Contained are three equations with rather interesting results… If you look at the first one they begin with a = b and end with 2 = 1 through some clever algebraic manipulations.
Now, while I have some math background, I’m no expert. It seems to me that this is incorrect, what jumps out to me immediately is, for instance, on the first equation, the third line results in 0 = 0, before they perform the actions required to yield 2 = 1.
Are these valid results? Or just a clever attempt to boggle the minds of those without a strong math background?
Even more importantly than 0=0 appearing in the middle if it, is that the step between the fourth and fifth lines relies on dividing by 0. (Since a = b -> a – b = 0).
The fact that you can’t do that is what invalidates the “proof.”
The bad step in the third equation is in going from -1/1 = 1/-1 to sqrt(-1/1) = sqrt(1/-1). For a, b > 0, sqrt(ab) = sqrt(a)sqrt(b), but not if a or b is < 0.
The funky step in the second equation has to do with adding the x[sup]2[/sup] term to both sides. Strange things can happen when you increase the degree of an equation to find solutions.
The mistake in the first is division by zero (line 4 to line 5).
In the second, the mistake is between lines 7 and 8. When they take the square root of both sides, it should be |3 - x| = |Pi - x|, which then becomes x - 3 = Pi - x.
In the third, the mistake is between lines 4 and 5. There are different ways to explain the mistake. One is the explanation that what you really have is one square root should be i, while the other is -i (different “branches” of the square root).
An “alternate” explanation is that, in general, sqrt(a/b) is not sqrt(a)/sqrt(b). In particular, it fails when a or b is negative. The reason it fails is the different “branches” explanation given earlier.
Others have answered while I typed, but maybe the same explanations in different words will help illuminate things.
In the first sequence of equalities, the move from the 4th line to the 5th line is invalid. They divide both sides by a - b, but since a = b, we have a - b = 0, so they’re dividing by zero. That’s never allowed.
In the second sequence, the error is in passing from the 7th to the 8th line. From
(3-x)^2 = (Pi-x)^2
you can only conclude
Either 3-x = +(Pi-x) or 3-x = -(Pi-x).
That page erroneously assumes the first option, when in reality it happens to be the second option that’s true. (Solve for x in the second option and you’ll see that you get the correct value.)
The third sequence is a little more difficult to explain precisely if you lack experience with imaginary numbers. Basically, when working with the sqrt function, you either need to treat it like a multivalued function, so that sqrt(1) = {1, -1}, or you need to consistently choose a particular so-called “branch” of the function. When passing from the 4th to the 5th line, that page uses different branches on the left and right-hand side of the equation. In short, they are not evaluating the same function on both sides, so it’s not surprising that they get different answers.
The middle one is a little tricker. There is nothing in the original equation that requiers “Pi” to equal 3.14… Therefore, “Pi” is just another variable. Let’s call it “y”.
Then you have x = (y+3)/2
or
y = 2x - 3
“y” has no unique solution, but is a function of x.
That step is perfectly valid. There’s nothing funky about it. Adding x^2 to both sides will never change the truth-value. It’s also a completely reversible step, because you can subtract off the x^2 no matter what value x has. The problem arises in the next couple steps.
Possibly, but I can still see an error in the math when “Pi” = 3.14159…
In the step where they take the sqrt of both sides, remeber, there is always a + and a - choice. Since the original equation was a first order, only one of those is garunteed. As mentioned above, you need to take the negative root of the left hand side, and you will get:
-(3-x) = Pi-x
which, when rearranged, gives you the second line.
Which basically means (on rereading), “what Tyrrell McAllister said three posts up.”
Actually, everyone’s missed it. ultrafilter is right that there’s tweakiness when you increase the degree of an equation, but adding x[sup]2[/sup] isn’t where the degree went up. John Mace is right that in this example “Pi” is just a variable, so let’s rename it y.
x = (y+3)/2
All well and good.
2x = y+3
Multiply each side by 2. Perfectly valid.
2x(y-3) = (y+3)(y-3)
AHA! We’ve multiplied by (y-3) which adds solutions (namely y=3) to the equation. The rest is just cancelling off the part that depends on x to get only this solution at the end.
You didn’t catch anything that others missed. You are just giving a different interpretation to the (admittedly terse) “proofs”. The rest of us have been reading the lines as connected by “implies” (or inclusion of varieties) while you are reading them as connected by “if and only if” (or equality of varieties). Under our reading, there is nothing wrong with that step, though it is certainly the point at which carelessness can lead to problems.
I guess I see what you’re saying to a point, which I understand to be that since Pi is transcendental, it is “algebraically independent” of rationals/radicals, so in that sense it behaves just like the variable “x”.
On the other hand, if we replace “Pi” in the proof with “4”, this seems equivalent to saying, “‘4’ is just a variable…”
You might say that “4” has a well-defined value but, then again, so does Pi (it’s just not expressed decimally, of course).
I still say the real mistake is where the square root is taken.
Indeed, in the given proof, the 3 may as well be considered a variable. Read this way, the proof “shows” that x = (y+z)/2 implies that z = y (or “is equivalent to”, under Mathochist’s reading).
No, I mean there’s nowhere in the “proof” that any actual value of “Pi” is used.
The square root is problematic, but it’s just what separates the solutions of the original equation from the solutions introduced when you multiply by (Pi-3). If you made the other choice of signs you’d be back to the first equation again.