1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3
1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2
1/2 + 1/4 + 1/8 + 1/16 + ... = 1
1 + 2 + 3 + ... + n = n * (n+1) / 2
1 + 3 + 5 + ... + (2n − 1) = n2
a2 + b2 = c2
Related posts:
Math visualization: (x + 1)2
Further reading:
Proof without Words: Exercises in Visual Thinking
Q.E.D.: Beauty in Mathematical Proof
Also, my thanks go out to fellow redditor cnk for improvements made to the 1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2 graphic.
69 comments:
> 1 + 3 + 5 + ... + 2 * (n − 1) = n^2
I believe this should be
1 + 3 + 5 + ... + (2n − 1) = n^2
Arnar,
You're right, I messed up the grouping on the last term. Thank you for the correction.
It took me a little while to see #6. I think it would help if you numbered or subtly color-coded the pairs of corresponding negative-space triangles.
That said, these are wonderful, brilliant proofs. Thanks for your effort. It was really neat to see!
(Meaning something like this.)
I love the elegance of these. I read in a book on learning recently that most of our brain is devoted to visual processing and only a small portion to language (i.e., symbol manipulation). I guess this might explain why visual proofs are so compelling. I suppose I admit a bias towards symbols when I say that I find it counter-intuitive that powers of 4 add up to 1/3, powers of 3 add to up 1/2, etc. Some sort of even/odd things I can't wrap my brain around. Any ideas on this? -- Barry
I don't get the "sum of first n integers" ... how does that image describe it?
@MaJJ The size of the entire rectangle is n by n-1 units. If you compare the empty space on the right side of the rectangle with the filled space on the left side of the rectangle, you can see that one is a perfect 180-degree rotation of the other. That is, exactly half of the rectangle (n * (n-1)) is filled. Does that help?
@ndef You mean "n by n+1", right? Then it would fit... Thanks for explaining this to me! :-)
@MaJJ Whoops! Yes, I meant "n by n+1" and "(n * (n+1))". Good catch!
@ndef At least I had to think about it harder :D
ndef,
I thought about labelling those triangles too, but then I noticed that all four of them are exactly equal, so it doesn't strictly matter how you assign them.
Also, thanks for answering questions while I was AFK today. I couldn't have explained it better myself. :)
Barry,
I think a lot of the bias you're talking about comes from the base 10 numbering system that we all grow up with. It's amazing what patterns emerge when you do something simple like change a sequence to binary notation, or any other base really. For example, someone on Hacker News pointed out that 1/2 + 1/4 + 1/8 ... is just a binary form of 9/10 + 9/100 + 9/1000 + ... = 1. The numbers are far from the same, but changing the first sequence to decimal notation reveals that it's really 0.1 + 0.01 + 0.001 + ... = 0.1111..., which is the binary equivalent of 0.9999..., which we know is the same as 1.
Using illustrations isn't exactly the same thing, but drawing a picture of a math or word problem is probably the oldest way of changing your frame of reference to gain new insight. Switching bases is just a different way of changing your frame of reference.
I realize now that I've rambled on a bit, and I hope I've answered your question. :) Thanks for reading.
Bill -- love your blog. I couldn't find you on Twitter and wanted to follow your blog (I pretty much use Twitter like a RSS reader these days), so I took the liberty of entering your Atom feed into TwitterFeed.com and outputted it to a new Twitter account: http://twitter.com/lizardbillblog.
Would love to give you the login info for that Twitter account if you want it...
Excellent post
yeeguy,
Thanks a lot for setting that up for me. I've been thinking about joining Twitter for awhile now, so I guess it's time to take the plunge. You can email me the account information at bill[dot]lizard[dot]blog[at]gmail[dot]com. (I wonder if this sort of obfuscation still fools address harvesting bots?)
Thanks again.
Hi there
An interesting post, but you're mistaken on the 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... = 1.
