# Visually deceptive “proofs” which are mathematically wrong

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?

• 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 – MCT 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

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$$^1$$: • 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. – MCT 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
• I like this version better than the one at youtube.com/watch?v=kx41KG_wC_Y because in that version of the never ending chocolate bar, it's too obvious what's happening. Are you sorry you didn't see it when you were a kid because now it's too late for you to get tricked into thinking volume is not a fixed thing that adds when you take the union? Actually, in ZF, you can't prove that there exists a way to define volume that satisfies those properties as shown at youtube.com/watch?v=s86-Z-CbaHA. You might love that video. – Timothy Jan 9 '19 at 4:21

A bit surprised this hasn't been posted yet. Taken from this page: • I don't see from this explanation why $\pi = 24$ :) – MCT 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 supports Monica Apr 7 '14 at 13:33
• @MJD Interestingly this has a lot to do with Taxicab geometry (en.wikipedia.org/wiki/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. – MCT 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*}

• 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
• I just came across this in Tao's notes on Measure Theory :-) ... the rearranging being invalid if a sequence involves the negatives – Mateen Ulhaq Dec 20 '16 at 1:07

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

• 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

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 $E$. 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.

• 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
• This fallacy was invented by Charles Dodgson (otherwise known as Lewis Carroll), and published in the section Curiosa Mathematica.of The Lewis Carroll Picture Book (ed. Stuart Dodgson Collingwood, pub. 1899) – Rosie F Feb 17 at 20:19
• @RosieF - I was not aware of that, but you are correct – Henry Feb 17 at 22:02

Another example : From "Pastiches, paradoxes, sophismes, etc." and solution page 23 : http://www.scribd.com/JJacquelin/documents

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. • @ 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? – NibblyPig 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 http://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes... – Did Apr 8 '14 at 13:59 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 $BA=CA$.

• Of course, ASS congruence isn't a thing. – user2357112 supports Monica 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. – Nick Matteo 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

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: 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.

• 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 :) – NibblyPig 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. – NibblyPig 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

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.

• -_- Now that's just silly. – Navin Apr 10 '14 at 5:47
• projecteuler.net/problem=33 – 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. isn't a mathematician Jun 28 '16 at 1:31
• I believe the more famous example is $\frac{19}{95}=\frac15$. – Simply Beautiful Art Feb 8 '17 at 0:13

Squaring the circle with Kochanski's Approximation1 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. I found the image at a blog belonging to Greg Ross.

Spoilers

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 $11\times 10$.

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

• 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"? – user856 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. – user2055 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

That $\sum_{n=1}^\infty n = -\frac{1}{12}$. http://www.numberphile.com/videos/analytical_continuation1.html

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).

• 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. – user85798 Apr 7 '14 at 20:15
• I think "seemingly nonsense results" is a little harsh. Divergent series methods often give "right" answers when applied to real problems. It's no different or more suspect than extending any other arbitrary linear, bounded functional. There wouldn't be so many different divergent series methods if nobody considered it useful. – Tim Seguine Apr 10 '14 at 21:06

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

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*}

• 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

I might be a bit late to the party, but here is one which my maths teacher has shown to me, which I find to be a very nice example why one shouldn't solve an equation by looking at the hand-drawn plots, or even computer-generated ones.

Consider the following equation: $$\left(\frac{1}{16}\right)^x=\log_{\frac{1}{16}}x$$

At least where I live, it is taught in school how the exponential and logarithmic plots look like when base is between $0$ and $1$, so a student should be able to draw a plot which would look like this: Easy, right? Clearly there is just one solution, lying at the intersection of the graphs with the $x=y$ line (the dashed one; note the plots are each other's reflections in that line).

Well, this is clear at least until you try some simple values of $x$. Namely, plugging in $x=\frac{1}{2}$ or $\frac{1}{4}$ gives you two more solutions! So what's going on?

In fact, I have intentionally put in an incorrect plots (you get the picture above if you replace $16$ by $3$). The real plot looks like this: You might disagree, but to be it still seems like it's a plot with just one intersection point. But, in fact, the part where the two plots meet has all three points of intersection. Zooming in on the interval with all the solutions lets one barely see what's going on: The oscillations are truly minuscule there. Here is the plot of the difference of the two functions on this interval: Note the scale of the $y$ axis: the differences are on the order of $10^{-3}$. Good luck drawing that by hand!

To get a better idea of what's going on with the plots, here they are with $16$ replaced by $50$: 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.

• 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.

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. :-(

Solution

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 $\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.

This is essentially the same as the chocolate-puzzle. It's easier to see, however, that the total square shrinks. 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.

• 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

My favorite is: enter image description here Is the area of both shapes are equal? enter image description here It looks that $$area=\frac{44×86}2=22×86$$ But, the triangle is impossible!

• There is no fallacy here. The rectangle $ABCD$ can be dissected into two triangles $AFD, BCE$ and a trapezium $ABEF$ with parallel sides $AB=86$ and $EF=43$, then the triangles rotated so as to produce the pictured triangle. The pictured triangle is acute, not right-angled, but the picture doesn't claim that it's right-angled, only that the perpendicular to $AB$ is 44, which it can be. – Rosie F May 6 at 8:24 This is a fake visual proof that a sphere has Euclidean geometry.

I do not think this really qualify as "visually intuitive", but it is definitely funny I don't know about you but to me, it looks like the hexagons are stretched horizontally. If you also see it that way and you trust your eyes, then you could take that as a visual proof that $$\tan\frac{7}{4} < 60^\circ$$. If that's how you saw it, then it's an optical illusion because the hexagons are really stretched vertically. Unlike some optical illusions of images that appear different than they are but are still mathematically possible, this is an optical illusion of a mathematically impossible image. The math shows that $$\tan^{-1} 60^\circ = \sqrt{3}$$ and $$\sqrt{3} < \frac{7}{4}$$ because $$7^2 = 49$$ but $$3 \times 4^2$$ = 48. It's just like it's mathematically impossible for something to not be moving when it is moving but it's theoretically possible for your eyes to stop sending movement signals to your brain and have you not see movement in something that is moving which would look creepy for those who have not experienced it because your brain could still tell by a more complex method than signals from the eyes that it actually is moving.
To draw a hexagonal grid over a square grid more accurately, only the math and not your eye signals can be trusted to help you do it accurately. The math shows that the continued fraction of $$\sqrt{3}$$ is [1; 1, 2, 1, 2, 1, 2, 1 ... which is less than $$\frac{7}{4}$$, not more.