Visually stunning math concepts which are easy to explain

Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

• It looks like mathpop or demand for math entertainment) – rook Apr 2 '14 at 12:59
• Something that has always annoyed me is irrational numbers. They are easy to understand (like the proof for $\sqrt{2}$), but not being able to have a rational representation just irks me. And don't even get me started on transcendentals like $\pi$ and $e$. – Cole Johnson Apr 4 '14 at 16:12
• @ColeJohnson the 'transcendentality" is the beauty of it! – Guy Apr 4 '14 at 17:07
• There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words – Martin Brandenburg Jun 20 '14 at 22:13
• @TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site. – user147263 Aug 29 '14 at 18:37

Riemann integration has always amazed me. Its simple yet extraordinary.

Proof that the area of a circle is $\pi r^2$ without words: Proof Without Words: The Circle

• but he has to prove $C=2\pi r$ first – zinking Apr 8 '14 at 6:23
• @zinking No, $\pi$ is defined to be the constant that goes in that place in that equation to make it hold. However, I was rather dissatisfied with this "proof". There's lots of distortion involved with deforming the area of the circle into the area of the triangle, that one would have to know calculus to understand why the distortion doesn't matter (at which point, why not just calculate the integral). This is more of a memorization technique to remember the formula. – Travis Bemrose Apr 13 '14 at 20:02

Check out the "Proofs Without Words" gallery (animated) here:

http://usamts.org/Gallery/G_Gallery.php

And the related proofs here:

http://www.artofproblemsolving.com/Wiki/index.php/Proofs_without_words

Many of these are similar to the ones already listed here, but you get a bunch in one place.

A visual display that $0^0=1$. The following is a tetration fractal or exponential map with a pseudo-circle shown in orange. The red area is period $1$ and contains $1$. Example is $1^1=1$. The orange pseudo-circle which contains $0$ is period two. Example is $0^0=1, 0^1=0$.

• 0^0 = 1 is not even proven, it was just defined that way to make things easier, right? – Michael Apr 4 '14 at 18:59
• It depends on the context, but in the grand scheme of mathematics, it is considered undefined. This is because there are completely logical steps that point to it being zero, and equally logical ones that point to it being one. We can't call it both zero and one, so we call it undefined. – recursive recursion Apr 4 '14 at 21:41
• There is one problem with this. I have no idea how you make that image mean what you say it means... Visually stunning? Yes. Easy to explain? Maybe. Explained? No. – daviewales Apr 5 '14 at 12:31
• So, er, how does this show that $0^0=1$? – David Richerby Apr 5 '14 at 17:55
• $x^0 = 1$ because the multiplicative identity is one, $0^x = 0$ because zero is the multiplicative fixed point. That is, if and only if we never, ever, ever tussle with fractional powers; then and only then exponentiation is shorthand for repeated multiplication and therefore $0^0 = 1$. If we at any point use fractional powers, then exponentiation is shorthand for it's natural definition: $a^b = e^{b \ln a}$ and we all know that $\ln 0$ is undefined. There is no dispute, once we make the assumptions explicit the problem goes away. – Karl Damgaard Asmussen Apr 13 '14 at 9:16

A nationwide math contest in Germany recently came up with a task that I found beautiful to explain, because of two points.

1. You can get an idea, what the proof is, without applying mathematically accurate theory and this intuitive proof is most likely the right way.

2. At any given point of this intuitive proof, you can chime in and ask yourself: But how would I say this in mathematical terms? When you find these terms, eventually you get the proof you were looking for.

So here you go: Lea gets the task to write down 2014 numbers. These numbers have to fulfill a specification. For every set of three numbers from that whole set, the arithmetic average of these three must also be within the whole set of $2014$ numbers.

Your task is to proof, that Lea has to write down the same number $2014$ times. Every set of 2014 numbers with any variation in it would not fulfill the specifications.

So since we are talking about layman maths here, I'll go with the intuitive way. We have to find a reason, why choosing a set with different numbers would violate the specifications and we have to proof that always taking the same number would not violate them. The later one is rather easy. Take any arbitrary number three times. The arithmetic average will be the same number, which is in your set already. That wasn't too bad, right?

But what about sets with not all the same numbers? We are mathematicians, so we'll just do what we always do: Chop the problem into pieces we can solve. The first piece is where we have two equal numbers and one other number in our set. Let's assume, the single number is bigger than the two equal numbers. What will that do to our arithmetic average? Right, it will be below the middle between the bigger and the smaller number. We can write that arithmetic average down and specifications are ok. But now we have created another set of three numbers. The single, big number (I'll call it a), one of the two equal numbers (that would be b) and the arithmetic average of a, and b (I'll call that one c). So now we would have to also add the arithmetic average of a, b and c. A quick sketch will show you, that this new number is also slightly below the middle between a and b.

And like that we will always have to add a new number. The arithmetic average of a, b and the new number will never reach the middle. Something, that you can also verify with a few sketches. So we would have to add infinitely more numbers, but we wanted only 2014. Apparently, no two numbers can be equal.

