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2007-04-23, 08:46   #12
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by mfgoode It is not a riddle! Its just a maths curiosity which looks odd and that it is wrong but it is right. Mally
?????????

(ab)2=a2b2

Last fiddled with by davieddy on 2007-04-23 at 08:52

2007-04-23, 08:47   #13
drew

Jun 2005

2·191 Posts

Quote:
 Originally Posted by mfgoode The version you have given is totally wrong
I'm continually amazed at how someone can be so consistently wrong about so many things.

You've outdone yourself, Mally.

 2007-04-23, 08:53 #14 michaf     Jan 2005 479 Posts More generally: $a = \sqrt{a^2}$ and $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ thus $2 \cdot\sqrt{\frac{2}{3}} = \sqrt {2^2 \cdot \frac{2}{3})} = \sqrt{ 4 \cdot\frac{2}{3}}$ (This is merely an excercise in TeX for me, been quite a while since last usage...and I sure hope nobody saw me struggling :>) Last fiddled with by michaf on 2007-04-23 at 08:59
 2007-04-23, 10:37 #15 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Mally: is this something you tried on a calculator?
 2007-04-23, 12:11 #16 fetofs     Aug 2005 Brazil 36210 Posts Of course, there is the calculator approach to it: 2*sqrt(2./3) = 1.6329931618554521 sqrt(4/3) = 1.1547005383792515
2007-04-23, 12:58   #17
R.D. Silverman

Nov 2003

164448 Posts

Quote:
 Originally Posted by michaf More generally: $a = \sqrt{a^2}$ False. and $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ False.
Your results (often stated) are false without carefully stating the
domain in which they apply. They are certainly NOT true over C
without stating which branch cut you are taking.

2007-04-23, 13:00   #18
davieddy

"Lucan"
Dec 2006
England

11001010010102 Posts

Quote:
 Originally Posted by fetofs Of course, there is the calculator approach to it: 2*sqrt(2./3) = 1.6329931618554521 sqrt(4/3) = 1.1547005383792515
But can Mally mangle the syntax sufficiently to
arrive at equality?

His resourcefulness in this respect is boundless:)

Last fiddled with by davieddy on 2007-04-23 at 13:05

2007-04-23, 13:30   #19
m_f_h

Feb 2007

24·33 Posts

Quote:
 Originally Posted by drew I'm continually amazed at how someone can be so consistently wrong about so many things. You've outdone yourself, Mally.
!

I think the first part (maybe not the second part) of R.D.Silverman's post gives an indication to the intended solution.

PS: well, there are several solutions... at least (a trivial) one is easy to see.

Last fiddled with by m_f_h on 2007-04-23 at 14:03

2007-04-23, 13:42   #20
davieddy

"Lucan"
Dec 2006
England

647410 Posts

Quote:
 Originally Posted by m_f_h I think the first part (but not the second part) of R.D.Silverman's post gives an indication to the inteded solution.
Could you enlighten us as to this "solution" please.

David

 2007-04-23, 13:50 #21 akruppa     "Nancy" Aug 2002 Alexandria 2,467 Posts Fields of finite characteristic? In characteristic 2, the two are indeed identical, as both are equal to 0. But if someone is asking such a question and has a domain other than the run-of-the-mill R or maybe C in mind, he must say so, or the question is ill-stated and quite meaningless. Alex
2007-04-23, 14:04   #22
m_f_h

Feb 2007

1B016 Posts

Quote:
 Originally Posted by akruppa Fields of finite characteristic? In characteristic 2, the two are indeed identical, as both are equal to 0.
This was the trivial one...
PS: and in an extension of R which has an absolute value, the equation is certainly wong, that's why it seems improbable that it's a question of branch cut.

Last fiddled with by m_f_h on 2007-04-23 at 14:09

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