POTW Returns!

Square root and square are not always inverses. Was that cryptic enough?

I think that the flaw is the last step. If a^2 = b^2, that does not necessarily mean that a = b; a = -b still satisfies that equation. In this case, it should be trivial to find an s such that e - s is still positive while f - s is negative, as e is presumably considerably greater than f.

Mind you, I first had to convince myself that it was legal to multiply to equations together. I've never come across that before; of course, a little experimentation did indeed convince me that it was legal.

s is the mean of e and f. If you subtract s from e and f, you get a positive and negative number with the same absolute value. So taking the square root of both sides is indeed where it goes awry.

The best incorrect prove I've ever found results by using partial integration $int u'*v dx = u*v - int u*v' dx$ to solve $int tan(x) dx = int sin(x)/cos(x) dx$. Using $u'(x) = sin(x) => u(x) = -cos(x)$ and $v(x) = 1/cos(x) => v'(x) = sin(x)/cos^2(x)$, it follows that $int sin(x)/cos(x) dx = -cos(x)/cos(x) - int -cos(x)*sin(x)/cos^2(x) dx = -1 + int sin(x)/cos(x) dx$. Subtracting the integral on both sides yields $0 = -1$.

After multiplying the equations together, you get:

e^2 - 2es = f^2 - 2fs

But since 2s = e + f,

e^2 - e(e+f) = f^2 - f(e+f)

Or:

e^2 - e^2 - ef = f^2 - f^2 - ef

Or:

-ef = -ef

From there, one shouldn't get different values on either side of the equal sign if one adds s^2 to both and then takes the square root of both.

I could have used that first fallacy (cents and dollars) when I was teaching physics.

Others have already found the correct fallacy, namely that a^2=b^2 does not necessarily imply a=b. In this case, it's easy to show that (e-s)^2 = (f-s)^2 indeed does not imply e-s=f-s. By construction, e+f = 2s. Thus, e+f = s+s. Rearranging e-s = s-f, or e-s = -(f-s). Therefore, one or the other of e-s or f-s must be a negative value. Mathematically, it's not possible to determine which, but of course it would be f-s based on the physical situation described in the problem.

The square root isn't a problem (or negative) if you do the simplification I showed in #5, above, first. Adding s^2 results in both sides being 4e^2 + 7ef + 4f^2, which will never be negative, so taking the square root would be fine.

Instead, the fundamental problem is that one continues to do algebra with symbols that should have been canceled out of the equation. Just like in this famous example, where step six should really read "0=0":
https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

-ef = -ef

From there, one shouldn’t get different values on either side of the equal sign if one adds s^2 to both and then takes the square root of both.

I agree that Jason could've been a bit clearer in introducing e(e-2s) and f(f-2s), as you are right, you get these by the intermediate step of saying "I'm going to calculate -ef in these two ways, then add some things to it"

However I don't think you avoid the square problem just by pointing out this equivalency. Even phrased as s-ef, the basic error remains the same: for (s-ef [version 1])^2 = (s-ef [version 2])^2 you cannot deduce that the numbers for version 1 and version 2 are the same. Their absolute values must be the same, but one can be positive and the other negative.

@Dave: But that gives you a result that a = a, which is true but not exactly earth-shattering. The point of this problem is that, when y = x^2, there is always exactly one value of y for any given x, but there are almost always two values of x for any given y (if complex values for x and y are allowed). I say "almost always" because everything is kosher if the elephant and fly happen to be weightless.

The "1/4 dollar = 25 cents" example is a nice bit of foreshadowing.

Cthulhu--

One of the problems later in the semester is based on precisely the fallacy you mention. Stay tuned!

I screwed up the math earlier. I'll correct it in a moment.

Eric Lund, of course it's true but not exactly earth shattering. The earth-shattering part is the *wrong* stuff, like concluding e=f. We're supposed to figure out where the mundane turned into earth-shattering, and why it shouldn't have.

But okay, so there are two values for x. That doesn't mean that taking the square root itself was the wrong thing to do. Specifically, it means that the sentence "Taking square roots now gives us that e - s = f - s" is factually wrong. Taking the square root at that point instead gives us |e - s| = |f - s|, which is true. So the precise point at which things went off the rails algebraically is that the result of the square root as written was incorrect.

Other eric, the point is that s^2-ef doesn't have an unknown sign. The signs on both sides of the = are going to be identical, because e and f have the same values throughout. We know those values are positive, too, and that e > f, and so sqrt(s^2-ef) reduces to (e-f)/2.

