# What was the first bit of mathematics that made you realize that math is beautiful? (For children's book)

I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of Mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.

I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.

However, I myself am mathematically unsophisticated.

What was the first bit of mathematics that made you realize that math is beautiful?

For the purposes of this children's book, accessible answers would be appreciated.

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For me Euclid's proof of the infinitude of primes was the first thing that made me realize the beauty of mathematics. –  Manjil P. Saikia Mar 7 '13 at 7:02
Wow. Just last night I had a fierce argument with one of the bartenders of my usual watering hole who is a mechanical engineering student. He insisted that he has a better idea than me of what is mathematics. I am so going to print him a copy of Lockhart's text. Thank you for that link! –  Asaf Karagila Mar 7 '13 at 7:59
I can’t remember a time when I didn’t think that mathematics was beautiful and fascinating. –  Brian M. Scott Mar 7 '13 at 15:06
Although I don't know if it's what you are looking for, try looking up "vihart" on youtube--Even if it's not helpful, I guarantee you will appreciate it. –  Bill K Mar 8 '13 at 2:57
I think it's a shame that this question was voted closed... –  Will Mar 10 '13 at 19:50

Parallel lines. I was amazed to find out that they would never, ever meet.

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At about 10 or 11 I discovered that the area of a circle was half the circumference multiplied by the radius.

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The number of pennies stacked in a triangle $(1,3,6,10,\cdots)$ is along one diagonal line of Pascal's Triangle. The number of spheres stacked in a tetrahedron $(1,4,10,20,\cdots)$ is the line next to it. The next line is the number of hyperspheres in a pentachoron.

I was about $10$ and living in a hotel and home sick from school, stacking up pennies and "red hots" in pyramids, etc. I made a table of these numbers. Noticing the simple addition rule in the table, I extrapolated to the $4$th, $5$th, dimensions. Later when I learned of Pascal's triangle that moment was probably my biggest joy of mathematics, realizing I'd run into that years before.

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Back in School i noticed that each row of pascals triangle is a multiple of 11: 11^0, 11^1, 11^2 etc. –  Nils Werner Mar 8 '13 at 12:14
@vermiculus: I think what Nils Werner is saying is that the $n$th row consists of the numbers $\binom{n}{k}$, and that $\sum \binom{n}{k} 10^k = (10 + 1)^n = 11^n$. Of course some of the $\binom{n}{k}$ will have more than one digit. –  Jesse Madnick Mar 9 '13 at 22:25
Mark odds and evens in Pascal's triangle and you get something nice :) –  Manishearth Mar 10 '13 at 7:27
@vermiculus Shift multiples of 10 one position to the left, so 1 5 10 10 5 1 becomes 1 6 1 0 5 1 –  Nils Werner Mar 11 '13 at 9:01

For me it was the Times Table of $9$.

We are usually forced to memorize the multiplication tables in school. I remember looking at the table for $9$, and seeing that the digit in ten's place increased by one, while the digit in the one's place decreased by one.

$$\begin{array}{r|r} \times & 9 \\ \hline 1 & 9 \\ 2 & 18 \\ 3 & 27 \\ 4 & 36 \\ 5 & 45 \\ 6 & 54 \\ 7 & 63 \\ 8 & 72 \\ 9 & 81 \\ 10 & 90 \end{array}$$

After this, I realized that I could always add $10$ and subtract $1$ to get the next result. For a $7$ year old, this was the greatest discovery ever made.

And that your hands could give you the answer immediately: $7 \times 9$ = hold down your $7$th finger, leaves $6$ fingers on left of held down finger, and $3$ on right: $63$.. works all the way up to $9\times10$, beautiful.

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Fo me, it was the "flipping" of the digits along 9's times table. (e.g. $9 \times 9 = 81$ and $9 \times 2 = 18$ or $9 \times 3 = 27$ and $9 \times 8 = 72$. –  paraxor Mar 7 '13 at 11:50
Reminds me of when I realized that "skip counting" (like $3, 6, 9, 12, \ldots$) is the distributive law. –  Jesse Madnick Mar 7 '13 at 12:05
Add 10 and subtract 1 to get the result. I like that :) –  0a -archy Mar 7 '13 at 16:22
Also, both digits of the result, sum 9. (18, 27, 36...) –  Francisco R Mar 7 '13 at 17:13
I love this example, because it shows how the abstract side — finding patterns, understanding the reasons for them — arises directly out of the concrete side. Too often (as in the Lockhart) the two sides are presented as in opposition to each other — really, they’re intimately connected the whole way. –  Peter LeFanu Lumsdaine Mar 7 '13 at 19:53

I used to love naughty $37$.