What is happening here is you are taking a half of the remaining space and adding it to what you already have. Logically, if you do this ad nauseum, you will still have something left over - the other half of what you have just added.
A picture is not ideal for a proof such as this, because you cant see, after a short iteration of the pattern, the remaining space. Not because the pattern equates to a whole, but because whats left is too small to see.
Yeah, but isn't that why the "+ ..." is there? If you're doing the sum _infinitely_, the remaining space gets _infinitely_ small, or am I wrong?
I guess it's something like 0.999... = 1. If there is _infinite_ amount of 9s, you can't find the difference.
I think this helped me see more with these few pictures to figure out all those puzzeling multiple choice questions than anything I have ever seen. Apparently I needed a visual to "get it". Even though according to some of the other comments, there maybe errors, it still did what you intended and made it click for me. Thank you.
MaJJ,
You're absolutely right. They key to the first three proofs is that "+ ..." at the end of a series indicates that it is an infinite series. Not all infinite series will converge on a discreet value, but this one does.
This particular case is exactly like the proof that 0.999... = 1, which is even given as an example in the Wikipedia article for that series.
Most people have a greater aptitude with visual memory. I mean, when you're doing even a simple addition mentally do you more often visualise a 24 and a 17 and add them or simply get the sum 41 by an uberfast by thought-processes too fast (through rigorous practice) to label as 'now i am imagining a numeral 17 in my head' ?
On the other hand, what if it was the product? Then we would keep 68 floating in our head before adding 35*10 to it to get the answer. So, in a way, we actually seem to deal with pictures of symbols of numbers in our heads than abstract concepts of pure numerals.
This might be a bit off track but the way the pictorial representation seemed to delight people could just be an effect of our fundamental pictorial thought processes. Also, the degree of visualising ease varies widely. I've known people to solve engineering mechanics problems entirely in the margins (no free body diagrams needed). Whereas I draw one almost every time.
Bill, great work! See what I do with young people with infinite series:
My website: http://www.mathman.biz/
Sample problems from my worksheet book:
Questions (ch. 1) http://www.mathman.biz/html/prob1.html
Answers http://www.mathman.biz/html/ans1.html
Infinite series = area of a triangle
http://www.mathman.biz/html/infser.html
Java applet (need GeoGebra)
http://www.mathman.biz/html/GGB_infsertri/
ggb_InfiniteSeriesTriangle.html
Alex (a second grader) sees an infinite series:
http://www.mathman.biz/html/alexs.html
Don Cohen-The Mathman
To those wondering about e.g. 1/2 + 1/4 + 1/8 + ... adding up to one: Technically we cannot add an infinite number of values. If we are strict, we can only talk about the limit of the finite partial sums as we add more and more terms.
The equation 1/2 + 1/4 + 1/8 + ... = 1 above is really just a shorthand for saying: As we add up the numbers 1/2, 1/4, 1/8 etc. the limit of the sum is 1, which in turn means that only by adding more terms, we can get as close to 1 as we like, although we can never reach it.
I guess what I'm saying is that anonymous was partly right there. Only after we define what a sum over an infinite set means (in our case, the limit) - we can write equations like those above.
Yea, your right, to find the limit of the infinite series, considering the series is equal to (1)\(2^N). It is a geometric series. it is basically equal to (1\2)^n. All u do is use the geometric formula to find the limit. which is a\(1-r) so it would be (1\2)\(1-(1\2)) which is equal to one therefore the limit is equal to one.
Arnar, are you Icelandic?
> 1 + 3 + 5 + ... + (2n - 1) = n^2 is still wrong.
Insert, say, 3 for example. 2 * 3 is 6, - 1 is 5. Now, unless I failed years of math, 3^2 = 9, and 5 does not equal 9.
Bill, there is just something off about that one. Sadly, I'm just not seeing it. From the picture even, the formula is flawed, because in the section for 3, the answer should be 4; not 5, and not 9.
Hæ Axel! Yub, I am Icelandic :)
>> 1 + 3 + 5 + ... + (2n - 1) = n^2 is still wrong.
