Where is the mistake?

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Level 40
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Yes it is, DonDustin, the square root of -1 is termed i, the imaginary number.

gan_ws, that formula, which derives from

1 = √1 = √(-1)(-1) = √-1√-1 = i2 = -1

Is generally considered a fallacy in itself due to the fact that expanding a square root as two negatives is not equal to the solution of the square root. (Though a negative and a positive or two positives are allowed).
 
Level 20
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More like pure lameness.

I had a fun one that turned out to be simply a restriction error...

eπi = -1
(eπi)2 = -1*-1 = 1
e2πi = 1
ln e2πi = ln 1 = 0

But logarithmic formulae state that:
loga ab = b

And therefore, 2πi = 0, which doesn't make sense at all.

Turns out though, loga ab = b only if b is a real, not a complex -.-'
 
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I once upon a time came across an equation that was algebraically impossible to solve :)O!!!)

It had something to do with logarithms, I don't remember it :cry:

EDIT

Wait, Hindy, I'm probably a few years behind you with the algorithms, but what does the "ni" stand for? Either one of those could always be zero (I am taking a shot in the dark here)

EDIT 2

I don't remember complex #'s ><, I can't ever remember my chart, I think its like all reals > 0 or something like that
 
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Aye, yeah there are alot of fallacies of this type that stem from dividing by zero

x = 2x
-x | -x

x = 0

x = 2x
x/x = 2x/x
1 = 2 :)O)

Another...

a=b
a-b=b-a

a-b / a-b = b-a / a - b
1 = -1

Why? a-b = 0
 
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