$37 \times 3 = 111;$

$37 \times 6 = 222;$

$37 \times 9 = 333;$

$37 \times 12 = 444;$

$37 \times 15 = 555;$

$37 \times 18 = 666;$

$37 \times 21 = 777;$

$37 \times 24 = 888;$

$37 \times 27 = 999;$

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That is naughty! –  Loki Clock Mar 7 '13 at 12:57
But, there isn't a naught in any of that... –  Gary S. Weaver Mar 7 '13 at 14:05
Wow - 37 is my new favourite number! –  Steve Chambers Mar 8 '13 at 11:21
If you like 37, you'll love double-37: 37037*3=111111, 37037*6=222222, 37037*9=33333... –  MiniQuark Mar 8 '13 at 14:00
And in general any repetition of 037. For the infinite case, take 1/27. –  Joe Z. Mar 8 '13 at 20:05

Beremiz, an Arab mathematician, arrives on foot to a bedouin camp, accompanied by a friend that rides a camel, only to find three boys having a dispute about the testament of their recently deceased father. In the will, the father had left his herd of camels to be divided among his sons, in the following proportion: the older son should receive half of the camels, the middle son should receive a third of the camels, the younger son should receive one in each nine camels. The problem is that his herd was composed of 35 camels... So, the boys reckoned that, in order not to fail their father's instructions, they would have to reserve at most 17 camels to the older one, at most 11 camels to the middle one, and at most 3 camels to the younger one. This looked bad, not only because the divisions all left remainders, but because in fact there would still be a few camels attributed to no one after the three sons received their share! To help them find a fair solution to their predicament, Beremiz then asks his friend to give his own camel to the brothers, a suggestion to which his friend reluctantly assents. As a result, the division now goes smoother, and the three boys are happier, having respectively received 18, 12 and 4 camels. They are so happy, indeed, that they give the 2 remaining camels to Beremiz and his friend!

I once had to teach young children (and some not so young!) to sum fractions, and it was easy to generate a whole set of problems based on the same approach (a collection of fractions that do not sum to 1). I believe this will give rise to straightforward illustrations for your book, Liz.

Malba Tahan's book "The Man Who Counted" may be claimed to be the way in which any Brazilian younger than 80 has first gotten in touch with the beauty of math. Its romanticized recreational presentation makes it easily accessible and great fun for children already in their early school years. "Splitting 35 camels" is the first mathematical problem which the readers are confronted with, in this book.

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Realising why Zero is not Nothing, and Understanding Numbers

I first understood the difference between zero and none when thinking about thermometer readings. If you had a ton of thermometers scattered around the world, and you collected their readings periodically and put them in a database, what would you do if any thermometer was broken? If you just put a zero reading, you'll screw up your averages, but if you put a null value, you can handle broken thermometers easily.

That made me realise what a number is.

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One of my most memorable moments in mathematics was when I was attempting to prove the formula for the volume of a sphere on my own. I hadn't been taught calculus yet and had no idea about it, but I was convinced I could solve the problem. I used an infinite amount of small disks and added their volume ( essentially the limit of a riemann sum, an integral, but I didn' know that at the time) I made the disks a certain height, worked out the sum using sums of consecutive squares and then made the height equal zero. And voila, I got the right volume! Later I found out I had re-discovered a part of calculus. The realisation that different people can independently discover mathematical truths and techniques was beautiful to me.

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Tupper's self-referential formula is a self-referential formula defined by Jeff Tupper that, when graphed in two dimensions, can visually reproduce the formula itself. It is used in various math and computer science courses as an exercise in graphing formulae.

The formula was first published in his 2001 SIGGRAPH paper that discusses methods related to the GrafEq formula-graphing program he developed.

The formula is an inequality defined by: $${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$$ where $\lfloor \cdot \rfloor$ denotes the floor function and $\mathrm{mod}$ is the modulo operation.

