On a rectangular strip of paper, draw 13 straight lines as illustrated below.

Now carefully cut the shape along the following diagonal line, from the top of the first line to the bottom of the last line.

Take slide the upper piece along the diagonal up and left.

You now have a new shape with only 12 lines!

Now explain that you have taken 13 lines and turned them into exactly 12 lines. So clearly 13 lines equal 12 lines, and you have “proven” that 13 = 12.

But 13 is obviously not equal to 12, so something else is going on. How did 1 line just vanish? Why is there a missing line in the new shape?

Most people know something is going on, but very few people can exactly identify the source of the illusion. Watch the video for an explanation.

**“Prove” 13 = 12 With Geometry. The Vanishing Line Paradox**

Or keep reading.

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**Answer To The Vanishing Line Puzzle**

The source of the illusion is the lines in the new shape appear to be the same length as those of the original shape. But that is the trick.

There is of course no missing line. What is happening is the length of the “missing line” gets distrusted evenly to each of the lines in the new shape.

If you measure carefully, each of the 12 lines in the new shape are long, by exactly 1/12, compared to the lines of the original shape. Since each line is increased by only a small amount, most people do not notice the difference and perceive the new lines to be the same length as the original lines.

A key part of the illusion is that each line gets increased by a very small amount. If you try this illusion starting with 4 vertical lines, each of the 3 new lines would be longer by 1/3 each, and that difference would be easily noticeable.

**Source**

Baumback, Randall R. Mathematical puzzles for the secondary mathematics teacher: a collection, classification, and evaluation. Diss. 1980. Puzzle 69. http://csueastbay-dspace.calstate.edu/bitstream/handle/10211.3/7572/Randall.BaumbackThesis.pdf

]]>Alice and Charlie can complete the same job in 3 hours.

Bob and Charlie can complete the same job in 4 hours.

How long will the job take if Alice, Bob, and Charlie work together?

Assume each person works at a constant rate, whether working alone or working with others.

One college professor explained nearly all seniors were unable to solve the problem because they set up incorrect equations. I admit I stumbled a bit before setting up the correct equations. Can you figure it out? Watch the video for the solution.

**Can You Solve A Math Problem That Stumps US College Students? The Working Together Riddle**

Or keep reading.

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**Answer To A Working Together Problem**

Here is the wrong way many people approach the problem. They write the equations:

*a* + *b* = 2

*a* + *c* = 3

*b* + *c* = 4

Now adding up all 3 equations gives:

2(*a* + *b* + *c*) = 9

So then:

*a* + *b* + *c* = 9/2 = 4.5

This would mean it takes 4.5 hours to complete the job when all 3 people are working.

But this answer is clearly wrong! If Bob and Charlie complete the job in 4 hours, it should take *less* than 4 hours to do the job when Alice is working.

Similarly, if Alice and Bob can do the job in 2 hours, it should take less than 2 hours when Charlie helps; and if Alice and Charlie can do the job in 3 hours, it should take less than 3 hours when Alice helps.

So what went wrong? Well the above equations do not really make sense. What are the variables *a*, *b*, and *c* supposed to represent?

Many students blindly translate names into variables and then set up equations that do not capture the heart of the problem.

**The correct approach**

We know Alice and Bob will take 2 hours to complete a job. This means if we add up the percentage of the job Alice did, and the percentage of the job Bob did, the result will be 100% or 1.

(% job Alice does in 2 hours) + (% job Bob does in 2 hours) = 1

Since each works at a constant rate, the amount each does in 2 hours is 2 times as much work as would be done in 1 hour. So we have the equation:

2(% job Alice does in 1 hour) + 2(% job Bob does in 1 hour) = 1

Now let’s set up variables:

*a* = % job Alice does in 1 hour

*b* = % job Bob does in 1 hour

We then have the equation:

2*a* + 2*b* = 1

We can similarly write:

*c* = % job Charlie does in 1 hour

Now we can set up the other equations. Alice and Charlie complete the job in 3 hours, and Bob and Charlie complete the job in 4 hours. We have the equations:

3*a* + 3*c* = 1

4*b* + 4*c* = 1

We have a system of equations:

2*a* + 2*b* = 1

3*a* + 3*c* = 1

4*b* + 4*c* = 1

We could solve the system of equations, but let’s remember what we wanted to solve. We are interested in how long it takes all 3 people to do the job working together. So we need to solve for *t* in the equation:

*t*(*a* + *b* + *c*) = 1

We can get the same number of terms of each variable if we multiply each equation so the coefficient is the least common multiple of 2, 3, and 4. The l.c.m of 2, 3, 4 is 12. So we multiply the first equation by 6, the second by 4, and the third by 3.

