Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Related: Visually stunning math concepts which are easy to explain

Beside the wonderful examples above, there should also be counterexamples, where visually intuitive demonstrations are actually wrong. (e.g. missing square puzzle)

Do you know the other examples?

share|cite|improve this question
all of the answers below rely on a slight bend in a diagonal line which accounts for the missing area – ratchet freak Apr 7 '14 at 8:07
@ratchetfreak Actually I don't believe that is true for the chocolate puzzle. See my update – Michael Tong Apr 7 '14 at 11:48
@MichaelT The point is that if you had moved the piece congruently, you would have the bent-diagonal that the others use; you don't get it only because the animation 'fills it in' along the way. – Steven Stadnicki Apr 7 '14 at 17:53
Also, interestingly (well, interestingly-to-me), most of the variants rely on the fact that $F_{n+1}F_{n-1}-F_n^2 = (-1)^n$; presumably this makes for a more appealing false-dissection than $n^2-(n+1)(n-1)=1$ because the 'aspect ratio' of the rectangular side is more skewed. – Steven Stadnicki Apr 7 '14 at 17:56
Just like magic tricks, these concepts rely entirely on the tricking of the human senses (which tend to be easily fooled). – Domi Apr 8 '14 at 12:43

20 Answers 20

The never ending chocolate bar!


If only I knew of this as a child..

The trick here is that the left piece that is three bars wide grows at the bottom when it slides up. In reality, what would happen is that there would be a gap at the right between the three-bar piece and the cut. This gap is is three bars wide and one-third of a bar tall, explaining how we ended up with an "extra" piece.

Side by side comparison:


Notice how the base of the three-wide bar grows. Here's what it would look like in reality$^1$:


1: Picture source

share|cite|improve this answer
This - and both other answers - are completely equivalent to the one mentioned in the OP. – Jack M Apr 7 '14 at 6:22
Actually I'm not so sure -- in the case of the missing square puzzle, the trick is that the diagonals are not straight. With this puzzle it's that the chocolate is cut in a way that, in reality, a full row of chocolate would not be completed around the right edge. I've edited the post to explain that. – Michael Tong Apr 7 '14 at 11:36
This is the Banach-Tarski paradox of candybars! – Asaf Karagila Apr 7 '14 at 16:21
I love this one! – Integral Apr 13 '14 at 16:03

A bit surprised this hasn't been posted yet. Taken from this page:

enter image description here

share|cite|improve this answer
I don't see from this explanation why $\pi = 24$ :) – Michael Tong Apr 7 '14 at 11:58
I like this one because it shows that Archimedes really does have a problem: he has to explain why his limiting process (circumscribed and inscribed $n$-gons) approaches $\pi$ in the limit, while this one doesn't. There is an explanation, because the technique Archimiese used does actually work, but I don't think the explanation was available to Archimedes. They key point is that although the zigzag converges pointwise to the circle, the slopes of the segments don't converge, and the arc length is a function of the slope. – MJD Apr 7 '14 at 12:50
@MJD: Archimedes' technique does not have this problem. He used a postulate that if one convex curve encloses another, then the outer is longer than the inner. By constructing sequences of inscribed and circumscribed polygons, he was able to produce upper and lower bounds for the ratio of circumference to diameter. There was no assumption that a sequence of curves approaching the circle must have length approaching the length of the circle. The sequence of polygons in the picture only shows that $\pi$ is less than 4. – user2357112 Apr 7 '14 at 13:33
@MJD Interestingly this has a lot to do with Taxicab geometry ( Where this logic is actually correct. This shows that this is not fundamentally wrong, it is more a fairly arbitrary decision of how we define distance in classic geometry. – Vality Apr 7 '14 at 18:31
@Awesome Oh. Well I don't see how two people saying that $4! = 24$ implies plagiarism. The delivery were quite different. – Michael Tong Apr 9 '14 at 12:15

Visualization can be misleading when working with alternating series. A classical example is \begin{align*} \ln 2=&\frac11-\frac12+\frac13-\frac14+\;\frac15-\;\frac16\;+\ldots,\\ \frac{\ln 2}{2}=&\frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac1{12}+\ldots \end{align*} Adding the two series, one finds \begin{align*}\frac32\ln 2=&\left(\frac11+\frac13+\frac15+\ldots\right)-2\left(\frac14+\frac18+\frac1{12}+\ldots\right)=\\ =&\frac11-\frac12+\frac13-\frac14+\;\frac15-\;\frac16\;+\ldots=\\ =&\ln2. \end{align*}

share|cite|improve this answer
the problem here is that, since the series does not converge absolutely, then you can't change the order of summation? – Ant Apr 7 '14 at 9:20
@Ant Yes, exactly - one cannot rearrange the terms. – Start wearing purple Apr 7 '14 at 10:28
@Ant, this is also known as the Riemann series theorem. – David Apr 12 '14 at 0:25

