Is there a "simple" mathematical proof that is fully understandable by a 1st year university student that impressed you because it is beautiful?

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    I would argue for Euler's proof that $$\sum_{p \text{ prime}} \frac{1}{p}$$ diverges. – Alex Becker Mar 26 '14 at 21:45
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    If $M$ is a set, there is no surjective map from $M$ to the power set of $M$. Also: $\mathbb{Q}$ is countable. – Oliver Braun Mar 26 '14 at 21:48
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    Was going to say the same as Oliver. The proof is short: take $f:\;X\to\wp (X)$ then $\{x\in X\;|\;x\not\in f(x)\}\not\in \text{Im}f$, done. – Julien Godawatta Mar 26 '14 at 21:50
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    Check out this book: Proofs from The Book, it is basically meant to be a collection of particularly beautiful and not too-difficult mathematical proofs. – UncleZeiv Mar 27 '14 at 9:44
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    I think if the answer isn't "yes, I encountered more than one a week during my first year at university" then you're in the wrong field. You should be impressed by what you're learning ;-) But then, maybe I'm applying a low bar, I think something can be impressively "beautiful" without having to be the single most beautiful proof of all time, or even in the top 1000. IIRC for me the first one was in week 1 introduction to algebra, the proof that right inverses are also left inverses in a group using the lemma $z \cdot z = z \implies z = e$ – Steve Jessop Mar 27 '14 at 11:25

28 Answers 28

Here's a cute and lovely theorem.

There exist two irrational numbers $x,y$ such that $x^y$ is rational.

Proof. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$

(Nowadays, using the Gelfond–Schneider theorem we know that $\sqrt2^{\sqrt2}$ is irrational, and in fact transcendental. But the above proof, of course, doesn't care for that.)

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    I beat you by 16 seconds )) But since you wrote much more I will delete my answer ) – Artem Mar 26 '14 at 22:21
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    @Gankro: Although now we know that it is not the first case. An algebraic number to the power of an algebraic number is always transcendental (except for the trivial case when the base is $0$ or $1$). – Asaf Karagila Mar 27 '14 at 15:09
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    @AsafKaragila, I assume you intend one of those two algebraic numbers to be irrational? I'm pretty sure $2^2$ isn't transcendental. – Gankro Mar 27 '14 at 15:17
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    @Gankro: Of course, the one in the exponent has to be irrational. Yes, thank you. (My excuse: just woke up!) – Asaf Karagila Mar 27 '14 at 15:20
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    @Gankro you're pretty sure. you made my day. thanks. – Guy Apr 3 '14 at 12:22

How about the proof that


I remember being impressed by this identity and the proof can be given in a picture:

enter image description here

The picture is taken from Brian Sears' webpage. See also here.

Edit: Substituted $\frac{n(n+1)}{2}=1+2+\cdots+n$ in response to comments.

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    Seen this picture a couple of times, never understood it. – G. Bach Mar 26 '14 at 22:02
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    I've no idea what the strips to the left and right are for. But, aside from that, the first diagram shows squares of length 1, 2, 3, ..., $n$. The second diagram shows one $1\times1$ square, two $2\times 2$ squares, three $3\times3$ squares and so on; it's easy to see how to generalize the diagram up to $n$ $n\times n$ squares, which shows an area of $1^3+2^3+3^3+\dots+n^3$, arranged in an area of $(1+2+\dots+n)^2 = (\tfrac12n(n+1))^2$. – David Richerby Mar 26 '14 at 23:22
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    The strips at bottom and right illustrate how the grid pictured is $(1+2+\cdots+n)^2$. The bottom picture shows, for each $n$, $n$ squares (for even $n$ one of these squares is cut in half), each square $n\times n$; thus, for instance, the total light-blue area is $3\times (3\times 3) = 3^3$ and the total area is $1^3+2^3+\cdots+n^3$. – Steven Stadnicki Mar 27 '14 at 0:39
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    @ColinK: I don't know about US or CA high school curricula, which is part of the reason I can't argue about them ;-) I'm certain we proved $n(n+1)/2$ by drawing pictures and also by induction in high school in the UK. I've also just found a summation symbol on one of my end-of-first-year examination papers, on a question about the radius of convergence of $\sum a_n x^n$. I've said elsewhere though, it may be that regional differences in the degree of specialization of first-year university students means that this question has no baseline. – Steve Jessop Mar 27 '14 at 14:13
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    ... UK university studies are extremely specialized. I studied literally only mathematics and no other subject from day 1 of my degree for 4 years resulting in a master's. There were some topics that arguably are computer science or physics as well as being maths, and students can study multiple subjects in certain combinations (although I didn't). But basically if you're a maths student you're studying only maths, and so of course you'd see results like that quite soon. – Steve Jessop Mar 27 '14 at 14:28