Because I won't stop to consider animals with negative mass, or flies that are more massive than elephants. Just like if I'm presented with a 9 ft^2 square window, I don't spend even a microsecond pondering whether or not I should buy -12 feet of lumber to frame it (or zero feet, if only two opposite sides have negative length).

And from that practical perspective (rather than a strictly algebraic view), the problem went off the rails as soon as "e+f=2s" was written down. Because neither that nor any of the algebra that followed has any real purpose for, say, zookeepers or comparative biologists.

Taking the square root at that point instead gives us |e – s| = |f – s|, which is true. So the precise point at which things went off the rails algebraically is that the result of the square root as written was incorrect.

Other eric, the point is that s^2-ef doesn’t have an unknown sign.

In hindsight my comment is far more confusing than it needed to be. My point was the same as yours above: (e-s)^2 = (f-s)^2 does not mean e-s = f-s, and therefore does not mean e = f.

the problem went off the rails as soon as “e+f=2s” was written down. Because neither that nor any of the algebra that followed has any real purpose for, say, zookeepers or comparative biologists.

Well, yes and no. As a made-up problem where Jason probably intentionally added some obfuscatory steps, we can say that those steps have no good mathematical purpose. However, if someone is trying to solve a real problem where the answer is unknown to them, they might very reasonably start substituting in equalities and adding terms to both sides, and doing other steps like what Jason did in order to solve their problem. So I think Jason's problem might reasonably resemble a student's attempt to solve an algebra problem. In that respect (i.e., thinking of this as a teacher helping a student see their error), it is correct to say that they went off the rails when they took the square root of two sides and set them equal; they did not go off the rails by trying to find square terms, because we want students to do stuff like that! Right?

Just to elaborate a little on previous answers. Squaring is a one-to-one operation. Taking the square root is one-to-two. There are two possible square roots, and we have to be consistent in which one we pick, across the two sides of the equation. The arguer has been inconsistent. If he'd been consistent, he would have got

e - s = s - f

which would have been correct, but wouldn't have given the arguer the result he wanted.

@Gazza

Yes, I was thrown at first by the idea of "multiplying equations". What does that even mean, I thought. In fact, I almost posted that the argument had gone wrong there, and that the move wasn't truth-preserving. Fortunately I came to my senses just in time to avoid a major embarrassment. ;-)

Since Jason asked for "crystal clarity", I'll elaborate a little more on my answer (at #14).

First, I mentioned that taking the square root gives two alternative values, but I should have said more specifically that those values are positive and negative values with the same absolute value, e.g. 2 and -2.

Second, let's note that s is the arithmetic mean of e and f, i.e. it lies exactly half way between e and f. With that in mind, take another look at the arguer's equation:

(e - s)^2 = (f - s)^2

Since s lies half way between e and f, it follows that (e - s) and (f - s) have the same absolute value but opposite sign, i.e. one is the negative of the other. As long as we're equating the squares of those values, the equality holds. But it should now be clear that the equality would not be preserved if we simply removed the squaring and equated (e - s) with (f - s). We would then be equating a value with its negative. But that's just what the arguer does.

To put it another way, when the arguer factored the quadratic equation, he was free to choose either the positive root or the negative root. He chose the positive root for one side and the negative root for the other. That didn't matter as long as the roots were being squared. But it did matter once the squaring was removed, at the next step.

Sorry for 4 posts in a row, but I'm on a roll. ;)

To be honest, I'm generally more interested in fallacies committed in ordinary (natural language) arguments than those in mathematical proofs. I'm fascinated by natural language, the many ways in which it can used and misused (particularly in philosophy). One of the most common and easily-committed fallacies is the fallacy of equivocation, which is an inconsistency in the use of language.

Jason's room allocation problem commits a kind of equivocation:

"Start by temporarily putting the first and last person in room one. Then put the third person in room two,..."

The term "last person" here is rather vague, and vagueness is equivocation's best friend. When Jason says "last person", I imagine 13 people awaiting allocation of rooms in turn, so that the "last person" is the last one due to be allocated a room. But the so-called "last person" was in fact the second person to be allocated a room. He jumped the queue, and Jason is now calling him (by implication) the second person. The argument "works" by treating him as both the last person and the second. The last will be second, and the second will be last.

You could even interpret the elephant and fly problem as committing a kind of fallacy of equivocation, equivocating between the positive square root and the negative square root, under the ambiguous term "square root".

There have been a couple of comments about the "multiplying the two equations together" step.