Another example :

enter image description here

From "Pastiches, paradoxes, sophismes, etc." and solution page 23 :

A copy of the solution is added below. The translation of the comment is :

Explanation : The points A, B and P are not on a straight line ( the Area of the triangle ABP is 0.5 ) The graphical highlight is magnified only on the left side of the figure.

enter image description here

share|cite|improve this answer
Link doesn't have the solution that I can see... Curious about this one! – SLC Apr 8 '14 at 12:27
@ SLC : I am surprised that you cannot see the solution which is on the same page (23) of the referenced document. In order to avoid any further difficulty, I will add a copy of the solution to my first answer. – JJacquelin Apr 8 '14 at 13:38
Maybe you have some cookies or something affecting the page, there are no pages on your link, just 17 items none of which are the puzzle above. I see it's your page - perhaps it's set to private? – SLC Apr 8 '14 at 13:47
I understand the misunderstanding. In fact the 17 items are 17 different papers. One of them is entitled "Pastiches, paradoxes, sophismes, etc." It's open by clicking on it. Then you have acces of the pages. But you no longer need it, since the solution is now visible on my answer above. – JJacquelin Apr 8 '14 at 13:55
Why not use the URL – Did Apr 8 '14 at 13:59

Construct a rectangle $ABCD$. Now identify a point $E$ such that $CD = CE$ and the angle $\angle DCE$ is a non-zero angle. Take the perpendicular bisector of $AD$, crossing at $F$, and the perpendicular bisector of $AE$, crossing at $G$. Label where the two perpendicular bisectors intersect as $H$ and join this point to $A$, $B$, $C$, $D$, and $E$.

enter image description here

Now, $AH=DH$ because $FH$ is a perpendicular bisector; similarly $BH = CH$. $AH=EH$ because $GH$ is a perpendicular bisector, so $DH = EH$. And by construction $BA = CD = CE$. So the triangles $ABH$, $DCH$ and $ECH$ are congruent, and so the angles $\angle ABH$, $\angle DCH$ and $\angle ECH$ are equal.

But if the angles $\angle DCH$ and $\angle ECH$ are equal then the angle $\angle DCE$ must be zero, which is a contradiction.

share|cite|improve this answer
Very nice. (Hint to anyone confused: Only the final sentence is wrong, plus the diagram in a way.) – aschepler Apr 8 '14 at 0:23
A true classic, and inherently 'visual' in its deception. IMHO the best answer yet to this question. – Steven Stadnicki Apr 8 '14 at 0:26
On my screen it looks obvious that GH as drawn is not perpendicular to AE. Maybe if you rotate the entire diagram counterclockwise by ~1 degree, so that AD and AE are both slightly angled, it would obscure this and make it more convincing. – Mechanical snail Apr 8 '14 at 13:10

I think this could be the goats puzzle (Monty Hall problem) which is nicely visually represented with simple doors.

Three doors, behind 2 are goats, behind 1 is a prize.


You choose a door to open to try and get the prize, but before you open it, one of the other doors is opened to reveal a goat. You then have the option of changing your mind. Should you change your decision?

From looking at the diagram above, you know for a fact that you have a 1/3rd chance of guessing correctly.

Next, a door with a goat in is opened:

enter image description here

A cursory glance suggests that your odds have improved from 1/3rd to a 50/50 chance of getting it right. But the truth is different...

By calculating all possibilities we see that if you change, you have a higher chance of winning.