Cantor's Diagonalization Argument, proof that there are infinite sets that can't be put one to one with the set of natural numbers, is frequently cited as a beautifully simple but powerful proof. Essentially, with a list of infinite sequences, a sequence formed from taking the diagonal numbers will not be in the list.

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    The example sequence that is not in the list needs to be nowhere equal to the sequence formed from taking diagonal numbers. – aschepler Mar 27 '14 at 0:25

I would personally argue that the proof that $\sqrt 2$ is irrational is simple enough for a university student (probably simple enough for a high school student) and very pretty in its use of proof by contradiction!

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    Also, the proof that the set of real numbers is uncountable is quite strikingly simple yet Cantor's diagonal method can be used in many different areas. – InsigMath Mar 26 '14 at 21:53
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    Strictly speaking, what you have is a proof that there is no rational number whose square is $2$. The evidence that it's reasonable to nevertheless talk about something called $\sqrt 2$ has to come from elsewhere. – Jack M Jul 10 '14 at 1:40

Prove that if $n$ and $m$ can each be written as a sum of two perfect squares, so can their product $nm$.

Proof: Let $n = a^2+b^2$ and $m=c^2+d^2$ ($a, b, c, d \in\mathbb Z$). Then, there exists some $x,y\in\mathbb Z$ such that

$$x+iy = (a+ib)(c+id)$$

Taking the magnitudes of both sides are squaring gives

$$x^2+y^2 = (a^2+b^2)(c^2+d^2) = nm$$

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    Incidentally, a more general thing holds: for any fixed $N$, if two numbers are of the form $x^2 + Ny^2$, then so is their product. (See Brahmagupta's identity.) Brahmagupta (7th century) and Bhāskara (~1150) used this to solve Pell's equation. – ShreevatsaR Mar 27 '14 at 13:59
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    Also, another note: You don't need complex numbers for this - it can be done vis-a-vis $$(a^2 + b^2)(c^2 + d^2) = (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2 .$$ This can be written $(ac)^2 + 2abcd + (bd)^2 + (ad)^2 - 2abcd + (bc)^2$, which can then be factored as $$ (ac + bd)^2 + (ad - bc)^2 .$$ – izœc Mar 28 '14 at 5:10
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    @izœc That is what you get if you solve for $x$ and $y$ in the formula. Complex numbers simply make the process more elegant and intuitive. – Argon Mar 28 '14 at 21:11
  • @Argon I see what you mean - I suppose my thought was that it is very elegant that it could be shown only in integers. Either way, I had never seen or thought of the slick approach using complex numbers, so thanks! I will definitely use that with my tutee. – izœc Mar 29 '14 at 18:23

I would go for the proof by contradiction of an infinite number of primes, which is fairly simple:

  • Assume that there is a finite number of primes.
  • Let $G$ be the set of all primes $P_1,P_2,...,P_n$.
  • Compute $K = P_1 \times P_2 \times ... \times P_n + 1$.
  • If $K$ is prime, then it is obviously not in $G$.
  • Otherwise, none of its prime factors are in $G$.
  • Conclusion: $G$ is not the set of all primes.

I think I learned that both in high-school and at 1st year, so it might be a little too simple...

