Proof That 3 = 0?

Date: 10/28/2006 at 03:10:46
From: Alexander
Subject: 3=0? where is the mistake?

Let's say I have this: x^2 + x + 1 = 0
I can rewrite it like this: x + 1 = -x^2
and also like this: x(x + 1) + 1 = 0
So far so good, but when I use the second equation and the third I get
this: x(-x^2) + 1 = 0  --->  -x^3 = -1  --->  x = 1.
Now I use x = 1 in the first equation and I get 1 + 1 + 1 = 0 or 3 = 0

I must have done something incorrect but I just can't find the mistake.

Date: 10/28/2006 at 15:50:18
From: Doctor Rick
Subject: Re: 3=0? where is the mistake?

Hi, Alexander.

Let's see if I can express your reasoning more fully.  I think you're 
saying:

  (1) x^2 + x + 1 = 0
  (2) x + 1 = -x^2     by subtracting x^2 from each side of (1)
  (3) x(x+1) + 1 = 0   by factoring x from the first two terms of (1)
  (4) x(-x^2) + 1 = 0  by using (2) to replace (x+1) with -x^2 in (3)
  (5) -x^3 + 1 = 0

We see by inspection that one solution to (5) is x = 1; however, when 
we check this solution by substituting it in (1), we find that it is 
*not* a solution.  Why not?

That's a good question!  Perhaps we can learn something by finding 
the correct solutions to (1) and substituting them into each step of 
the work.  Using the quadratic formula, I find that the true 
solutions are -1/2 + i*sqrt(3)/2 and -1/2 - i*sqrt(3)/2.  You're 
familiar with complex numbers, right?

When I said above that "*one* solution to (5) is x = 1", I had in mind 
that the cubic has *three* solutions; the other two are complex.  In 
fact, those complex solutions are exactly the true solutions to (1)! 
So somehow, on the way to (5), we have introduced one extraneous 
solution, x = 1, in addition to the two valid solutions.

This leads me to try substituting not the valid solutions, but the 
extraneous solution, into each step of your work:

  (1) x^2 + x + 1 = 0  ==>  1^2 + 1 + 1 = 0 (3 = 0, not true)
  (2) x + 1 = -x^2     ==>  1 + 1 = -1^2    (2 = 1, not true)
  (3) x(x+1) + 1 = 0   ==>  1(1+1) + 1 = 0  (3 = 0, not true)
  (4) x(-x^2) + 1 = 0  ==>  1(-1^2) + 1 = 0 (0 = 0, TRUE!)

This tells us that the extraneous solution was added at step 4.  How? 
Well, we can see that we replaced x+1 = 2 with -x^2 = -1, which 
isn't the same value, so we don't have an equivalent equation.

Now let's look for a general principle we can take away from this.  I 
think the lesson is this.  In solving an equation, we transform it 
into EQUIVALENT equations, step by step until we have an equation 
that can be solved by inspection (or by known methods).  However, an 
EQUIVALENT equation is not merely an equation that is TRUE for 
values of the variable that make the original equation TRUE.  It also 
must be FALSE for values of the variable that make the original 
equation FALSE.

The reasoning in your solution meets only the first condition.  It 
assumes that the original equation is TRUE, and makes a substitution 
that is only valid under that condition.  In the process, it produces 
an equation that is true when the original equation is true, but is 
ALSO true for a value that makes the original equation false!  It 
isn't completely equivalent to the original equation; it introduces 
an extraneous solution.

Sometimes it is useful to transform equations in ways that introduce 
extraneous solutions.  For instance, we might square both sides of an 
equation involving square roots, in order to put it in a form we can 
deal with.  Extraneous solutions are not a disaster, as long as we 
are aware that what we have done may introduce one!  Then when we're 
done, we are careful to go back and check our solutions to see 
whether any of them is extraneous.

Looking at your problem in this light, after finding the solution 
x = 1, we go back (as you did) and see that it is not a solution to 
the original equation.  Our conclusion now, though, is simply that 
this solution was extraneous, not that 3 = 0, or that everything we 
did was wrong.  In fact, using techniques of complex-number algebra, 
we can find all three solutions to the cubic, check them, and find 
the two correct solutions by your method.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/