Let k equal the following 543-digit integer:

960 939 379 918 958 884 971 672 962 127 852 754 715 004 339 660 129 306 651 505 519 271 702 802 395 266 424 689 642 842 174 350 718 121 267 153 782 770 623 355 993 237 280 874 144 307 891 325 963 941 337 723 487 857 735 749 823 926 629 715 517 173 716 995 165 232 890 538 221 612 403 238 855 866 184 013 235 585 136 048 828 693 337 902 491 454 229 288 667 081 096 184 496 091 705 183 454 067 827 731 551 705 405 381 627 380 967 602 565 625 016 981 482 083 418 783 163 849 115 590 225 610 003 652 351 370 343 874 461 848 378 737 238 198 224 849 863 465 033 159 410 054 974 700 593 138 339 226 497 249 461 751 545 728 366 702 369 745 461 014 655 997 933 798 537 483 143 786 841 806 593 422 227 898 388 722 980 000 748 404 719

If one graphs the set of points $(x, y)$ in $0 \le x < 106$ and $k \le y < k + 17$ satisfying the inequality given above, the resulting graph looks like this (note that the axes in this plot have been reversed, otherwise the picture comes out upside-down):

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The $\sqrt{-1}$, complex analysis, and how real world problems could be solved "by these objects which don't exist."

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The fact that Gabriel's horn has infinite surface area, but finite volume, hence you can "fill it with paint, but you can never cover the whole surface".

Gabriel's horn (also called Torricelli's trumpet) is the graph of $y =1/x$ for $x\geq 1$ rotated around the $x$ axis.

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The realization that you can go on counting forever.

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This is a new interesting 4x4 Magic Sqaure, which I believe will be interesting to School Children. Here each element of the square is a square number. This was provided by Dr. Geoffrey Campbell

509020 is the sum of rows and columns

  29^2 |  191^2 |  673^2 |  137^2 || 509020
-------+--------+--------+--------++--------
71^2 |  647^2 |  139^2 |  257^2 || 509020
-------+--------+--------+--------++--------
277^2 |  211^2 |  163^2 |  601^2 || 509020
-------+--------+--------+--------++--------
653^2 |   97^2 |  101^2 |  251^2 || 509020
=======+========+========+========++--------
509020 | 509020 | 509020 | 509020 || 509020

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A story which I heard when I was in Primary School motivated me to understand the Power of Exponentials. The story goes like this--..........A Brahmin Priest presented the King with the Chess Board and explained to him how to pay War Games on this board . The King was pleased and asked the priest , what he wanted as a reward ..........The priest asked the King , that as he was very poor ,he needed some grains to feed his family .He asked the King to put one grain of rice in the first square of the chess board. , then put two grains in the second square ,four grains in the third square----and continue this way doubling the number of grains in the next square --till he reaches the end--the 64th square ."I will take whatever grains are there on the Chess Board..that will be sufficient for my needs"..said the Brahmin...........The king tried to satisfy the needs of the Brahmin ,but soon found out that all the grains in the Kingdom will still fall short of his needs ....... . The King was pleased with the priest's intelligence and appointed him as the Royal Astronomer & Astrologer .

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As a child, the Fibonacci numbers $$1,\; 1,\; 2,\; 3,\; 5,\; 8,\; 13,\; 21,\; 34,\; 55,\;\ldots$$ were very fascinating to me. They are named after the the Italian mathematician Fibonacci, who described these numbers in his 1202 book Liber abaci modeling a growing rabbit population:

Formally, the Fibonacci numbers $F_n$ are defined recursively by $$F_1 = 1, \quad F_2 = 1, \quad F_{n+2} = F_{n+1} + F_n$$ It was a lot of fun to compute them, one after the other, and to collect the results in ever-growing tables: $$F_3 = F_2 + F_1 = 1 + 1 = \mathbf{2}\\F_4 = F_3 + F_2 = 2 + 1 = \mathbf{3}\\F_5 = F_4 + F_3 = 3 + 2 = \mathbf{5}\\F_6 = F_5 + F_4 = 5 + 3 = \mathbf{8}\\F_7 = F_6 + F_5 = 8 + 5 = \mathbf{13}\\\vdots$$

At some point, I asked myself the question: To compute $F_{10}$, do I really have to compute all the Fibonacci numbers up to $F_9$ beforehand? So I tried to figure out some formula where you can plug in $n$, do some basic arithmetics, and get $F_n$ as a result. I've spent a lot of time on this. However no matter how hard I tried, I didn't succeed.

After a while I found the closed form $$F_n = \frac{1}{\sqrt{5}} \left(\left(\frac{1 + \sqrt{5}}2\right)^{\!n} - \left(\frac{1 - \sqrt{5}}{2}\right)^{\!n}\right)$$ in some book. I was paralyzed.

How can it happen that such an easy recurrence formula needs to be described by such a complex expression? Where do the square roots come from, and why does the expression still always evaluate to an integer in the end? And, most importantly: How on earth can one find such a formula??