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    @InsigMath You cannot show that $K$ must be a prime number, because it isn't always true. $2\cdot3\cdot5\cdot 7\cdot 11\cdot 13 + 1$ is divisible by 59. – MJD Mar 26 '14 at 22:43
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    This was already implicit in @InsigMath's comment, but this proof is not by contradiction. – Brenin Mar 26 '14 at 23:36
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    @Brenin Why does everyone get so hung up on "Euler's proof is/is not by contradiction"? Does it matter? The content of the theorem is clear. – Mario Carneiro Mar 27 '14 at 0:51
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    Anyway InsigMath's approach works. You don't show K is prime, just that it has a prime factor not in G. Then as Brenin says you avoid using contradiction on the existence of G. To some people this improves beauty, which after all is the question and so is worth getting hung up on :-) – Steve Jessop Mar 27 '14 at 12:31
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    @ShreevatsaR (The Euler thing was a slip of the fingers.) It just perplexes me why so many people get fixated on the status of Euclid's proof in particular as being by contradiction or not. Lots of proofs can be considered to be contradiction or not depending on how you look at them, but it's very rare to see Euclid's proof mentioned without someone needing to point out that it is not by contradiction. There is a formulation of the proof that is by contradiction, and another that is not. For the most part, it's just plain irrelevant to the topic. – Mario Carneiro Mar 27 '14 at 16:09

By the concavity of the $\sin$ function on the interval $\left[0,\frac{\pi}2\right]$ we deduce these inequalities: $$\frac{2}\pi x\le \sin x\le x,\quad \forall x\in\left[0,\frac\pi2\right].$$ enter image description here

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    What a "nice answer"! – amWhy Mar 29 '14 at 12:07

The first player in Hex has a winning strategy.

There are no draws in hex, so one player must have a winning strategy. If player two has a winning strategy, player one can steal that strategy by placing the first stone in the center (additional pieces on the board never hurt your position) then using player two's strategy.

  • Huh? "There are no draws, therefore there exists a way one player can force a victory"? That doesn't appear to follow. – Jack M Mar 27 '14 at 9:34
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    He forgot to add "by symmetry of the board and move: as if player one has a winning strategy player two could steal it." – phcoding Mar 27 '14 at 10:00
  • I guess it's OK to use determinacy of finite, tie-free two-player games of perfect information as a lemma. The strategy-stealing argument is simple and beautiful but I don't think it counts as an answer to the question because you've omitted the proof that there are no ties in Hex. This is essential to conclude that one of the players has a winning strategy and it is non-trivial to prove -- it takes about a page, in this presentation. – David Richerby Mar 27 '14 at 15:14
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    @DavidRicherby How about the proof that the first player has a winning strategy in Chomp? – bof Mar 27 '14 at 23:43
  • @bof I think that would be a better example, yes. – David Richerby Mar 27 '14 at 23:50

You cannot have two dice (with numbers $1$ to $6$) biased so that when you throw both, the sum is uniformly distributed in $\{2,3,\dots,12\}$.

For easier notation, we use the equivalent formulation "You cannot have two dice (with numbers $0$ to $5$) biased such that when you throw both, the sum is uniformly distributed in $\{0,1,\dots,10\}$."

Proof: Assume that such dice exist. Let $p_i$ be the probability that the first die gives an $i$ and $q_i$ be the probability that the second die gives an $i$. Let $p(x)=\sum_{i=0}^5 p_i x^i$ and $q(x)=\sum_{i=0}^5 q_i x^i$.

Let $r(x)=p(x)q(x) = \sum_{i=0}^{10} r_i x^i$. We find that $r_i = \sum_{j+k=i}p_jq_k$. But hey, this is also the probability that the sum of the two dice is $i$. Therefore, $$ r(x)=\frac{1}{11}(1+x+\dots+x^{10}). $$ Now $r(1)=1\neq0$, and for $x\neq1$, $$ r(x)=\frac{(x^{11}-1)}{11(x-1)}, $$ which clearly is nonzero when $x\neq 1$. Therefore $r$ does not have any real zeros.

But because $p$ and $q$ are $5$th degree polynomials, they must have zeros. Therefore, $r(x)=p(x)q(x)$ has a zero. A contradiction.