>Insert, say, 3 for example. 2 * 3 is 6, - 1 is 5. Now, unless I failed years of math, 3^2 = 9, and 5 does not equal 9.
You are forgetting the 1 and the 3. If you insert n=3 you get (2*1 -1) + (2*2 - 1) + (2*3 -1) = 1 + 3 + 5 which indeed equals 3^2 = 9.
Anon,
Arnar hit it right on the head. If n = 3, then the last number in the sequence is (2n - 1) = 5, and the whole sequence is 1 + 3 + 5 = 9.
What may be causing confusion is that the first few proofs involve infinite series, but this sequence terminates.
Bill t L and all. This is wonderful. I am a non mathematician who late in life discovered I really love math. If math had been taught this way to me, who knows what worlds would have opened.
Anon, references a book re how the brain processes visual information vs. text or symbols. I am writing a paper RIGHT now on why we need more visual imagery in world agricultural training. A reference (from anyone) would be helpful. There are over 4800 living languages in the world and most textbooks/training manuals are written in just a tiny fraction of those languages. The FAO uses just 6. Is there any wonder that technological advances are mostly confined to less than 20-30% of the worlds 195 +/- nations?
I love your proofs. All "debunking" is a good thing, though the Maths Purists seem reluctant to take these as proofs.
Not visual, but explanatory thought on Achilles and the tortoise:
Scale 10, 10 = 9.999... = 9 +9/10 + 9/100.....
Scale 2, 2 = 1.1111.... = 1 +1/2 +1/4 +1/8...
Hence the pyramids.
Fred
Biocides.LT,
I can't really give a good reference on the specific topic of visual imagery in world agricultural training, but I can point you to a few really good books on better approaches to teaching math. Innumeracy and Beyond Numeracy, both by John Allen Paulos, and How to Solve It by Polya all had a profound effect on how I think about math. Good luck with your paper.
FredStotle,
I would have to side with the purists and say that these aren't rigorous proofs at all, but visual evidence that the backing proofs are correct.
I just thought "Six Visual Proofs" made a much catchier title than "Some Very Convincing Visual Evidence That the Proofs of Six Mathematical Formulas Are In Fact Correct." :)
Biocides LT.
You could try University of Reading(UK!), Department of Agricultural Development. I retired sometime ago so don't know exactly what they are up to these days, but they should know, if anyone does!
Fred
Don Cohen,
I just got a chance to visit all the links you provided, and I really like your site. I'm very impressed with Alex for illustrating an infinite series at such a young age! In second grade most of us (in the US) were just leaning our multiplication tables. Congratulations to his teacher!
Barry,
I don't know if you're still wondering about the even odd thing, but the way I see it, you split the whole into a certain number of equal parts (say 4), and 'color' one (1/4) and then you take one of them (leaving 1/3 colored) and repeat the process. The you would then take one of those 4 and so on. This means that 1/n + 1/n^2 + 1/n^3 + ... will always equal 1/(n-1).
Re 1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2
I appreciate the value of this effort in a general sense. But it's a definite no go on the visual for that one.
If we state each of these portions in terms of the available LCD we get 27/81 + 9/81 + 3/81 + 1/81 + ... = 40/81. So even after just a few terms, we're clearly approaching 1/2, and are plainly left to accept that the formula itself is a correct one.
The accompanying visual however does not match this formula accurately. I should think this obvious. Brought to completion, the visible results (represented in light green) do not consume half of the square. To be visually accurate the combined results would have to fill half of the square, even if that half is along the diagonal. It does not. So it can't possibly be accurate.
To spot the breakdown, take note of the difference between the 1/81 and the 1/27 as they're laid out here. The 1/81 is presented as a square. And since we know that 1/27 is precisely three times larger than 1/81, we surmise that the 1/27 should be large enough to house precisely three times the area attributed to 1/81 or, if you will, three such squares. It is not. It plainly accomodates only two-and-a-half of them.
