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# Can You Find the Mistakes? Five MORE "Faulty" Puzzles to Try

Updated: Jul 5

Welcome back to GLeaM! It seems like my last post was pretty well-received, so we're back at it again with a part 2!

If you missed the last post, let me give a brief introduction. One of my favorite types of

puzzles are those where you're supposed to discover an error in a proof—one of those bizarre arguments that ends in something completely absurd like 2=1 or "squares have three sides". The goal is to discover where that mistake is and where everything went so wrong. These false "proofs" are often called fallacies, and I invite you to scour them for errors. Consider it a contest—you vs the puzzle—and you want to come out on top.

To me, finding errors and discovering contradictions reveals something innate about the field of math itself. After all, we can't have all sorts of errors in a field based entirely on logic, so we have to avoid the gray areas! Many outsiders would argue that this rigid structure makes math super bland and boring, but it's actually what pushes us to new creative heights. We're inspired to tackle new angles and new directions to find new avenues around possibly wishy-washy statements, and that's what leads to the most brilliant sparks of intuition in math across the world.

So without further ado, here are five more of my favorite fallacies.

For some of the more introductory examples to fallacies, you'll want to read my last article, but this concept should explain itself just as well with the examples here too!

We'll range through a bunch of different topics, and you'll find hints (and occasionally solutions!) at the bottom of the article. Let's start!

1) The Root of All Errors

Source = MindYourDecisions: "Prove" 2 = 0 Using Square Roots

(Similar to an idea in Hexaflexagons and Other Mathematical Diversions by Martin Gardner)

Ok, maybe I went a little bit far with the pun in this title, but this error is certainly the "root" of a lot of others. This one requires a bit of knowledge of complex numbers, but essentially all you need to know is that i is the square root of -1. (Feel free to skip forward if that still seems foreign.)

See if you can figure out what might be wrong with this proof: 1 is the square root of 1, so we can replace it: 1 is the product of -1 and -1, so we can replace again: i is the square root of -1, so we're multiplying i and i together. i times i is -1:  Wait a second, that's impossible! Did you find the mistake?

If not, remember all hints and solutions are at the end of the article!

2) Probability and Presents

Source = Hexaflexagons and Other Mathematical Diversions by Martin Gardner

There are all sorts of fallacies in probability. Here, I'm going to take Martin Gardner's puzzle originally phrased in terms of neckties and tweak it a bit, but most of it is quoted pretty directly:

Say you and your friend Alice have just got presents for Christmas. You begin to argue over which of you received the more expensive present. You and Alice finally agree to settle this by visiting the store where both presents were bought and checking their value. Whoever wins (and has the most expensive present) must give their present to the loser as a consolation.

This is how you reason: "The chances that I will win the argument or lose it are equal. If I win, I will be poorer by the value of the present. But if I lose, I am sure to gain a more expensive present. Therefore, the contest is clearly to my advantage (because on average, I'll end up being richer).

To make this line of reasoning more clear, say your present is \$30. Half the time, you'll lose \$30, but the other half of the time you'll gain MORE than \$30.

But, Alice can reason in the same way! No bet can put both sides at an advantage, so what's the matter here?

Source = MindYourDecisions: "Prove" 90 Equals 100 (and PROMYS!)

For those of you who love geometry, I certainly have a problem for you!

As a bit of a backstory, it turns out that I'd actually seen this problem before: Back in March when I was applying to math summer programs, including the one called PROMYS, I solved this problem for a problem set. Flash forward to this series and I came across it again in more of the popular sphere, and I thought it would be a perfect problem to show you, so here goes!

Take a quadrilateral, ABCD. Two of its opposite sides, AD and BC, are the same length. Angle BAD is 90 degrees, and angle ABC is 100 degrees.

Draw in the perpendicular bisectors of side CD and side AB. Take their intersection and call it point E. Courtesy of MindYourDecisions

Because these two line segments are perpendicular bisectors, E is equidistant from point A and point B, and E is also equidistant from point C and point D. Thus, EA = EB and EC = ED.