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    I think this proof relies on the fact the each die has an even number of faces, so the polynomials get odd degrees. So this comes to my mind: Is it possible to have two heptahedral (seven-faced) dice, each numbered $0,1,\ldots,6$ such that the sum is uniform in $\{ 0,1,\ldots,12 \}$?? In this case, it will be possible to write the 12th-degree polynomial $r(x)$ as the product as two 6th-degree real polynomials. But can their coefficients be taken to be in $[0,1]$? – Jeppe Stig Nielsen Mar 28 '14 at 12:52

Given a square consisting of $2n \times 2n$ tiles, it is possible to cover this square with pieces that each cover $2$ adjacent tiles (like domino bricks). Now imagine, you remove two tiles, from two opposite corners of the original square. Prove that is is now no longer possible to cover the remaining area with domino bricks.


Imagine that the square is a checkerboard. Each domino brick will cover two tiles of different colors. When you remove tiles from two opposite corners, you will remove two tiles with the same color. Thus, it can no longer be possible to cover the remaining area.

(Well, it may be too "simple." But you did not state that it had to be a university student of mathematics. This one might even work for liberal arts majors...)

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    The really beautiful part is when you try to generalize this to other situations. For example, try proving that it is impossible to tile a 8x8 chessboard which is only missing one corner square with 3x1 dominoes. It took me a week to discover the solution, and it was glorious when I finally saw the light. – JeremyKun Nov 5 '15 at 5:00

One little-known gem at the intersection of geometry and number theory is Aubry's reflective generation of primitive Pythagorean triples, i.e. coprime naturals $\,(x,y,z)\,$with $\,x^2 + y^2 = z^2.\,$ Dividing by $\,z^2$ yields $\,(x/z)^2+(y/z)^2 = 1,\,$ so each triple corresponds to a rational point $(x/z,\,y/z)$ on the unit circle. Aubry showed that we can generate all such triples by a very simple geometrical process. Start with the trivial point $(0,-1)$. Draw a line to the point $\,P = (1,1).\,$ It intersects the circle in the rational point $\,A = (4/5,3/5)\,$ yielding the triple $\,(3,4,5).\,$ Next reflect the point $\,A\,$ into the other quadrants by taking all possible signs of each component, i.e. $\,(\pm4/5,\pm3/5),\,$ yielding the inscribed rectangle below. As before, the line through $\,A_B = (-4/5,-3/5)\,$ and $P$ intersects the circle in $\,B = (12/13, 5/13),\,$ yielding the triple $\,(12,5,13).\,$ Similarly the points $\,A_C,\, A_D\,$ yield the triples $\,(20,21,29)\,$ and $\,(8,15,17),\,$ enter image description here
We can iterate this process with the new points $\,B,C,D\,$ doing the same we did for $\,A,\,$ obtaining further triples. Iterating this process generates the primitive triples as a ternary tree

$\qquad\qquad$ enter image description here
Descent in the tree is given by the formula

$$\begin{eqnarray} (x,y,z)\,\mapsto &&(x,y,z)-2(x\!+\!y\!-\!z)\,(1,1,1)\\ = &&(-x-2y+2z,\,-2x-y+2z,\,-2x-2y+3z)\end{eqnarray}$$

e.g. $\ (12,5,13)\mapsto (12,5,13)-8(1,1,1) = (-3,4,5),\ $ yielding $\,(4/5,3/5)\,$ when reflected into the first quadrant.

Ascent in the tree by inverting this map, combined with trivial sign-changing reflections:

$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$

$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$

$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$

See my MathOverflow post for further discussion, including generalizations and references.

I like the proof that there are infinitely many Pythagorean triples.

Theorem: There are infinitely many integers $ x, y, z$ such that $$ x^2+y^2=z^2 $$

Proof: $$ (2ab)^2 + ( a^2-b^2)^2= ( a^2+b^2)^2 $$

One cannot cover a disk of diameter 100 with 99 strips of length 100 and width 1.

Proof: project the disk and the strips on a semi-sphere on top of the disk. The projection of each strip would have area at most 1/100th of the area of the semi-sphere.

