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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

Here's a simple "proof" of something impossible that one of my friends came up with in high school. Someone who really understands second year high school algebra ought to be able to see the mistake in it, but at least half the math teachers at school couldn't figure it out! Here goes.

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Sorry, I'll stop making equation images unless I need them for something complicated. This was just I test to see if I understood the system.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,568

**igloo** **myrtilles** **fourmis**

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

You're on the right track

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Not entirely.

What do we know about roots of a 2nd degree equation? i.e. how many are there?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

I assume you mean a second degree polynomial equation here? There are two (though once in a while they are the same as each other). I think I see where you're going with this, but it is also quite possible to find the problem with this proof using only what we know about the properties of complex numbers. These are properties that I believe are taught in the second year of Algebra at most high schools.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I agree with what Ricky is implying.

In that second step, where you square root the -1's, they need to have a ± in front of them.

And don't worry about making those equations as pictures. If I remember, they take up very little memory.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Sorry I was so cryptic. I just like to let others trying to get it without being told the answer.

mathsyperson, I believe that the pictures are generated on the fly. So they take up only as much memory as their text does.

I assume you mean a second degree polynomial equation here? There are two (though once in a while they are the same as each other).

Well, technically, there are always two. Just because they have the same value doesn't mean they aren't two roots.

I think I see where you're going with this, but it is also quite possible to find the problem with this proof using only what we know about the properties of complex numbers.

Can you be more specific? Besides the ± which mathsyperson pointed out, I see no other incorrect steps in the proof.

Edit:

Ah, on further review, yes, there is another mistake. Square root properties (specifically of a/b) don't apply to negatives.

*Last edited by Ricky (2006-03-04 11:15:06)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

Now I see what you were saying. When solving algebra equations, if you apply a square root to both sides it is then necessary to be careful to choose the correct + or - sign when you apply it, because the correct solution to your equation might only be valid in one of those cases.

For this problem, it really doesn't matter which square root you do in line 2 (+sqrt) or (-sqrt) as long as you do the same one to both sides because you aren't trying to solve for a variable from line 1. Line 2 is a fully correct statement by itself: the sqaure root of -1 does equal the square root of -1. It probably would have been clearer if I had just started the proof from line 2.

It is the second mistake that Ricky found that I claim is the real problem with this proof. You can't go from root(1/-1) to root (1)/root(-1). The first equals i, and the second equals i raised to the minus one power. The cyclical property of imaginary numbers was one thing I was hoping to get to here:

.

.

i^0 = 1

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

i^5 = i

.

.

.

And so on in both directions. So by splitting the square root on the right side in line 4, you are effectively saying i^1 = i^-1, which is not true, it's two off on the sequence.

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

When I first went through this exercise a long time ago, it also got me thinking about more general problem solving strategies. Let's say you have a proof like this that you know is wrong but you can't figure out where the problem was introduced. If the proof was really long then it could take forever to go through it line by line to find the mistake. A good trick is to evaluate it at key points throughout your work.

For example, if your proof is 100 lines long and you know you got the wrong answer at the end, then instead of scanning back line by line for your mistake, try going back to line 50 and evaluating both sides of your equation. If they come out to the same number, then you made a mistake after line 50, otherwise, you must have a mistake before line 50. You can narrow down where your mistake occured in this way.

In this proof, you could eventually evaluate line 3 and see that what is says is i = i. That's ok.

Line 4, however, says that i = i^-1 which is the same as -i. That's not correct.

So in this case the mistake must have been introduced between lines 3 and 4.

Just trying to get a little useful problem solving strategy out of this puzzle.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You have to be careful with that method though because you can be right for the wrong reason. But that is what I used as well.

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**kimrei****Member**- Registered: 2006-04-05
- Posts: 8

Hi,

Try this:

if x = 1

then

x = x²

therfore

x-1 = x²-1

factorise:

(x-1) = (x-1)(x+1)

remove LCD

1 = x+1

take accross the one

0 = x

And x = 1 in which case:

0 = 1

If you want multiply both sides by 2 and subtract 1 to get

0*2-1 = 1*2-1

which is the same as

-1 = 1

See if you can spot the problem, there is one

*Last edited by kimrei (2006-04-05 11:07:54)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

Ummm... it seems that whenever you dip your toe into "0" that kind of thing happens, and x-1=0.

Or, to put it another way, you shouldn't divide by (x-1) after the "(x-1) = (x-1)(x+1)" step.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**kimrei****Member**- Registered: 2006-04-05
- Posts: 8

There's also this one:

0=0

x*0=0

divide both sides by 0

x=1

and 0/0 is often accepted to be equal to 1 (as there's exactly one 0 in 0)

but likewise there are an infinite number of 0s in 0 so it's often not accepted too

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

You Cannot Divide By 0.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Or,when you want to solve this parametric euation, for example:

x/a=10, then here's two cases:

1 .a=0, no solution;(division by 0)

2. a!=0, x=10 a

IPBLE: Increasing Performance By Lowering Expectations.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

to the first post

i believe the falicy he was looking for was not the +/- one, but to do with the square root expansion

its often taught as a law, that sqrt(a/b) = sqrt(a)/sqrt(b)

but this is only true, when b is a positive real number

in the case of -1, this is non positive, the falicy he was looking for is that while sqrt(-1/1) = sqrt(-1)/sqrt(1), the sqrt(1/-1) =/= sqrt(1)/sqrt(-1)

that is because while i/1 = i, 1/i = -i

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You got it. In general, the fallacy that was committed was using a theorem while not meeting its preconditions first.

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**liuv****Member**- Registered: 2006-05-14
- Posts: 29

or

or

or

So we should NOT consider

*Last edited by liuv (2006-05-19 16:44:29)*

I'm from Beijing China.

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