If anyone is still wondering about this step, one can look at it this way.

There are two equations

[1] e - 2*s = -f

and

[2] e = -f + 2*s

Multiply both sides of [1] by e, giving

e^2 - 2*s*e = -e*f

We can substitute the value of -f + 2*s for e in the right side of this equation (from [2]), giving

e^2 - 2*s*e = -(-f + 2*s)*f

which gives

[3] e^2 - 2*s*e = f^2 - 2*f*s

More generally, if F = G and H = J, where F, G, H, and J are expressions, then F*H = G*J by the logic above, with the intermediate step of F*H = F*H.

There's another thing we can do here, which is solve for s. It's very close to half the weight of the elephant. I leave it to the other readers to demonstrate that without looking up any actual fly and elephant weights and cranking through the numbers. :)

Besides seeing just where the argument goes wrong, it may be worth taking a moment to note what clever tricks the setter has employed to mislead us. In this case he needed to stop us from too easily seeing that one of the roots was positive and the other negative. That would have given the game away. Introducing s, the arithmetic mean of e and f, guarantees that (e - s) and (f - s) will have opposite signs, without making it too obvious. It's worth noting that the setter didn't mention the word "mean", or use the usual formula for a mean. If he'd done so, we would have been more inclined to have in the back of our mind that e > s > f, and that would have made the opposite signs easier to see.

So, nice trick, but I think we saw through it without too much difficulty. I'm looking forward to the greater deceptions to come. ;)

There are three cases possible. Case1: e=f; in this case e-s and f-s are 0. Case 2: e>f, thus case e-s>0 and f-s<0. Case 3: e<f, thus e-s0.

For Case 1 and Case 2 one simply gets the original equations back again.

Case 4 is not a real case as it requires that both e-s>0 AND f-s>0 and it happens this is just the solution to the square roots chosen in this FALSE proof.

oops Case 3 did not come through correctly should be

Case 3: e<f, thus e-s0

Still - somehow Case 3 does not like to publish correctly. Last try.

Case 3: e<f, thus (e-s)0

POTW #2 is up!
Where it goes wrong (spoilers! Stop reading if you want to do it yourself *********************************In the step between equation #3 and equation #4, the student has multiplied the left side by -1/-1 but multiplied the right side by -1/1.*************************

Oops, my answer #24 is WRONG. I think I just lost track of the negative sign flips. The problem lies in "cancelling the 4x-40"s. You can't do that, you have to multiply each side by the other's denominator, leading to (4x-40)(13-x-7+x) = 0 or simplified (4x-40)(6) = 0. So, x = 10. Plugging that back into the original equation shows that it works: 15/3 - 5 = 0/3

As a rule you can absolutely cancel common numerators by dividing by that shared value. In this case, though, because x=10 is the solution, dividing each side by 4x-40 in the "cancellation" step is division by zero.

Right. Which in hindsight would've lead me to a much more elegant solution to the problem.
1. Realize he must be dividing by zero.
2. Calculate x based on that fact.

Yep, that's how I did it. It's clever in the choice of constants, too, but 4x-40 is a little too suspicious. But, you know, good for the beginning of a semester.

The key here is that you can't cancel a common factor of zero, especially if you are going to take a reciprocal afterwards. The final equation allows you to cancel 4x - 40 only if x != 10.

Of course this is an equation with a single unknown, so it is possible to simply solve it. If you do so, you will find that x is indeed 10. So the final equation becomes 0/-3 = 0/3, which is correct, but hardly impressive.

You all beat me to it! I was going to put up a separate link for this week's problem.

In general the problem's get harder as they go along, though as it happens I think problem five is actually the hardest one of the term. Stay tuned!

Easier than the last problem, I think, because dividing by zero is such a common trick in this sort of problem.

There's a slight misdirection in the language. "Canceling the 4x − 40" sounds more innocuous than "dividing by 4x - 40", which should immediately cause us to think "but what if that term is zero?".

The main problem i have with the POTW2 is that it is not a "prove" at all. The starting equation is not universally true, so if the final simplification yields 7=13, then the conclusion should be that the starting equation simply has no valid solutions. The only error here is then that one misses the valid solution x=10 by not considering the zero of 4x-40 in the division.

On a side note, the 8-digit example reminds of an interesting fact I once figured out. The largest possible m-digit number to any base n, using every digit only once (example: 9876 would be m=4 to base n=10), divided by the smallest such number (example: 1234) is always n-2 with a remainder of m (9876 = 1234*(10-2)+4).