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    How do you do this projection? How do you calculate the area of the strips' projection? – Paŭlo Ebermann Mar 27 '14 at 0:44
  • Project vertically from the plate onto the lid. Calculating the strip's projection area is a straightforward freshman calculus exercise. – Michael Mar 27 '14 at 0:56
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    @PaŭloEbermann The sphere has the remarkable property that the area is uniformly distributed along strips of constant $z$, that is, the area of the portion of the sphere between $z=a$ and $z=b$ is proportional to $b-a$. The theorem follows. – Mario Carneiro Mar 27 '14 at 0:58

how about deriving Binets formula for Fibonacci numbers based on the characteristic polynomial:

$$ f_n = \frac{\phi^n - \psi^n}{ \sqrt{5}} $$

  • I saw the derivation using before knowing that the characteristic polynomial is actually a widely used method. When I first saw it, the beauty of the proof was that you could simply obtain a guess for the solution using a Stetson-Harrison method, and you instantly know that it has to be the only solution once you check it because the Fibonacci numbers are uniquely determined. – JiK Mar 28 '14 at 10:48

If you have any set of 51 integers between $1$ and $100$, the set must contain some pair of integers where one number in the pair is a multiple of the other.

Proof: Suppose you have a set of $51$ integers between $1$ and $100$. If an integer is between $1$ and $100$, its largest odd divisor is one of the odd numbers between $1$ and $99$. There are only $50$ odd numbers between $1$ and $99$, so your $51$ integers can’t all have different largest odd divisors — there are only $50$ possibilities. So two of your integers (possibly more) have the same largest odd divisor. Call that odd number $d$. You can factor those two integers into prime factors, and each will factor as (some $2$’s)$\cdot d$. This is because if $d$ is the largest divisor of a number, the rest of its factorization can’t include any more odd numbers. Of your two numbers with largest odd factor $d$, the one with more $2$’s in its factorization is a multiple of the other one. (In fact, the multiple is a power of $2$.)

In general, let $S$ be the set of integers from $1$ up to some even number $2n$. If a subset of $S$ contains more than half the elements in $S$, the set must contain a pair of numbers where one is a multiple of the other. The proof is the same, but it’s easier to follow if you see it for a specific $n$ first.

  • A similar interesting proof is that if you pick 51 integers from 1-100 there must exist a coprime pair, because n and n+1 are alway coprime. – user85798 Mar 27 '14 at 8:44
  • with $15$ and $21$, $d=3$, but both factorizations include more odd numbers. (Also, neither factorization is "(some $2$s)$\cdot d$".) – Eric Towers Mar 27 '14 at 23:47
  • The largest odd factor of 15 is 15, and the largest odd factor of 21 is 21. I should have mentioned that (some 2s) might be (no 2s at all), since it's possible that when you have two numbers with the same largest odd factor, one of them is odd. The argument still holds, and the factorization when there are no 2s is $2^0\cdot d$. For example, the (some 2s)$\cdot$(largest odd factor) factorization of 15 is $2^0\cdot15$. – Steve Kass Mar 28 '14 at 1:13

The proof that an isosceles triangle ABC (with AC and AB having equal length) has equal angles ABC and BCA is quite nice:

Triangles ABC and ACB are (mirrored) congruent (since AB = AC, BC = CB, and CA = BA), so the corresponding angles ABC and (mirrored) ACB are equal.

This congruency argument is nicer than that of cutting the triangle up into two right-angled triangles.

Parity of sine and cosine functions using Euler's forumla:

$e^{-i\theta} = cos\ (-\theta) + i\ sin\ (-\theta)$

$e^{-i\theta} = \frac 1 {e^{i\theta}} = \frac 1 {cos\ \theta \ + \ i\ sin\ \theta} = \frac {cos\ \theta\ -\ i\ sin\ \theta} {cos^2\ \theta\ +\ sin^2\ \theta} = cos\ \theta\ -\ i\ sin\ \theta$

$cos\ (-\theta) +\ i\ sin\ (-\theta) = cos\ \theta\ +i\ (-sin\ \theta)$


$cos\ (-\theta) = cos\ \theta$

$sin\ (-\theta) = -\ sin\ \theta$


The proof is actually just the first two lines.