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I first discovered the Fibonacci sequence by playing with an 8-digit calculator. There's a bug in most standard 8-digit calculators (the kind you get as promotional gifts, for example) so that if you press "1 + = + = + = ...", it will give you the Fibonacci sequence. –  Joe Z. Mar 7 '13 at 21:16
In order to find this formula, you usually will use linear algebra, and the property that linear recurrences can always be represented in matrix form. In particular, the two exponential terms correspond to the eigenvalues of the matrix $\left [ \begin{matrix}1 & 1 \\ 1 & 0\end{matrix} \right ]$, which, as you could probably guess, has a characteristic polynomial of $x^2 - x - 1$. –  Joe Z. Mar 7 '13 at 21:23
JoeZeng: Wow, that's a nice bug-abuse :) Nowadays I know, of course, how to derive the formula. Besides the linear algebra method you indicated (my favorite), there is also a standard way to do it by generating functions. –  azimut Mar 7 '13 at 21:32
There's a way to derive this using only plain algebra, which would be nicer for a children's book. Let a = (1+sqrt(5))/2 and b = (1-sqrt(5))/2. Show that a^2 = a + 1 (and similarly for b). Multiply by a and simplify to get a^3 = 2a + 1, and show that you can repeat this to get a^n = F(n)*a + F(n-1) (and similarly for b). Take a^n - b^n, simplify and rearrange and you're done. –  Michael Shaw Mar 8 '13 at 5:55
A bug? I thought it was a feature? –  Raskolnikov Mar 15 '13 at 20:56

Maybe the fact that the homotopy category of a model category is equivalent to the full subcategory of fibrant-cofibrant objects with homotopy classes of morphisms.

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Congratulations for finding that out as a child. –  azimut Jun 21 at 13:26
Dry humor? (Hope so.) –  Did Aug 21 at 10:10

Not an experience of mine, but I'm currently reading The Greeks by H. D. F. Kitto and I think this page deserves to be here:

But let us not be too superior to those Greeks who "shut their eyes." They kept something else wide open, namely their minds, and although the eye-shutting retarded the growth of science, the mind-opening led to things perhaps equally important, metaphysics and mathematics.

Mathematics are perhaps the most characteristic of all the Greek discoveries, and the one that excited them most. We shall be more understanding of those who shut their eyes to facts if first of all we keep in mind the Greek conviction that the Universe is a logical whole, and therefore simple (despite appearances) and probably symmetrical, and then try to imagine the impact of their minds on elementary mathematics.

It happens that I myself—if I may be personal for a moment—was enabled to do this by an insomnia-beguiling piece of mathematical research that I once did myself. (Mathematical readers are permitted to smile.) It occurred to me to wonder what was the difference between the square of a number and the product of its next-door neighbors. $10 \times 10$ proved to be $100$, and $11 \times 9 = 99$—one less. It was interesting to find that the difference between $6 \times 6$ and $7 \times 5$ was just the same, and with growing excitement I discovered, and algebraically proved, the law that this product must always be one less than the square. The next step was to consider the behavior of next-door neighbors but one, and it was with great delight that I disclosed to myself a whole system of numerical behavior of which my mathematical teachers had left me (I am glad to say) in complete ignorance. With increasing wonder I worked out the series to $10 \times 10 = 100$; $9 \times 11 = 99$; $8 \times 12 = 96$; $7 \times 13 = 91$… and found that the differences were, successively, $1, 3, 5, 7, \ldots$, the odd-number series. Even more marvelous was the discovery that if each successive product is subtracted from the original $100$, there is produced the series $1, 4, 9, 16, \ldots$. They had never told me, and I had never suspected, that Numbers play these grave and beautiful games with each other, from everlasting to everlasting, independently (apparently) of time, space, and the human mind. It was an impressive peep into a new and perfect universe.

(original source image)

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Draw any triangle. On each side of the triangle, draw an equilateral triangle such that the new equilateral triangle shares a side with the original triangle. Connect the midpoints of your three new triangles - the result is another equilateral triangle!

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My first think of infinity was going from one corner of square to opposite corner. Start with route (0,0) -- (0,1) -- (1,1), then take (0,0) -- (0,½) -- (½,½) -- (1,½) -- (1,1) and so on. Path will come visually closer to diagonal, but lenght will stay at 2.

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I first discovered that math was beautiful upon learning the divisibility rules. At that point I was just like "IT WORKS IT WORKS! HOW DID PEOPLE KNOW THAT?!" I remember I once stayed up to test the divisibility rule of dividing by $8$ (if the last three numbers in the dividend are divisible by $8$ then the whole number is divisible by $8$).