Needless to say this occurs at each level of the diagram, starting with 1/9 and 1/3. The formula's correct, but the corresponding visual is not true to the formula.
Tom Raywood
A couple of points. Using numbers 1 and 3, does this mean that 1/8 + 1/32 + 1/128 + 1/512...= 1/6?
Secondly 0.999 recurring is the same as 1. It can be proved by the 10n-n=9n equation where substituting 0.999 (recurring) for n this would give us 9.999-0.999=9 meaning 9n=9 and n=1
Just something else to think about
Bob C
Thomas,
It's not intended that the light green elements consume half the square, but all the green ones do. The dark green triangles in the upper right corner of each green rectangle is part of the rectangle. These elements and the diagonal line bisecting the square were added as a result of a discussion on reddit.
After reading your comments, I'm thinking of removing the differently colored triangle elements and just making the diagonal a dotted line. I think this will be the best compromise between the current image and the original one.
Bill the Lizard,
Nice work. Owing [no doubt] to a dash of obtuseness on my part, the way you've explained it makes it now perfectly clear. I'd have to say then that the visual at its core level is a perfectly good one. The diagonal line is bound to throw a dog or two off the hunt but, I suspect, an equal number of viewers may have needed an additional line of reckoning so to speak. I assume that's why the diagonal was added. It may be that a dotted line is a good compromise.
Another option, perhaps, is to work with a circle rather than a square. Section it off in thirds, a la Mercedes Benz, and rotate it so that one radius is at zero degrees. From there it's easy to visualize how 1/81 beside 1/27 beside 1/9 beside 1/3 approaches 1/2 at the 180 degree position or, more to the point, as it approaches the implied diameter of the circle.
Tom Raywood
Thomas,
If you (and many others) needed the written explanation, then the graphic just wasn't doing its job. My whole goal with this post is to show the truth of these equations without using any words at all. I really appreciate your taking the time to explain in detail exactly what wasn't working for you.
What wasn't working in the original graphic (just as it is now, only without the diagonal) was that many people were having trouble seeing how the green blocks added up to exactly half the area of the whole. As you said, hopefully the dotted diagonal line is a good compromise.
Using a circle to show sections that add up to 1/2 is a great idea. I'll draw it up soon, but I'll probably save it for a future post along similar lines. Thanks for sharing your ideas and your critiques.
By the way, there is a proof for the first three. They're geometric sequences.
In the first one, the first term (a) is 1/4 and the ratio (r) is 1/4. Using a/1-r you get 1/4 divided by 3/4, which equals 1/3.
In the second, r=1/3 and a=1/3. 1/3 divided by 2/3 is 1/2.
In the third, r=1/2 and a=1/2. 1/2 divided by 1/2 is 1.
I noticed that no matter what fraction you start with as a, if r=a than the sum will be a with the denominator reduced by one.
Anon,
Right on both counts.
1/8 + 1/32 + 1/128 + 1/512... does converge to 1/6. This may also show up in a future post (see my reply to Thomas above). However, to follow the general form of the others, it should be 1/(8^1) + 1/(8^2) + 1/(8^3) + ... which is the same as 1/8 + 1/64 + 1/512 + ..., and converges to 1/7.
0.999... = 1 has been hotly debated on the internet for years, but it's long been accepted by mathematicians that the two represent the same real number.
Avram,
"I noticed that no matter what fraction you start with as a, if r=a than the sum will be a with the denominator reduced by one."
Good observation. It's possible to make an infinite number of series of this form, that all converge to a different rational number. Fascinating.
Bill the Lizard,
Very well then. Grab this to use if you'd like:
http://4.bp.blogspot.com/_9jC8jB1_lgg/SnItK3jgs_I/AAAAAAAAAAs/WCLNOxFY05E/S220-h/unit+circle.png
Tom Raywood
great post!!!
What !!