12*a* + 12*b* = 6

12*a* + 12*c* = 4

12*b* + 12*c* = 3

Now we add up all of the three equations. We get:

24(*a* + *b* + *c*) = 13

In 24 hours, all three of them can complete the job 13 times. So we divide by 13 to find out how long it would take to complete a single job.

(24/13)(*a* + *b* + *c*) = 1

The correct answer is all three working together would take 24/13 hours, or about 1 hour and 51 minutes, to complete the job.

**Sources**

Toom, Andrei. “A Russian teacher in America.” Journal of Mathematical Behavior 12.2 (1993): 117-139. See the “Tom, Dick and Harry problem” (page 134 of text, page 18 of the following pdf). https://faculty.utrgv.edu/eleftherios.gkioulekas/OGS/Misc/ARUSSIAN.PDF

Jerome Dancis, Associate Professor Emeritus, University of Maryland, Algebraic word problems. www.math.umd.edu/~jnd/Algebraic_word_problems.pdf

]]>Let’s do the integral of 1/*x* with the following substitutions.

We use the integration by parts formula.

Now we simplify the right hand side and cancel like terms.

And voila, we have proved 0 = 1. This is obviously an absurd conclusion. So where is the mistake in this proof? Check out the video for an explanation.

**“Prove” 0 = 1 Using Calculus Integrals**

Or keep reading.

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**Answer To False Proof 1 = 0 Using Integration By Parts**

The mistake in the proof is forgetting the constant of integration. The indefinite integral on the left equals a function plus a constant *c*, and the one on the right equals the same function plus a different constant *C*. We can cancel out the function, and then we get *c* = 1 + *C*. So we have proved the constant on the left is 1 more than the constant on the right.

In other words, the mistake is cancelling the indefinite integrals and expecting equality–the two sides might differ by a constant.

Students often complain about writing the constant of integration. But as you can see, it is vitally important. If you forget the arbitrary constant “plus *C*,” you could quite easily prove 0 = 1 and break math!

An alternate resolution is to consider the definite integral from *a* to *b*. Here is what happens after using integration by parts.

Notice the term of 1 evaluated from *a* to *b* vanishes making for the boring statement that the definite integral equals the definite integral.

So when you’re doing integrals, always make sure to include the arbitrary constant for indefinite integrals, or make sure to work with limits of integration using definite integrals.

**Sources**

https://math.stackexchange.com/questions/806254/using-integration-by-parts-results-in-0-1

https://math.stackexchange.com/questions/1104958/fake-0-1-integral-examples

https://math.stackexchange.com/questions/424854/1-0-by-integration-by-parts-of-tanx

http://mathforum.org/library/drmath/view/62380.html

Su, Francis E., et al. “One Equals Zero: Integral Form.” Math Fun Facts. https://www.math.hmc.edu/funfacts/ffiles/10002.3-8.shtml

**Similar topic**

There is also a false proof that 2 = 1 using derivatives.

See my blog post: Prove 2 = 1 With Calculus. Can You Spot The Mistake?

Or watch the video.

]]>The major reasons are:

–I need more time to make math videos.

–Over the last 10 years game theory has become part of mainstream news coverage, decreasing the uniqueness and need for this column.

I still want to cover game theory, but for reasons explained below, I need another strategy besides the weekly column.

**Longer explanation**

In the last 10 years I have posted over 500 articles on my weekly column game theory Tuesdays. I have also made several game theory videos on YouTube.

I have received many positive comments and emails from students, professors, and many others who love the game theory column, which also became the basis for my book The Joy of Game Theory. I want to thank you for supporting the site all of these years.

I have also happily shared game theory content from other amazing people discussing game theory. For starters, I have compiled an annotated bibliography of the best game theory books which has helped many people learn about game theory. I have also shared many resources on the web.

