Complex numbers trick (3 Viewers)

[Riviet]

Here is something interesting that my friend showed me the other day:
If 1 = sqrt(-1x-1)
then 1 = sqrt(-1) x sqrt(-1)
1 = i x i
1 = -1
2 = 0
Haha, can you explain it?

 

[Wackedupwacko]

simple when dealing with complex numbers the normal rules regarding the square root are ignored... thus u cant expand to root(-1) x root(-1) rather u would hafta multiply the terms insides first...

 

[webby234]

sqrt(ab) does not equal sqrt(a) x sqrt (b) when dealing with complex numbers (when a and b are negative).

 

[insert-username]

x squared = x + x + x.... x times
deriving both sides
2x = 1 + 1 + 1... x times
2x = x
2 = 1

You can't divide through by a variable because the variable may be 0. In this case, 2 x 0 = 1 x 0. Taking the derivative f both sides is also inaccurate, since (x² - x - 6) = (x - 6) when x = 0, but when x is 0 the derivative of (x² - x - 6) = -1, and the derivative of (x - 6) = 1.


I_F

 

[gman03]

Here is something interesting that my friend showed me the other day:
If 1 = sqrt(-1x-1)
then 1 = sqrt(-1) x sqrt(-1)
1 = i x i
1 = -1
2 = 0
Haha, can you explain it?

How is this odd? composition of functions dude

 

[Riviet]

I never said it was odd. I said it was something interesting.
Anyway, i showed it to my maths teacher and he said it also had something to do with the arguments of i.

 

[Trebla]

Remember that:

√(x²) = |x|

Since x can either be negative or positive

 

[Wackedupwacko]

i still think the best one is this

(x² - x²) = (x² - x²) now lets factorize
x(x-x) = (x+x)(x-x)
x = 2x
1 =2

(see if you can spot where this is wrong

 

[SeDaTeD]

Are we working in an integral domain?

 

[Templar]

Reminds me of a proof -

x squared = x + x + x.... x times
deriving both sides
2x = 1 + 1 + 1... x times
2x = x
2 = 1

RHS only work for integral x, you can't differentiate a function with only point values.

 

[PiGMAN]

i still think the best one is this

(x² - x²) = (x² - x²) now lets factorize
x(x-x) = (x+x)(x-x)
x = 2x
1 =2

(see if you can spot where this is wrong

dividing by 0....

 

[Riviet]

dividing by 0....

Precisely!

 

[airie]

haha seen it somewhere else that someone mentioned that '1 can equal to 2 and it's proven mathematically...' except they never said how. reading this thread, it looks like their conclusion is baseless! ... or does anyone know how it can be proven otherwise? ... don't really think so, 1 doesn't equal to 2 after all...or does it?

 

[bananasmoothy]

Drat, 1 doesn't equal 2???

There goes all my hard work...

 

[Riviet]

This isn't related to complex numbers, but still another number trick:
1 = 9/9
= 9 x 1/9
= 9 x 0.111111111111111111111111111111111111......
= 0.999999999999999999999999999999999999999......
=/= 1

Soo infinitly close to 1 but not exactly equal to it.

 

[Trebla]

I've heard that one before. There is no number between 0.99999.... and 1.0, so they can be considered practically equal.

 

[SeDaTeD]

How about a surreal number haha?

The decimal representation 0.99999999.... is just an infinite series , whose limit is 1.
0.9 + 0.09 + 0.009 + ....

 

[Riviet]

Yeah it's basically just a limiting sum.

 

[Sober]

Every rational number can be represented as a repetition, for example:
0.5 = 0.49999...

Some mathematical fallacies: http://www.math.toronto.edu/mathnet/falseProofs/fallacies.html

 

[KeypadSDM]

I've heard that one before. There is no number between 0.99999.... and 1.0, so they can be considered practically equal.

Under the basic defenitions of set theory & analysis you can prove that 0.99999... = 1

They're not practically equal, they ARE equal.

If they weren't, then limiting sums don't apply, and henceforth all of calculus is a fallacy.

Just use the cauchy convergence criteria on the summation of 9/(10^n) from n = 1 to infinity, and you'll see it converges to 1.