I believe Gauss was tasked with finding the sum of all the integers from $1$ to $100$ in his very early schooling years. He tackled it quicker than his peers or his teacher could, $$\sum_{n=1}^{100}n=1+2+3+4 +\dots+100$$ $$=100+99+98+\dots+1$$ $$\therefore 2 \sum_{n=1}^{100}n=(100+1)+(99+2)+\dots+(1+100)$$ $$=\underbrace{101+101+101+\dots+101}_{100 \space times}$$ $$=101\cdot 100$$ $$\therefore \sum_{n=1}^{100}n=\frac{101\cdot 100}{2}$$ $$=5050.$$ Hence he showed that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}.$$

  • This is not a rigorous proof of the fact but I think we can cut him some slack considering that he was 10 years old. – user124862 Mar 28 '14 at 1:40
  • Why is not rigorous? – Sawarnik Apr 12 '14 at 9:37
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    This is perfectly rigorous, it is a finite sum so there's nothing wrong with the proof here – CameronJWhitehead Apr 27 '14 at 9:53
  • Well I just was noting that he only proved it by proving it is true for $n=100$ and not for an arbitrary $n$. – user124862 Apr 28 '14 at 11:11

If $H$ is a subgroup of $(\mathbb{R},+)$ and $H\bigcap [-1,1]$ is finite and contains a positive element. Then, $H$ is cyclic.

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    How many first year university students know group theory though? – InsigMath Mar 26 '14 at 21:51
  • Well, most of my classmates did. – Swapnil Tripathi Mar 26 '14 at 21:52
  • What do you mean by $H \cap [-1, 1]$? – dani_s Mar 26 '14 at 21:53
  • I am so sorry.. (R,+) – Swapnil Tripathi Mar 26 '14 at 21:54
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    Yes. We assume an element 'a' to be the smallest postive element and claim that H = <a>… We'll prove using contradiction. We suppose there is an element x of H which is not a power of a. Then, ma<x<(m+1)a. We have , 0<x-ma<a. Thus, contradicting our choice of a. – Swapnil Tripathi Mar 26 '14 at 22:10

Euler's solution to the Basel problem using the (unjustified at the time) Weierstrass factorization theorem. Eventually led to reading Hardy's Divergent Series with understanding coming in fits and starts.

Fermat's little theorem from noting that modulo a prime p we have for $a\neq 0$:

$$1\times2\times3\times\cdots\times (p-1) = (1\times a)\times(2\times a)\times(3\times a)\times\cdots\times \left((p-1)\times a\right)$$

Proposition (No universal set): There does not exists a set which contain all the sets (even itself)

Proof: Suppose to the contrary that exists such set exists. Let $X$ be the universal set, then one can construct by the axiom schema of specification the set

$$C=\{A\in X: A \notin A\}$$

of all sets in the universe which did not contain themselves. As $X$ is universal, clearly $C\in X$. But then $C\in C \iff C\notin C$, a contradiction.

Edit: Assuming that one is working in ZF (as almost everywhere :P)

(In particular this proof really impressed me too much the first time and also is very simple)

  • How come there are set theories (which are sufficiently strong to develop at least basic set theory and logic) in which the existence of a universal set is provable? – Asaf Karagila Mar 27 '14 at 23:21
  • @AsafKaragila: But assuming that one is working in something like Zermelo-Frankel the universal set is not possible (and almost everyone assume ZF). Thanks for the remark I've edited and added than we assume ZF or something similar. – Jose Antonio Mar 28 '14 at 1:53

Most proofs concerning the Cantor Set are simple but amazing.

The total number of intervals in the set is zero.

It is uncountable.

Every number in the set can be represented in ternary using just 0 and 2. No number with a 1 in it (in ternary) appears in the set.

The Cantor set contains as many points as the interval from which it is taken, yet itself contains no interval of nonzero length. The irrational numbers have the same property, but the Cantor set has the additional property of being closed, so it is not even dense in any interval, unlike the irrational numbers which are dense in every interval.