When I started the column in 2007, no one was blogging about game theory regularly, and there were only a handful of videos on game theory on YouTube. Today, things like the prisoner’s dilemma and Nash equilibrium are commonplace in mainstream news articles and YouTube videos. There are even excellent free video courses such as William Spaniel’s Game Theory 101 (now a University of Pitt professor), Game Theory with Ben Polak (Yale professor), or Coursera’s game theory course (two professors from Stanford + one from University of British Columbia).

It’s great progress that most people have heard of the prisoner’s dilemma and other game theory concepts.

But unfortunately there are still big misconceptions about game theory. For example, a video on SciShow, which now has over 700,000 views, defines the Nash equilibrium incorrectly! The video instead used the definition for a “dominant strategy.” How is this possible?! SciShow is one of the most popular YouTube education channels with ample funding and access to academics. And they still got it wrong! All it would have taken is reading Wikipedia or sending an email to me to get the definition correct.

Sadly my post to explain the error did not receive enough attention. But I did leave a comment on the video. I never heard back from SciShow about the mistake. But I think they got the message: the video now does have an annotation to indicate the error. You can see it at 3:30 in the video SciSchow game theory annotation at 3:30 (you can only see annotations on desktop, so you cannot see it off your phone’s app).

The incident made me realize: something has to change. There is an audience that wants to learn about game theory, but I need a better way to reach the audience.

I have over 400 saved ideas for game theory columns in my drafts, so I could easily write posts for the next 10 years without any trouble. But game theory has taught me to evolve with the times. As I wrote in an early post on this blog, if you aren’t winning the game you are playing, consider changing the game.

I am thinking about the best way to reach people, whether it be by making videos about game theory, creating a game theory course, writing another book about game theory, or re-launching the column in another way.

I’m putting a the weekly column on pause until I can figure out a more effective strategy.

In the meantime I might post about game theory if I see a newsworthy topic, watch a good YouTube video, or read a game theory book worth sharing.

I’m open to suggestions and happy to hear feedback if you have any ideas (leave a comment or email me: presh@mindyourdecisions.com). I have gotten many requests to do a podcast, but I currently do not have the time to be able to make a good podcast.

Let me know! Thank you for supporting Game Theory Tuesdays for a decade!

]]>Mr. Lars then asked each person, including his wife, “How many distinct people did you shake hands with?” Each person answered honestly, and surprisingly each person gave a different answer!

How many distinct people did Mrs. Lars shake hands with? And how many distinct people did Mr. Lars shake hands with?

The problem seems impossible, but it includes just enough information. Watch the video for a solution.

**“Impossible” Riddle – The Handshake Logic Puzzle**

Or keep reading.

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**Answer To Seemingly Impossible Handshake Logic Puzzle**

Let’s work out a smaller example to understand the logic. Imagine a smaller meetup with Mr. and Mrs. Lars and 1 other couple. The group has 4 people in total. Since no one shakes hands with his or her spouse, and no one shakes hands his or her own hand, each person can possibly shake hands with 0, 1, or 2 people. If Mr. Lars heard 3 different answers from the 3 people he asked, then each person gave one of the 3 possible answers.

The person who shook 2 hands must have shaken hands with everyone else except his or her own spouse. This implies everyone else at the meetup shakes hands with at least 1 person. The only person who could have answered 0 is then the spouse of the person who answered 2. The answers of 2 and 0 are paired and belong to the other couple.

By process of elimination, the remaining answer of 1 must be from Mrs. Lars. Thus, Mrs. Lars shook hands with 1 person.

What about Mr. Lars? He did shake hands with the person who answered 2, he could not have shaken hands with his wife who answered 1, and he could not have shaken hands with the person who answered 0. Thus, Mr. Lars must have shaken hands with 1 person, exactly the same as his wife.

**Graphical Approach (1 Other Couple)**

We can draw 4 points for the group, and we can draw an edge between two points if the two people shook hands. I will label the points for Mr. and Mrs. Lars and then call the other couple A and A’.

One person answered 2, meaning some point has 2 edges. If we draw 2 edges from Mrs. Lars, then both of the other points end up connected by an edge. But this is impossible since some person answered 0.

Thus, it has to be one of the other points connected by 2 edges. As that person cannot be connected to his or her spouse, the 2 edges have to connect to Mr. and Mrs. Lars.

