Still waiting for DMcC's answer to post #14

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.

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

have a beer while you're waiting.

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.

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.

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

One of these?

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



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.

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

A simple problem solved.



[/thread]

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

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.

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!

Dear God, not the man flu? Campy, we hardly knew ye!

Better man flu than bird flu!

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?

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

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

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"?

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.

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