The Menger sponge which is a 3d extension of the Cantor set simultaneously exhibits an infinite surface area and encloses zero volume.

The derivation of first principle of differentiation is so amazing, easy, useful and simply outstanding in all aspects. I put it here:

Suppose we have a quantity $y$ whose value depends upon a single variable $x$, and is expressed by an equation defining $y$ as some specific function of $x$. This is represented as:


This relationship can be visualized by drawing a graph of function $y = f (x)$ regarding $y$ and $x$ as Cartesian coordinates, as shown in Figure(a).

curves of the given funtion Consider the point $P$ on the curve $y = f (x)$ whose coordinates are $(x, y)$ and another point $Q$ where coordinates are $(x + Δx, y + Δy)$.

The slope of the line joining $P$ and $Q$ is given by:

$tanθ = \frac{Δy}{Δx} = \frac{(y + Δy ) − y}{Δx}$

Suppose now that the point $Q$ moves along the curve towards $P$.

In this process, $Δy$ and $Δx$ decrease and approach zero; though their ratio $\frac{Δy}{Δx}$ will not necessarily vanish.

What happens to the line $PQ$ as $Δy→0$, $Δx→0$? You can see that this line becomes a tangent to the curve at point $P$ as shown in Figure(b). This means that $tan θ$ approaches the slope of the tangent at $P$, denoted by $m$:

$m=lim_{Δx→0} \frac{Δy}{Δx} = lim_{Δx→0} \frac{(y+Δy)-y}{Δx}$

The limit of the ratio $Δy/Δx$ as $Δx$ approaches zero is called the derivative of $y$ with respect to $x$ and is written as $dy/dx$.

It represents the slope of the tangent line to the curve $y=f(x)$ at the point $(x, y)$.

Since $y = f (x)$ and $y + Δy = f (x + Δx)$, we can write the definition of the derivative as:

$\frac{dy}{dx}=\frac{d{f(x)}}{dx} = lim_{x→0} [\frac{f(x+Δx)-f(x)}{Δx}]$,

which is the required formula.

This proof that $n^{1/n} \to 1$ as integral $n \to \infty$:

By Bernoulli's inequality (which is $(1+x)^n \ge 1+nx$), $(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2} $. Raising both sides to the $2/n$ power, $n^{1/n} <(1+n^{-1/2})^2 = 1+2n^{-1/2}+n^{-1} < 1+3n^{-1/2} $.

Can a Chess Knight starting at any corner then move to touch every space on the board exactly once, ending in the opposite corner?

The solution turns out to be childishly simple. Every time the Knight moves (up two, over one), it will hop from a black space to a white space, or vice versa. Assuming the Knight starts on a black corner of the board, it will need to touch 63 other squares, 32 white and 31 black. To touch all of the squares, it would need to end on a white square, but the opposite corner is also black, making it impossible.

  • It would be already interesting to know if it is possible for knight (starting anywhere) to touch every space on the board exactly once (ending anywhere). I'm not sure the answer would be so easy though... – Surb Mar 11 '15 at 19:47

The Eigenvalues of a skew-Hermitian matrix are purely imaginary.

The Eigenvalue equation is $A\vec x = \lambda\vec x$, and forming the vector norm gives $$\lambda \|\vec x\| = \lambda\left<\vec x, \vec x\right> = \left<\lambda \vec x,\vec x\right> = \left<A\vec x,\vec x\right> = \left<\vec x, A^{T*}\vec x\right> = \left<\vec x, -A\vec x\right> = -\lambda^* \|\vec x\|$$ and since $\|\vec x\| > 0$, we can divide it from left and right side. The second to last step uses the definition of skew-Hermitian. Using the definition for Hermitian or Unitarian matrices instead yields corresponding statements about the Eigenvalues of those matrices.

I like the proof that not every real number can be written in the form $a e + b \pi$ for some integers $a$ and $b$. I know it's almost trivial in one way; but in another way it is kind of deep.

  • How do you prove it? – Babai Mar 14 at 18:49

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