The easiest way to think about it for me is, if you choose the car first, switching is guaranteed to be a goat. If you choose a goat first, switching is guaranteed to be a car. You're more likely to choose a goat first because there are more goats, so you should always switch.

share|cite|improve this answer
Of course the fact that your visuals include nice pointers along the path to the correct answer kind-of undermines your point, but I see where you're going. :) – Jules Apr 8 '14 at 4:26
Hehe, I had to put the solution and I had to use graphics because the question wanted visually deceptive proofs. I think the first two pictures are really visually deceptive because you see two doors and know one is a goat, so it seems blatantly to be 50/50 chance :) – SLC Apr 8 '14 at 12:23
The easiest way for me to think about it: instead of 3 doors let's say there's 1 million. You pick a door, and now we'll eliminate 999,998 doors with goats behind them. There are only two doors now, one of which was the door you picked randomly out of 1 million. What are the odds you guessed right on your first try? – 16807 Apr 9 '14 at 19:10
Nice, I like it. – SLC Apr 10 '14 at 8:26
There is another to think this question, which I consider the most obvious way to start with. When you choose a door, more probably you choose the goat. So, probably the car is one of the others doors, but changing won't make difference. Except when another goat is showed to you, if you were probably wrong about the first choice, probably the car is in the another door. Therefore changing is a good idea. This way do not gives you the probability, but its a more intuitive way to know that changing is better. – Integral Apr 13 '14 at 16:39

Here's how to trick students new to calculus (applicable only if they don't have graphing calculators, at that time):

$0$. Ask them to find inverse of $x+\sin(x)$, which they will unable to. Then,

$1$. Ask them to draw graph of $x+\sin(x)$.

$2$. Ask them to draw graph of $x-\sin(x)$

$3$. Ask them to draw $y=x$ on both graphs.

Here's what they will do :

enter image description here

$4$. Ask them, "What do you conclude?". They will say that they are inverses of each other. And then get very confused.

share|cite|improve this answer
I don't get it. – Cruncher Apr 7 '14 at 15:22
What bothers me most about the post is your instructions to "laugh at" students and/or "their futile attempts." That strikes me as being snobbish - which most students struggling with math would do better without. – amWhy Apr 7 '14 at 15:39
6. Laugh at the downvotes on this answer. – A little lime Apr 7 '14 at 17:47
@Alittlelime to be fair... I think it's a little heavily downvoted. – Cruncher Apr 7 '14 at 20:15
Actually these downvotes bother me more than the answer. The redaction or tone might have been unfortunate, but the example is not bad. – leonbloy Apr 8 '14 at 21:07

Every triangle is isosceles :


Proof : Let $O$ be the intersection of the bisector $[BC]$ and the bisector of $\widehat{BAC}$. Then $OB=OC$ and $\widehat{BAO}=\widehat{CAO}$. So the triangles $BOA$ and $COA$ are the same and $BA=CA$.

share|cite|improve this answer
Of course, ASS congruence isn't a thing. – user2357112 Apr 7 '14 at 13:20
@user2357112 ... to the disappointment of middle-schoolers everywhere. – ApproachingDarknessFish Apr 7 '14 at 16:12
The p0rblem is that the bisector of BC and the bisector of A intersect outside the triangle — on the circumcircle, in fact. – kinokijuf Apr 8 '14 at 18:54
@kinokijuf: No, the problem is the one mentioned by user2357112. Where the intersection happens doesn't change the argument. – Kundor Apr 9 '14 at 1:55
@kinokijuf is right. It is not about ASS congruence. $OF=OG$ because $AO$ is a bissector. So, $AFO=AGO$ and $OFB=OGC$ --- each pair because of cathetus and hypotenuse. That implies $AB=AC$ and $BOA=COA$. – sas Apr 9 '14 at 9:42

A favorite of mine was always the following:

\begin{align*} \require{cancel}\frac{64}{16} = \frac{\cancel{6}4}{1\cancel{6}} = 4 \end{align*}

I particularly like this one because of how simple it is and how it gets the right answer, though for the wrong reasons of course.

share|cite|improve this answer
-_- Now that's just silly. – Navin Apr 10 '14 at 5:47
3 – Amihai Zivan Apr 10 '14 at 12:58
Here's some mathematica that finds all of them: works[n_, d_] := n > 0 && d > 0 && n != d && First@IntegerDigits[n] == Last@IntegerDigits[d] && Last@IntegerDigits[n]/First@IntegerDigits[d] == n/d; Select[Tuples[Range[10, 99], 2], Apply[works]] – Leo Izen Aug 28 '14 at 12:07
A related thread. – J. M. Jun 28 at 1:31

One of my favorites:

\begin{align} x&=y\\ x^2&=xy\\ x^2-y^2&=xy-y^2\\ \frac{(x^2-y^2)}{(x-y)}&=\frac{(xy-y^2)}{(x-y)}\\ x+y&=y\\ \end{align}