Bob C,
Yes 1/8 +1/32 +1/128... but this is the same as 1/4 +1/16 + 1/64 divided by two. Since that converges to 1/3, then the same number divided by two, or 1/6.
I love these visual models. About 10-12 years ago, when I was homeschooling my youngest. We had wonderful worksheets that were similar to the 3rd example. It was a great way to really drive home how fractions work.
How pleased am I that she is now pursuing math in grad school!
Very nice.
Proofs 1-5 are accurate. Anyone disagreeing is just not misunderstanding something.
I do not see how 6 displays the proof however, but have seen another example using tiles), with a=3x3 (9 squares), b=4x4 (16 squares) and c=5x5 (25 squares).
I love how proof 1 clearly shows that the green area is 1/3 from the top left, bottom left and bottom right squares, with the same being true for the one left over and so on.
The confusion which some people are having with number 5 is that they are confusing the number (from the X axis of the graph) with n, where n is in fact the number minus 1 and divided by two.
ePlanner,
#6 proves the Pythagorean theorem by showing that the square of side c (the bigger, dark green square) is equal in area to the sum of the two smaller, lighter green squares. It does this by inscribing them in squares of equal size, and showing that the area left over (the four gray triangles) are equal.
There are literally hundreds of published proofs of the Pythagorean theorem, but this is my favorite due to its simplicity.
As for #5, I was hoping that the numbers at the bottom would help by associating the number of dots in each channel with the numbers 1 + 3 + 5 = ... in the formula. Maybe I should have flipped the image so those numbers were at the top?
#6) Yeah true. I just think the tile one leaves no doubt as to the size of each area, where as one could believe that the grey triangles in both the bottom images are different in size).
With #5, the formula is of course 100% correct, but I believe the confusion comes from people having to do a calculation in the heads to figure out the value of n (i.e. n = (final number + 1)/2). Instead, they are mistaking n as being the final number (as it is usually represented as such).
This image http://img217.imageshack.us/i/proofx.png/ uses a formula that has this step worked out. I also added a few rows at the bottom for further clarity, but these are not required.
Cheers for the interesting topic, been awhile since I have looked at this sort of stuff and #5 even confused me for a bit.
ePlanner,
I see your point about #6. I wonder if I should have labeled the a and b squares in the bottom left figure? At any rate, I started with the top figure and copied/pasted elements from it to construct the bottom two, so I can promise that the triangles are all equal. :)
On #5, it did seem to be that fact that n was not the last number in the sequence that confused a few people. Moving the calculations over to the RHS does seem to improve the clarity. Thanks for pointing this out. I'll keep it in mind next time I create a similar post.
while they are interesting, i must say they are all wrong. these all work off the assumption that 0.99999... means that the nines go on forever. they never end and they are "infinite".
however you cannot say that 0.99999... = 1, because there is no way that something infinite(0.999...) can equal something finite. one (1) is finite. we accept it as a real number, a rational number, and a whole number. the number one has a definite end and therefore cannot equal something that has no end
Anonymous,
No, you're wrong about that. 0.999... is definitely equal to 1. This has been accepted by professional mathematicians for a very long time, and is taught in text books. There are many proofs of it here.
Can anyone shed some light on the first one? I just don't see how you can see 1/3 in the green (or 2/3 in the grey)?
Bob: When looking at the first iteration (four squares, each with area 1/4), try to ignore the right-top one. It will be dealt with in the second iteration. Now you see that the green square takes 1/3 of the three squares. And it's like that with all the other iterations.
(I hope I wrote it understandably :) )
Bob T. Builder,
Just to add to MaJJ's explanation, there are exactly 3 of each size of the smaller squares. One of each size is shaded green, so 1/3 of the whole figure is green, and the remaining 2/3 is grey.
Another way of looking at it is if you focus on the green squares, look to the right of each one. The grey square to the right of each green square is the same size. Now look directly up from each green square and you'll notice the same thing. So the area to the right of the green squares and the area above the green squares are each equal to the green area.
