Where is the mistake?

[gan_ws]

1+1 = 1+ √1
= 1+ √-1*1
= 1+ √-1 * √-1
= 1+(-1)
= 0????

I found this funy.. hahaha

 

[D a w n]

the √ of a negative number is not possible ( in this case -1)

 

[PurplePoot]

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).

 

[Technomancer]

The square route of a negative number is possible... you just have to break out the good ol' imaginary units.

However, that still wouldn't work in gan's equations.

EDIT: Dang it, Poot. You and your sneaky posting.

 

[alfredx_sotn]

1 * 0 = 0
2 * 0 = 0

From here we have:

1 = 0/0
2 = 0/0

We sub 0/0 = 1 into 2 = 0/0 and we get:

1 = 2

PURE GENIUS

 

[HINDYhat]

More like pure lameness.

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

e^πi = -1
(e^πi)² = -1*-1 = 1
e^2πi = 1
ln e^2πi = ln 1 = 0

But logarithmic formulae state that:
log_a a^b = b

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

Turns out though, log_a a^b = b only if b is a real, not a complex -.-'

 

[Herman]

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

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

 

[PurplePoot]

i = The imaginary number (same as before, Herman)

The other thing is pi (π). VBulletin apparently only supports a capital version.

 

[alfredx_sotn]

More like pure lameness.

Lack of humor I see.

 

[Herman]

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