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I must have been very small, around three of four, when I suddenly dashed out of my room, full of excitement, wanting to show my dad something that had made a great impression on me.

I held a book, it's front cover facing me.

In a flash, I gave it two half-turns. One upside-down, the other left to right. This is what came out:

I held my breath, as the trick wasn't over yet. Sure enough, two same quick moves and -lo and behold- the front cover was facing me properly again, just as in the beginning.

That surely must have been my first conscious encounter with symmetry.

I held the memory dearly close for a number of years but then forgot about it completely. It came back to me, only very recently, after going through the first pages of Nathan Carter's Visual Group Theory and seeing this image:

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I have to admit that although I'd frequently been told that mathematics was "beautiful", I didn't really get that while I was in school - even high school. I enjoyed mathematics, and saw plenty of things that were fun, and even cute, but I never really understood any ideas with sufficient depth to think of them as beautiful.

When I did encounter ideas that I found beautiful, it was in my first year at college. In fact, there were two closely related ideas in quick succession. We were just being introduced to vector spaces. This was the first time I'd seen an abstract space, but it didn't really seem to mean much except as a fancy way to talk about high dimensional Euclidean spaces.

But then I saw my first example of a vector space that didn't just look like the vectors I'd seen in high school. It was the space of infinitely differentiable functions: $C^{\infty}$. We were shown the linear operators associated with two common differential equations (exponential growth and simple harmonic motion): \begin{eqnarray} &\frac{\textrm{d}\phantom{y}}{\textrm{d}t} - kI \\ &\frac{\textrm{d}^{2}\phantom{y}}{\textrm{dt}^2} + kI. \end{eqnarray} We saw the fairly routine proofs that these were linear operators on $C^{\infty}$, but then came the magical part: The solution sets to these differential equations were subspaces of $C^{\infty}$, the canonical solutions I was familiar with were basis sets for these solution spaces, and the solution spaces were actually the nullspaces of these operators!

Later (maybe even in that same lecture) we saw how linear regression - the hitherto tedious process of finding the "line of best fit" - could be understood as a linear projection $P$ operator onto a two dimensional subspace of the data space. Given a data vector $\mathbf{x}$, the projected vector $P\mathbf{x}$ represented the line that was closest to the data - the line of best fit - and the difference term $\mathbf{x}-P\mathbf{x}$ represented the error term. I was astonished at how much more elegant this was than the clunky formulas I'd had to memorize in high school.

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The fact that $\Bbb C$ is algebraically closed.

About 12 years old, after I just learned about quadratic equation such as $x^2=a$ may or may not have solutions, my mother told me about complex numbers: you attach the number $i=\sqrt{-1}$ to real numbers and after that $x^2=-1$ have solutions.

"Nah", I said, "that doesn't help much: although you now have solutions for $x^2=a$ for $a$ in the old number system, which are reals, you still don't have solutions of $x^2=a$ in the new number system, which are complex. You still don't have a solution of $x^2=i$, for example. And having complex solutions for some of the complex numbers is no better than having real solutions for some of the real numbers."

Then she showed me the roots of $x^2=i$ and explained that $x^2=a$ has complex solutions for any complex $a$. The ingenuity of the complex numbers impressed me a lot.

Then she told me about polynomials of degree higher than 2 and that they all have roots in the same field of complex numbers, that you don't need to "attach" $n^{th}$ root of $-1$ or any other number in order to have any polynomial of degree $n$ have roots, that $\sqrt{-1}$ is sufficient for them all. And I was impressed even further.

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The fact that you can add natural numbers successively in the order you prefer and that you can split subtraction:
...I remember that day when got taught this in class which made me really excited so I had to tell my Mum =D

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For me, it was the beauty of the number 1, how it can be multiplied with anything , and it won't change the number it is being multiplied with, also how it can be represented as any number divided by itself such as 4/4=1 I would also love to share this beautiful poem by Dave Feinberg that is titled "the square root of 3" and was also featured in a Harold and Kumar Movie, it renewed my love for math and is and always has been one of my favorite poems! :

I’m sure that I will always be A lonely number like root three

The three is all that’s good and right, Why must my three keep out of sight Beneath the vicious square root sign, I wish instead I were a nine

For nine could thwart this evil trick, with just some quick arithmetic

I know I’ll never see the sun, as 1.7321 Such is my reality, a sad irrationality

When hark! What is this I see, Another square root of a three

As quietly co-waltzing by, Together now we multiply To form a number we prefer, Rejoicing as an integer

We break free from our mortal bonds With the wave of magic wands

Our square root signs become unglued Your love for me has been renewed

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Here is my favorite classic illustrating the power and beauty of mathematical argument. Consider the question:

Question: Can an irrational number raised to an irrational number be rational?