Therefore, $1+1=1$

The error here is in dividing by x-y

share|cite|improve this answer
that's a good one, division by zero is cool trick. do you know any written source of more like this one? – lowtech Apr 8 '14 at 15:22
Is this a visually deceptive "proof"? – Rahul Apr 9 '14 at 7:24
@Rahul I would think so. Just looking at the equations makes it seem like it would be valid. – David Starkey Apr 9 '14 at 14:05
However, this works in the zero ring because zero is invertible there, and indeed in the zero ring 1 + 1 = 1 because 1 = 0. – Jason Polak Apr 9 '14 at 15:24
@Rahul A few years ago, me and my friend were confused by this and tried for days to figure out what was wrong. We went to one our math teachers and he was stumped too until he realized after 20 minutes the problem is dividing by 0. So, I'd say it's deceptive. – Cole Johnson Apr 11 '14 at 0:51

A recent example I found which is credited to Martin Gardner and is similar to some of the others posted here but perhaps with a slightly different reason for being wrong, as the diagonal cut really is straight.

enter image description here

I found the image at a blog belonging to Greg Ross.


The triangles being cut out are not isosceles as you might think but really have base $1$ and height $1.1$ (as they are clearly similar to the larger triangles). This means that the resulting rectangle is really $11\times 9.9$ and not the reported $11\times 10$.

share|cite|improve this answer

Here is one I saw on a whiteboard as a kid... \begin{align*} 1=\sqrt{1}=\sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=\sqrt{-1}^2=-1 \end{align*}

share|cite|improve this answer
result of the assumption that $\sqrt b \cdot \sqrt a = \sqrt{a\cdot b}$ for all reals $a$ and $b$ instead of only for positive reals – ratchet freak Apr 10 '14 at 9:38
I don't see anything visual about this; it is a left-hemisphere only affair. – Marc van Leeuwen Apr 13 '14 at 7:54
A more accurate way to write this would be $1=\sqrt{1}=\sqrt{e^{i\pi}\times e^{-i\pi}}$, from which you end up with $i\times -i=1$. – Glen O Apr 13 '14 at 8:53

That $\sum_{n=1}^\infty n = -\frac{1}{12}$.

The way it is presented in the clip is completely incorrect, and could spark a great discussion as to why.

Some students may notice the hand-waving 'let's intuitively accept $1 -1 +1 -1 ... = 0.5$.

If we accept this assumption (and the operations on divergent sums that are usually not allowed) we can get to the result.

A discussion that the seemingly nonsense result directly follows a nonsense assumption is useful. This can reinforce why it's important to distinguish between convergent and divergent series. This can be done within the framework of convergent series.

A deeper discussion can consider the implications of allowing such a definition for divergent sequences - ie Ramanujan summation - and can lead to a discussion on whether such a definition is useful given it leads to seemingly nonsense results. I find this is interesting to open up the ideas that mathematics is not set in stone and can link to the history of irrational and imaginary numbers (which historically have been considered less-than-rigorous or interesting-but-not-useful).

share|cite|improve this answer
There are series, like the above, for which such "calculation" works and makes sense. It's OK to do things like that as long as we afterwards are in pursuit of more tight and strict explanation of given phenomena. – user2622016 Apr 7 '14 at 8:17
Can you provide a citation or explanation for why "The way it is presented in the clip is completely incorrect"? – CodesInChaos Apr 7 '14 at 15:41
Ah numberphile, the worst ever attempt at mathematical education. – bwv869 Apr 7 '14 at 20:15
What's 'visual' about this, other than the fact that someone made a video explaining it? – Steven Stadnicki Apr 8 '14 at 0:25
For those interested in this result, here is a very interesting article from Terence Tao explaining how to derive this result using elementary (? - I don't know exactly what elementary means, I've seen it defined once as 'without using techniques that use complex numbers', but I think it's fair to call this elementary) methods. – Ruben Apr 9 '14 at 21:32

\begin{equation} \log6=\log(1+2+3)=\log 1+\log 2+\log 3 \end{equation}

share|cite|improve this answer
"that was legitness" -_- – David_Shmij Jun 4 at 23:09

There are two examples on Wikipedia:Missing_square_puzzle Sam Loyd's paradoxical dissection, and Mitsunobu Matsuyama's "Paradox". But I cannot think of something that is not a dissection.

share|cite|improve this answer
This is already mentioned on the original post. – Andrés E. Caicedo Apr 7 '14 at 6:05

Here is a measure theoretic one. By 'Picture', if we take a cover of $A:=[0,1]∩\mathbb{Q}$ by open intervals, we have an interval around every rational and so we also cover $[0,1]$; the Lebesgue measure of [0,1] is 1, so the measure of $A$ is 1. As a sanity check, the complement of this cover in $[0,1]$ can't contain any intervals, so its measure is surely negligible.

