 2016 9:01 pm

laklak wrote: I think I'll have one now, because it's 5 o'clock somewhere.


A simple problem solved.

:beer:

[/thread]
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Re: Simple. Problem. I think.

#29  Postby laklak » Feb 11, 2016 9:03 pm

I live to serve, m'lord. What will you be having?
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Re: Simple. Problem. I think.

#30  Postby Gareth » Feb 11, 2016 9:05 pm

DMc C and logical bob. (I like logical bob) :
Thank you. That's clear and true. I am enriched. But it does mean that any calculation using paradoxical numbers will have even more "right"answers than I thought.

I came over here after reading the metatron's question at [ http://www.rationalskepticism.org/mathe ... 51418.html ], thinking I might find a bit more Rat and a little less Skep than over in astronomy.

He was struggling with the problem: 8 - 6 + 4 = ?
He thought the answer was minus 2 and couldn't see why the book insisted it was 6.

Members patiently explained in a number of ways... with a certain amount of confusion... why the book was right, and his question was answered with civility. I confess I would have said "Look, if I gave you eight apples, took six of them away and then gave you another four, how many apples........?", but nobody insulted his intelligence.

As a result, the metatron has risen considerably in my estimation.
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.

To celebrate... Drinks all round!

Six stoups of your finest ale, there, barman, if you please.
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Re: Simple. Problem. I think.

#31  Postby campermon » Feb 11, 2016 9:06 pm

laklak wrote:I live to serve, m'lord. What will you be having?


Ah! Cheers!

I'm on the Fosters here. With paracetamol and ibuprofen for the man flu!

:drunk:
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Re: Simple. Problem. I think.

#32  Postby laklak » Feb 11, 2016 9:29 pm

Dear God, not the man flu? Campy, we hardly knew ye!
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Re: Simple. Problem. I think.

#33  Postby campermon » Feb 11, 2016 9:54 pm

:lol:
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Re: Simple. Problem. I think.

#34  Postby Thommo » Feb 12, 2016 5:31 am

Better man flu than bird flu! :wink:
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Re: Simple. Problem. I think.

#35  Postby ughaibu » Feb 12, 2016 5:47 am

DavidMcC wrote:Easy, the 4 4th roots of 1 are: +1, -1, +i -i, where i = square root of -1.
What are the eight 8th roots of 1?
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Re: Simple. Problem. I think.

#36  Postby logical bob » Feb 12, 2016 9:55 am

ughaibu wrote:
DavidMcC wrote:Easy, the 4 4th roots of 1 are: +1, -1, +i -i, where i = square root of -1.
What are the eight 8th roots of 1?

1, -1, i, -i, √ 2 /2 + √ 2 /2 i, √ 2 /2 - √ 2 /2 i, - √ 2 /2 + √ 2 /2 i, - √ 2 /2 - √ 2 /2 i.

In general nth roots of 1 are n evenly spaced points on the unit circle in the complex plane

i.e. cos ( 2mπ/n) + i sin (2mπ/n) for m = 0, 1, 2, ... , n-1
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Re: Simple. Problem. I think.

#37  Postby ughaibu » Feb 12, 2016 10:07 am

I see. So the eleven 11th roots are non-arbitrary.
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Re: Simple. Problem. I think.

#38  Postby logical bob » Feb 12, 2016 10:14 am

ughaibu wrote:I see. So the eleven 11th roots are non-arbitrary.

What do you mean? How could any the roots of any equation be "arbitrary"?
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Re: Simple. Problem. I think.

#39  Postby ughaibu » Feb 12, 2016 10:44 am

logical bob wrote:
ughaibu wrote:I see. So the eleven 11th roots are non-arbitrary.

What do you mean? How could any the roots of any equation be "arbitrary"?
We're talking about definitions, not equations.

ETA: Okay, I see what you mean, and retract my question.
Last edited by ughaibu on Feb 12, 2016 10:49 am, edited 1 time in total.
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Re: Simple. Problem. I think.

#40  Postby Thommo » Feb 12, 2016 10:49 am

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