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cheesy
Supreme tranceaddict
Registered: Feb 2004
Location: Los Angeles, CA / Seattle, WA / S.F. Bay Area, CA
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| quote: | Originally posted by Floorfiller
the square of a number cannot be negative so...
i ^2 =/= -1 |
i is an imaginary number, it's definition is i^2 = -1 (thus sqrt(-1) = i)
___________________
-tom
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Dec-06-2004 02:42
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mezzir
BEES?

Registered: Nov 2002
Location: assachusetts
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Dec-06-2004 03:50
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enferno
Penus Maximus

Registered: Jan 2004
Location: jesus land
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Dec-06-2004 03:51
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eulerfx
BELIEVE IN ME

Registered: Apr 2004
Location: Boulder, USA
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Dec-06-2004 04:17
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Flyboy217
Senior tranceaddict

Registered: Aug 2003
Location: In DEEP SPACE... Space... space... sp...
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Re: Amazing Proof :)
| quote: | Originally posted by kypez
1 = sqrt(1) = sqrt( -1 * -1 ) = sqrt(-1) * sqrt(-1) = i * i = i ^2 = -1
therefore 1 = -1
anyone know whats wrong |
The Fundamental Theorem of Algebra states that an nth degree polynomial has precisely n complex roots. In particular, this means that there are n nth roots of unity. For n = 2, the roots are +/-1. That is to say, both 1 and -1 are square roots of 1.
On the other hand, the function f(x) = √2 is known as the "principal square root," and is defined as a mapping from the set of nonnegative real numbers to itself--in other words, you can only feed it nonnegative reals, and only get back nonnegative reals.
In the middle of your series of equations, you abandon this convention by taking the principal square root of -1, which is not defined. To give a shorter example, we know that
1² = -1² = 1
However, the following series of equations is false, like yours:
√1 = 1 = -1
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Dec-06-2004 04:58
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fitom tiel
spectacular

Registered: Apr 2003
Location: down in it
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Dec-06-2004 05:40
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