Slavakion

I saw this and knew that it had to be TFP'd. This can't be possible, right? But here it is. Please, discuss for eight+ pages. Oh, and things written off to the right clarify the next step I took.

raveneye

When you divide by (b-a) you're dividing by zero.

Since division by zero is undefined, your result afterwards can be just about anything.

a-j

Yep, when you divide by b-a, you must state for (b - a) not equal to zero (b != a), since as raveneye pointed out, it is undefined. However you now have two contradicting conditions a = b, and a != b. So your conclusion is invalid.

KnifeMissile

To try to make this, otherwise trivial, thread a little more exciting, let me mention that it is possible to divide by the additive identity (zero) and that's in the case of the trivial field (a field that consists only a single element). In that case, 1 = 2 = 1 + 1, and so there is no contradiction.

Incidentally, if I recall correctly, there are simpler versions of this fallacy. Where did you find this, Slavakion?

Redlemon

I'd just like to note...
...66 minutes pass...
I'm surprised it even lasted that long, but it was early in the morning.

I also wanted to complement Slavakon on the nice formatting job on those equations.

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a-j

Ok I've given this more though, and I think you haven't actually proved that 1 = 2, you have "proved" that a = 2a. You can't just substitute whatever you want into a. You need to solve for a, and you end up with a=0 and hence b=0. Since you are assuming b > 0, then your proof is contradictory so either your assumptions were incorrect or your logic is incorrect.

Slavakion

On another forum. GamingWorld, which might surprise you if you know the type of people that frequent it.
Well, if you remember my other thread about .999... equaling 1 based on a sketchy formula, then you'd know why I said that. It got a whopping 2 pages of replies, when I had expected something more like this thread. Meh, 15 minutes of fame, eh?

Oh, and thanks about the formatting.

vinaur

Here's an example of -1=1 (incorrect of course, but I won't tell where, yet):

1) 1=1
2) 1/1=1/1
3) -1/1=-1/1
4) -1/1=1/-1
5) sqrt(-1/1)=sqrt(1/-1)
6) sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)
7) i/1=1/i
8) i^2=1
9) -1=1

Slavakion

Ooh, I can feel it. I know the answer, but I don't. It... I... aw, I'm gonna hurt myself if I keep thinking about it.

a-j

This one is fairly easy, when you take square roots you need to assume both values that the input could have produced (i.e. sqrt(1) = -1,1). So your reasoning should have been something like:

sqrt(-1/1)=sqrt(1/-1)
sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)
(±i)/(±1)=(±1)/(±i) => i^2=±1 (if you work out all combinations)
-1=1, or -1=-1 (the latter is correct)

When people end up with something like 2=1, it doesn't mean they actually proved 2=1, it means there is a contradiction which means the they did something incorrect or their initial conditions were wrong. Proof by contradiction is very strong tool if used correctly. However I understand this is all in fun.

Slavakion

>_<

That's it. I almost had it.

vinaur

You have some good reasoning there, however consider the following (I start wit step 5 from my previous post):

5) sqrt(-1/1)=sqrt(1/-1)
6) sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)
7) sqrt(-1)*sqrt(-1)=sqrt(1)*sqrt(1)
8) sqrt(-1)^2=sqrt(1)^2
9) -1=1

In this example you cannot get two answers for sqrt(-1) because you are not taking the sqrt. The mistake is somewhere else....

a-j

Good point.
The problem is going from sqrt(1/-1) to sqrt(1)/sqrt(-1)
1/i = -i, so in fact it should start out

5) sqrt(-1/1)=sqrt(1/-1)
6) sqrt(-1)/sqrt(1)=-sqrt(1)/sqrt(-1)
7) sqrt(-1)*sqrt(-1)=-sqrt(1)*sqrt(1)
8) sqrt(-1)^2=-sqrt(1)^2
9) -1=-1

The square root properties (sqrt(a/b) = sqrt(a)/sqrt(b)) don't exactly work the same for complex numbers.

ajji

Yes these two samples were incorrect because to get the result wrong operations have to be done and the complex number properties were not respected sqrt(-1) = i ...

but still in mathematics we can by taking it in other way (from the result) proof that an incorrect statement is correct here is an exemple I remember from my teacher at school.

Some one asked to a mathematician : "Do you admit that if 2 + 2 = 5 , then we can deduct that you are the pope?." "For sure the mathematician replied, think about that. if we suppose that 2 + 2 = 5. By doing -2 on each side we got 2 = 3. By symmetry we know that 3 = 2 if we do -1 on each side we then got 2 = 1. The pope and me are two but as 2 = 1, then the pope and myself are only one. So we can conclude that I am the pope ! "
This property is called ex-falso quodlibet and says that if a statement is false and true then all other statement is true...

If one day you prove that 0 = 1, as you know that 0 <> 1 then you can deduct using a proof by contradiction that whatever mathematics theorem you can imagine are true. So this will be the end of all the nice theorem that many mathematician has found.

By the way I am not the pope, are you ?

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vinaur

Good job! The problem is in fact in step 6. Here's a different explanation for it though.

More classic falacies!!

KnifeMissile

Actually, square roots work exactly the same for complex numbers. The problem is that squaring is not an invertable function and, thus, square root functions (note the plural) cannot be used to preserve equality.

So, in my esteemed opinion, the fallacy is the use of sqare roots...


PS. I'm sorry to bait the debate but I'm short on time at the moment. I'll clear this up, later...

0energy0

if i can be squared, then -1 cannot be reversed (as in square rooting it?) what theorem explains this?

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