This is of course wrong, as the set of all rationals has Lebesgue measure $0$, and sets with no intervals need not have measure 0: see the fat Cantor set. In addition, if you fix the 'diagonal enumeration' of the rationals and take $\varepsilon$ small enough, the complement of the cover in $[0,1]$ contains $2^{ℵ_0}$ irrationals. I recently learned this from this MSE post.

share|cite|improve this answer

This is my favorite.

\begin{align}-20 &= -20\\ 16 - 16 - 20 &= 25 - 25 - 20\\ 16 - 36 &= 25 - 45\\ 16 - 36 + \frac{81}{4} &= 25 - 45 + \frac{81}{4}\\ \left(4 - \frac{9}{2}\right)^2 &= \left(5 - \frac{9}{2}\right)^2\\ 4 - \frac{9}{2} &= 5 - \frac{9}{2}\\ 4 &= 5 \end{align}

You can generalize it to get any $a=b$ that you'd like this way:

\begin{align}-ab&=-ab\\ a^2 - a^2 - ab &= b^2 - b^2 - ab\\ a^2 - a(a + b) &= b^2 -b(a+b)\\ a^2 - a(a + b) + \frac{a + b}{2} &= b^2 -b(a+b) + \frac{a + b}{2}\\ \left(a - \frac{a+b}{2}\right)^2 &= \left(b - \frac{a+b}{2}\right)^2\\ a - \frac{a+b}{2} &= b - \frac{a+b}{2}\\ a &= b\\ \end{align}

It's beautiful because visually the "error" is obvious in the line $\left(4 - \frac{9}{2}\right)^2 = \left(5 - \frac{9}{2}\right)^2$, leading the observer to investigate the reverse FOIL process from the step before, even though this line is valid. I think part of the problem also stems from the fact that grade school / high school math education for the average person teaches there's only one "right" way to work problems and you always simplify, so most people are already confused by the un-simplifying process leading up to this point.

I've found that the number of people who can find the error unaided is something less than 1 in 4. Disappointingly, I've had several people tell me the problem stems from the fact that I started with negative numbers. :-(


When working with variables, people often remember that $c^2 = d^2 \implies c = \pm d$, but forget that when working with concrete values because the tendency to simplify everything leads them to turn squares of negatives into squares of positives before applying the square root. The number of people that I've shown this to who can find the error is a small sample size, but I've found some people can carefully evaluate each line and find the error, and then can't explain it even after they've correctly evaluated $\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.

share|cite|improve this answer

To give a contrarian interpretation of the question I will chime in with Goldbach's comet which counts the number of ways an integer can be expressed as the sum of two primes:

It is mathematically "wrong" because there is no proof that this function doesn't equal zero infitely often, and it is visually deceptive because it appears to be unbounded with its lower bound increasing at a linear rate.

share|cite|improve this answer
The lack of a proof doesn't mean that it's mathematically wrong; it just means we don't know it's right. (I don't know many folks who would take a finite section of a graph as a legitimate mathematical proof.) – Steven Stadnicki Apr 9 '14 at 8:01
@StevenStadnicki I collapsed my scare-quotes around "mathematically wrong" down to "wrong". But they are still kinda scary! – Dan Brumleve Apr 9 '14 at 8:56
An example for what @StevenStadnicki is saying is the Collatz Conjecture. Just because we don't have a proof for it doesn't mean it's wrong, it just means we haven't proven it. – Cole Johnson Apr 11 '14 at 0:58
What do the colors mean? – Martin Thoma Apr 11 '14 at 19:00
@moose From the Wikipedia article linked in the answer: "The function $g(E)$ is defined for all even integers $E>2$... The coloring of points in the above image is based on the value of $E/2$ modulo 3 with red points corresponding to 0 mod 3, blue points corresponding to 1 mod 3 and green points corresponding to 2 mod 3." – Travis Bemrose Apr 13 '14 at 14:40

I do not think this really qualify as "visually intuitive", but it is definitely funny

They do such a great job at dramatizing these kind of situations. Who cannot remember of an instance in which he has been either a "Billy" or a "Pa' and Ma'"? Maybe more "Pa' and Ma'" instances on my part...;)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.