[FFreak3]

Let us start with

1=1 (I)

Then divide each side by 1

1/1 = 1/1 (II)

Then multiply each side by negative 1

-1/1 = -1/1 (III)

Then switch the negative sign on one side to the bottom

-1/1 = 1/-1 (IV)

Now we find the square root of both sides

sqrt(-1) / sqrt(1) = sqrt(1) / sqrt(-1) (V)

Now we resolve the square roots

i/1 = 1/i (VI)

Now we cross multiply:

i^2 = 1^2 (VII)

Now we resolve the squares:

-1 = 1 (VIII)


And there you have it. -1 is the same thing as +1, and therefore negative numbers do not really exist. If you wish to prove me wrong, then show me which step was wrong.

This post has been edited by FFreak3: 18 September 2008 - 09:59 PM

[joshofalltrades]

Pah, hold on.

This post has been edited by joshofalltrades: 18 September 2008 - 11:11 PM

[joshofalltrades]

sqrt(-1) / sqrt(1) = sqrt(1) / sqrt(-1) (V)

Now we resolve the square roots

i/1 = 1/i (VI)

This is interesting to me. i = sqrt (-1), so what you did here was change your sqrt(-1) to i and those square roots didn't go away, or resolve as you put it. So where the hell did your sqrt (1) go?

This post has been edited by joshofalltrades: 18 September 2008 - 11:17 PM

[Dr Lecter]

This is basically... bullocks, using improper maths I've proved that infinite = anything, 0 = 1, and a bunch of other nonsense. That's all this is.

[Patch]

Like the one that proves 3=9 by dividing by 0.

[FFreak3]

sqrt (1) = 1, david.

1*1 =1, so the
sqrt(1) = 1

And if you believe it to be nonsense, then prove it.

[reiner]

Then switch the negative sign on one side to the bottom

-1/1 = 1/-1 (IV)

Now we find the square root of both sides

sqrt(-1) / sqrt(1) = sqrt(1) / sqrt(-1) (V)

This doesn't work. When you say the square root of both sides, it would be this:
sqrt(-1/1) = sqrt(1/-1)

Which isn't necessarily what you had iun your proof.

At that point, if you simplified it would be
sqrt(-1) = sqrt(-1)
i = i

Which is still correct.

However, you instead went:
-1/1 = 1/-1
sqrt(-1/1) = sqrt(1/-1)
sqrt(-1) / sqrt(1) = sqrt(1) / sqrt(-1)

This is incorrect at this point. You can only apply this breakdown if the numerator and denominator values are positive real numbers.

[Deucaon]

I could change the title and proudly proclaim "existence has no negative numbers" but it would mean the exact same thing.

[FFreak3]

This is incorrect at this point. You can only apply this breakdown if the numerator and denominator values are positive real numbers.

Very well, care to prove this?


edit: hint: you're not quite right

This post has been edited by FFreak3: 19 September 2008 - 11:51 AM

[Ninja Duck]

Reiner's right. It says so in Wikipedia.

"For all non-negative real numbers x and y,

sqrt(xy) = sqrt(x)sqrt(y)."

The proof of reiner's statement "You can only apply this breakdown if the numerator and denominator values are positive real numbers" comes from the contradiction that you showed us when you assume it's true for negative numbers. If you want to prove that this equation holds for non-negative real numbers, well, that's left as an exercise to the reader.

Edit: If reiner's not quite right, it's because the numerator doesn't have to be positive. It can be zero.

This post has been edited by Ninja Duck: 19 September 2008 - 03:01 PM

[FFreak3]

ergh, you guys aren't good for math trolling so I'll just tell you.

It's a property of i, that when it is created as a divisor, it becomes -i.

So really, we would have i/1 = 1/-i, and thus -i^2 = 1, which is 1=1.

The proof is obvious, but difficult to explain, so I will try to express it inexactly.

If i = sqrt(-1), then i^2 = sqrt (-1*-1), which resolves to sqrt(1), which resolves to +1, which is explicitly against the definition of i. This extends to a number of other operations with i that have to be handled specially to maintain the definition of i.

[reiner]

Ninja duck was right. If you apply values in an equation like that you still get 1 = -1 which disproves the theory, where if you correctly apply the root, you get 1 = 1. Thanks for clarifying that and pointing out my mistake.

Also did not remember that property FFreak3. How eclectic!

[FFreak3]

i has a lot of weird and useless properties, as do negative numbers in general. The problem is that the properties aren't axiomatic, and spring from the definition itself, while conflicting with existing algebra.
We had the same problem with infinity, but now we have special rules for it, so it's all good.

As far as I'm concerned, there is no true math that is right in all circumstances to all people, there is only math that is
a)useful
B)general
c)easy
and in that order of priorities.

[Ninja Duck]

Oh snap! I can't believe I didn't think of that. Thanks for reminding me. I actually feel pretty stupid now, since I'm taking the math GRE in a couple of months. Here's an alternate proof that 1/i=-i:

i^4 = 1, so i^{-1} = i^3. You can pull i^2 out to get i^3 = i * i^2 = -i. Woot.

FFreak3, I would contest your point that there is no "true math". There's only clumsy notation that makes math hard to work with. The natural law is still true, even if humans can't manipulate it like they want to.

[Patch]

I believe that there is no solid foundation in this world but science.
Go ahead and prove me wrong!