You're on the right track

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.

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 (2006-03-05 10:15:06)

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.

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 (2006-04-06 09:07:54)

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

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

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