From this graph, there is only one point not connected by an edge. This is the spouse of the person who answered 2. This is the only possible choice for the person who answered 0. Thus the answers of 2 and 0 are coupled.

Furthermore, we can see Mrs. and Mr. Lars both have 1 edge, so each of them shook hands with exactly 1 person.

**Solving The Puzzle (4 Other Couples)**

The same logic works for the original problem. There are 10 people in the group, and each person could shake hands with 0, 1, 2, 3, 4, 5, 6, 7, or 8 people.

The person who shook hands with 8 people shook hands with everyone except his or her own spouse. Since everyone else shakes hands with at least 1 person, that means this person’s spouse must be the person who shook hands with 0 people. The answers of 8 and 0 are paired and belong to a married couple.

Someone else shook hands with 7, which means shaking hands with everyone except his or her own spouse and the person who answered 0. Now everyone else, except the spouse and the person who answered 0, shakes hands with at least 2 people, meaning this person’s spouse must be the person who shook hands with 1 person. The answers of 7 and 1 are paired and belong to a married couple.

The same logic can be continued to show that the answers of 6 and 2 are a married couple, as are the answers of 5 and 3.

By process of elimination, the remaining answer must be the answer of Mrs. Lars who therefore shook hands with 4 people.

What about Mr. Lars? He would have shaken hands with the people who gave the answers of 8, 7, 6, and 5. He did not shake hands with the person who answered 4 (his wife). The answer of 3 shook hands with the answers of 8, 7, and 6; the answer of 2 shook hands with the answers of 8 and 7; and the answer of 1 shook hands with the answer of 8. The answer of 0 did not shake anyone’s hand. Therefore, Mr. Lars shook hands with exactly 4 people, just like his wife.

**Graphical Approach (4 Other Couples)**

We can draw 10 points for the group, and we can draw an edge between two points if the two people shook hands. I will label the points for Mr. and Mrs. Lars and then call the other couples A/A’, B/B’, C/C’, and D/D’.

One person answered 8, meaning some point has 8 edges. For the same reason as the 1 couple case, this point cannot be Mrs. Lars (as that would mean else everyone would answer at least 1–but someone answered 0), so it has to be some other point that connects to everyone else in the group. Suppose that is person A.

From this graph, there is only one point not connected by an edge, the spouse of the person who answered 8. This is the only possible choice for the person who answered 0. Thus the answers of 8 and 0 are coupled.

Now someone else answered 7, suppose that is B. In every other couple, each person is already connected by 1 edge. Now take some point and draw 6 more edges. This will connect to everyone in the group except the spouse of that point.

From this graph, there is only one point not identified already or connected by 2 edges, and that is the spouse of the person who answered 7. This is the only possible choice for the person who answered 1. Thus the answers of 7 and 1 are coupled.

We can continue the logic to find that the answers of 6 and 2, as well as the answers of 5 and 3 are coupled. Here are the graphs for those steps.

From the graph, we can see Mrs. and Mr. Lars both have 4 edges, so each of them shook hands with exactly 4 people.

Amazingly we could solve this puzzle by careful logical reasoning!

**Generalizing ( n Other Couples)**

The logic generalizes if the meetup has *n* other couples and the same setup. The answers would be the numbers 0, 1, …, 2*n*. We could then deduce the married couples would be the pairs 2*n* and 0, 2*n* – 1 and 1, and so on, with the unpaired answer of *n*.

This leads to the conclusion Mrs. Lars shook hands with *n* people, and we could further deduce Mr. Lars also shook hands with exactly *n* people (namely, the people who gave answers of *n* + 1, *n* + 2, …, and 2*n*).

Amazingly we figured out the handshakes of both Mr. and Mrs. Lars from this limited information!

**Thanks to all patrons! Special thanks to:**

Marlon Forbes

Shrihari Puranik

Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Sources**

The puzzle was published by Martin Gardner who credits Lars Bertil Owe.

https://puzzling.stackexchange.com/questions/11306/dr-and-mrs-smith-shake-some-hands

http://skepticsplay.blogspot.com/2009/05/handshakes-solution.html

https://alokgoyal1971.com/2013/11/09/solution-to-puzzle-38-famous-handshake-puzzle/

http://www.casact.org/newsletter/index.cfm?fa=viewart&id=6021

http://www.cut-the-knot.org/pigeonhole/FiveCouples.shtml#solution

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