Answer: One of the classic answer goes as follows. Consider the number $x=\sqrt{2}^\sqrt{2}$. If $x$ is rational, we are done. If $x$ is irrational, then consider $x^{\sqrt2}$, which is $2$ and now we are done.

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I love those "anyway" proofs. –  Joe Z. Mar 8 '13 at 1:00
The really amusing thing is that after the proof is given, we still don't know whether $\sqrt{2}^{\sqrt{2}}$ is rational or not! –  Jesse Madnick Mar 8 '13 at 9:29
@ted $(a^{b})^c = a^{bc}$. Hence, $(\sqrt2^{\sqrt2})^{\sqrt2} = \sqrt{2}^{\sqrt2 \cdot \sqrt2} = \sqrt2^2 = 2$. –  user17762 Mar 8 '13 at 20:49
How about $e^{\ln 2} = 2$? –  Nemis L. Feb 27 at 21:24
@SimenK. It is significantly more difficult to prove that either $e$ or $\log 2$ is irrational than it is to prove that $\sqrt 2$ is. –  Sam DeHority Mar 21 at 4:52

When I was a kid I realized that $$0^2 + 1\ (\text{the first odd number}) = 1^2$$ $$1^2 + 3\ (\text{the second odd number}) = 2^2$$ $$2^2 + 5\ (\text{the third odd number}) = 3^2$$ and so on...

I checked it for A LOT of numbers :D

Years passed before someone taught me the basics of multiplication of polynomial and hence that $$(x + 1)^2 = x^2 + 2x + 1.$$ I know that this may sound stupid, but I was very young, and I had a great time filling pages with numbers to check my conjecture!!!

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I started math reasonably late, so I'm not sure this is a perfect example for a children book. I was totally amazed that the number $\pi$ is encountered in completely unrelated situation. Of course, I knew it's a ratio of circumference of the circle to the diameter, but then I learnt of the Normal probability density function.

So if you stretch a bell-shaped curve (Gaussian function) from $-\infty$ to $\infty$ the area under it $converges$ to $\sqrt{\pi}$?? How is it possible that this area has something to do with the square root of the ratio of circumference of the circle to the diameter? To be honest, even now, after learning the related proofs and derivations I still find it quite baffling.

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When I was a kid, I remember that I used to make 'little discoveries', although after I got all excited I realised other people already knew this and I didn't discover anything new, I was always really proud when I made one of those discoveries, Here are a few that I made and approximately how old I was when I did make the discovery (just saying I am 13 years old right now).

• An odd number plus an odd number is even, an even number plus an even number is even, and odd plus even is odd (approximately age 4-5)
• The rule for the Fibonacci sequence (I remember I saw it on the board or something in my classroom and then I just found the pattern) (age 6, I remember this one accurately bc it was during the first year of my primary education)
• If you rip a piece of paper in half, and then both in half again, and again etc, the number of pieces of paper you have when you ripped it $n$ times is $2^n$ (age 10)
• The fact that you can split a triangle into two right triangles and apply trig ratios to them (I later learned the Sine and Cosine Laws so yeah) (age 11)
• I was learning differential calculus, and I didn't know integral calculus yet. My dad told me that a integral is the inverse of the derivative, Then I figured out the rule that offsets the power rule and I figured out that the integral of nothing is an arbitrary constant (When i officially learnt integral calculus I found the proper formulas and I was just using the wrong symbols) (age 12)
• I proved to myself that the cardinality of the set of Reals and the set of Complex numbers are the same (age 13)
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1) Modular arithmetic fascinated me. I could not believe that with just a few tools, I could find the remainder left when $3^{100}$ is divided by 8. ($3^2\equiv 1$ mod $8$ and hence the result.)

2) Euclid's proof of the infinitude of primes. (Let the number of primes be finite. Let them be $P=\{p_1,p_2,\dots,p_r\}$. Take $k=p_1p_2\dots p_r+1$. None of the primes in $P$ divides $k$, hence $k$ is a prime or divisible by a prime not in $P$, and so we have a contradiction.)

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## protected by Zev ChonolesMar 7 '13 at 22:43

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