Ah excellent, I get it now - both ways! Cheers MaJJ and Bill the Lizard.
The 6th visual proof is simpler. Einstein used a single line from the hypotenuse making two other, enclosed, right triangles. The area of each of the three triangles (two inner, one outer) is proportional to the square of its hypotenuse. It is the same proportion for all three similar triangles.
Visually it can then be seen that a^2+b^2=c^2.
This visual proof would require the "visualization" of the equation
kA1^2+kA2^2=kA3^2
where A is an area, and k the proportional constant.
This comes from the author Manfred Schroeder ("Fractals, Chaos, Power Laws"), who got it from Schneior who got if from Ernst Strauss, who got it from Albert who came up with it during a home schooling session with his Uncle Jacob.
Can you visualize a proof of the fact that
sum of reciprocals of squares of the positive integers equals pi squared by 6?
After looking at the fourth proof I set up a display of a triangle of golf balls. I wanted to know if this proof stays true when you increase the numbers. Even with replete rows and columns the proof stayed true. I was hoping you could further help my understanding as to the reasons behind the logic of the fourth proof.
Hi Josh, thanks for commenting. There's an old story about Gauss that might shed some light on the fourth proof. (I'm quoting from my own post about Gauss, but this story has been retold hundreds of times.)
"When Gauss was in school at age 10, his teacher, perhaps needing a quiet half hour, gave his class the problem of adding all the integers from 1 to 100. Gauss immediately wrote down the answer on his slate (this would have been in the year 1787). He had quickly noticed that the sum (1 + 2 + 3 + ... + 100) could be rearranged to form the pairs (1 + 100) + (2 + 99) + (3 + 98) + ... + (50 + 51), and that there were 50 pairs each equaling 101. This reduced the problem to the simple product 50 x 101 = 5050."
The fourth proof here is illustrating the same idea. It's just arranging the numbers in a different way than Gauss did over 200 years ago.
I hope that helps.
Thank you! That was extremely helpful. I am new to blogging but find it exciting and useful. I recently uploaded your proof to my blog and used it to help explain the problem.
I was wondering if you would take a look and possibly give me some pointers about a successful blog. Your's was easy to use and informational. I would like to improve mine and would appreciate any suggestions. Thank you in advance for your time.
Thank You,
Josh Koerner
Josh Koerner,
I love the golf ball photo you used. You can really make an abstract concept more accessible by 1) showing a picture of it, and 2) putting it in terms of a real-world problem.
As for blogging advice, a lot of people will tell you to write what you know. This is fine advice, but I tend to take a slightly different approach. I like to write about things as I'm learning them. If I can explain a concept good enough to put it in writing, I feel like I've come to a pretty good understanding of it.
I hope that's helpful, and good luck with your blog.
more evidence that the universe is an integer truth value of existence that's dividing over time into 'smallest discernably discreet parts' of universe...
...and all the 'smallest discernably discreet parts' add up to make-and-occupy-positions-within the qcd lattice... all derivable from number theory...
why not?
:P
See Alex's work, as a 3rd grader, showing 1/3+1/9+1/27...->1/2 Below his work on 1/2+1/4+... at http://www.mathman.biz/html/alexs.html. He said 1/3 + 1/2of1/3 = 1/2 which I didn't see until I looked carefully at his picture.
Don Cohen-The Mathman
The 1/2 + 1/4 + 1/8 + 1/16 + .. = the sum of 1/2^n - 1, where n ranges from 0 to infinity. By basic analysis, this is simply a geometric series, sum of r^n, where r = 1/2, therefore it is convergent, and the answer is simply derived either by knowing the geometric series theorem or by using taylor series expansion around 0, to get 1/(1-r) - 1 ( remember we had to subtract one for the series to work out right.) therefore you arrive at 2 - 1 = 1, the desired answer.
Really, really nice post!
I couldn't see #6 until I saw the image of ndef. So that color shading would indeed help.
Post a Comment