(This is just a basic fact of perpendicular bisectors. To convince yourself that this is true, let's label the point where AB meets the perpendicular bisector, X. (This is just the midpoint of AB.) The triangles AEX and BEX are congruent because AX = BX by definition and XE is a side of both triangles. Also, the angle between these two sides is 90 degrees for both. Thus, EA = EB) Courtesy of MindYourDecisions

If we consider this part of the diagram, Courtesy of MindYourDecisions

you'll notice that triangles AED and BEC are congruent because they have three equal sides. This means that angle DAE is equal to angle EBC.

Also, if we consider this part of the diagram, Courtesy of MindYourDecisions

you'll notice that triangles AEX and BEX are congruent (Here, X is once again the midpoint of AB) because they have three equal sides (and even a 90-degree angle) in common. This means that angle EAB is equal to angle EBA.

Thus, angle DAE + angle EAB = angle EBC + angle EBA.

But, angle DAE + angle EAB = 90 degrees, and angle EBC + angle EBA = 100 degrees, so

90 = 100.

This is absurd. What went wrong?

What's even crazier is that the fact we chose 100 degrees as our angle is kind of arbitrary. If we replace that with any other obtuse value, we have that every obtuse angle has measure 90 degrees, meaning every obtuse angle is right?? How is that even possible?

4) All numbers are... interesting?

Source = Hexaflexagons and Other Mathematical Diversions by Martin Gardner

Don't worry, this isn't a philosophical debate.

We're going to be looking at another situation where logic can be too strong. Remember, in the last article, we looked at a couple situations where applying logic seems to go astray, showing that if we can eat some amount of cake without getting full, we can eat any amount of cake, and talking all about the famous unexpected hanging paradox. This is such a fascinating topic to discuss, and I'd encourage you to check out the discussion in the appendix at the end of that article, but for now, here's an interesting argument for you along the same lines.

Let's say any number that has a kind of unique property is "interesting".

1 is interesting because it's the only number for which multiplying by any number spits the same number back out. (It's the identity: 1 * x = x for any x.)

2 is interesting because it's the smallest prime.

3 is interesting because it's the first odd prime.

1729 is interesting because it's the smallest number that can be expressed as the sum of two cubes two ways: 12^3 + 1^3 = 10^3 + 9^3 = 1729. (The great Indian mathematician Srinivasa Ramanujan loved this one.)

602,200,000,000,000,000,000,000 is interesting because it's an approximation for a very important concept in chemistry. (Look up Avogadro's Number.)

And so on...

There are so many interesting properties, but every number can't be that special right. Like what about 38901398123489 or 129087102349707? That's just me typing down random digits.

There's no way to presume any number has some sort of property that is entirely unique to it. In fact, we'd be hard-pressed to come up with these unique properties for the first 100 numbers.

Well, here's a proof that every number is interesting (and not "dull"), courtesy of Martin Gardner:

If there are dull numbers, we can then divide all numbers into two sets—interesting and dull. In the set of dull numbers, there will be one number that is the smallest. Since it is the smallest uninteresting number, well that's an interesting property, so it becomes, ipso facto, an interesting number. We must, therefore, remove it from the dull set and place it in the other. But now there will be another smallest uninteresting number. Repeating this process will make any dull number interesting.

How would you resolve this paradox? Why doesn't this argument make a lot of sense?

You could brush this aside with a "logic doesn't apply here" but think deeper. What about this generates such a bizarre conclusion?

5) Repeated Exponents

Source = MindYourDecisions: "Prove" 4 = 2

One thing mathematicians love more than anything is to take things to infinity. Let's try making exponent towers that go to infinity and beyond!!

Say you were asked to find an answer x to the following problem: Since this goes to infinity, we can replace the expansion at the top of the tower (the power the lowest x is raised to) with 2 as well. After all, both the overall expression and the power contain infinite x's! This gives: And then: Which tells us: Guess what? We can do the exact same thing with 4.

Say you were asked to find an answer x to the following problem: Similarly, we have: And then:

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