So what if all numbers are different? There is one special case. Let's call our numbers a, b and c again. If b is equally far away from a and c (so b could be 3, a could be 1, then c would be 5). In that case, b is the arithmetic average. But we have to have 2014 different numbers. As soon as we add a fourth number d, it's spoiled. d could be 7, to be still in a distance of 2 to c, but then the set a, b and d would not contain its own arithmetic average. So we know, that within a set of 2014 numbers, we would have sets of three, where these three numbers don't include their own arithmetic average, no matter what we do.

And now we look back at our idea about the set with two equal numbers. We see: As soon as we have a bigger and a smaller number and the number in between those is not exactly in the middle, we can once again start with our endless construction of arithmetic averages. We always replace the number between the bigger and the smaller one by the arithmetic average of the three and we can never reach the middle, but it will always get closer to the middle (thus be another number).

And as I said, making this proof mathematical will not alter it. It will be all the same, but with more equations and sequences. Since we excluded the option of making anything infinite, it is correct as it stands here. This one made me realize: Proofs are not the miracles or the magic they seemed to be for me during high school. Of course, there are hard proofs (and things you can't proof, there is a proof for that), but often you only have to think clearly and to chop the problem into the right pieces.

• Hmm, I don't think "induction by contradiction" ("contradiction by induction"?) is valid even though our intuition would like it. In other words, you've shown that you can't inductively build a set with the desired property, having smaller sets along the way that also have the property. But this doesn't exclude the possibility that the property might not hold for all smaller sets, yet 2014 is the first time you have a set large enough for the property to hold. – Travis Bemrose Apr 13 '14 at 20:36
• Sort the list in non-decreasing order as $a_1, a_2, \ldots, a_{2014}$. Since the average of $a_1$, $a_2$, and $a_3$ is among these, it must equal $a_2$, so $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$ for some $d$. Since the average of $a_2$, $a_3$, and $a_4$ is among the list, is must equal $a_3$. Proceeding in this manner we see that $a_k = a_1 + (k-1)d$ for some $d$. Now note that the average of $a_1$, $a_2$, and $a_4$, which is $a_1 + \frac{4}{3} d$, is also in the list. It follows that $d = 0$. – Daniel McLaury Aug 28 '14 at 1:36

In plane geometry Morley’s theorem is a stunning fact in my opinion: In any triangle, the points of intersections of adjacent trisectors of the angles form an equilateral triangle :

In analytical geometry: The generalization of triangles or tetrahedra in n-dimensions is simpleces. And the formula for the simplex volume is a beauty, for example, the volume of four-dimensional simplex which is called pentachoron, pentatope or 5-cell (using the coordinates of its vertices): $$\text{Four dimensional volume} = \pm\frac{1}{4!}\;\begin{vmatrix} \;x_1-x_5 && y_1-y_5 && z_1-z_5 && w_1-w_5\;\\ \;x_2-x_5 && y_2-y_5 && z_2-z_5 && w_2-w_5\;\\ \;x_3-x_5 && y_3-y_5 && z_3-z_5 && w_3-w_5\;\\ \;x_4-x_5 && y_4-y_5 && z_4-z_5 && w_4-w_5\;\\ \end{vmatrix}$$ In the case of triangles we get $$A=±\frac{1}{2}\;\begin{vmatrix} \;x_1-x_3 && y_1-y_3\;\\ \;x_2-x_3 && y_2-y_3\;\\ \end{vmatrix}$$

Of course, we can write it even shorter if we use vectors.

And 4-dimensional spheres are truly amazing, as are tesseracts

And some things are truly arcane (Ramanujan summation formula): $$1+2+3+4+5\,+\,...= -\frac{1}{12}$$ You can get a bit of information about it in this and this Wikipedia articles, but divergent series like this (please refer to Wikipedia again) are not an elementary topic to easily understand. (It's usually met with misunderstanding and downvotes. It's currently 'attacked' in Wikipedia because it is not understood by lay mathematicians. The simplest way to get at least some sort of idea is probably to treat it (it's not completely arbitrary but actually one of the most beautiful concepts in math) as an abstraction, something abstract like, say, square roots of negative numbers. That's a bad comparison, but Ramanujan's idea is slightly tougher than pentachorons or tesseracts. It's not arbitrary assigning $$-1/12\,$$ to the zeta function. This arcane formula found its way into physics (Casimir force). Euler is considered the first to derive this formula more than two hundred years ago.

As to calculus, my vote would go to the beauty of Euler’s formula already posted in this thread ($$\,\boldsymbol{e^{ix}=\cos{x}+i\sin{x}}\,$$)

• **WARNING!** $$-\frac1{12}=\zeta(-1)\ne1+2+3+4+\ldots$$ – Simply Beautiful Art Jul 2 '17 at 22:52
• I would remove my downvote if concerns in the previous comment was addressed. – Frenzy Li Nov 13 '17 at 18:16

In group theory, "visual" explanation of the group $D_4$ (or sometimes called $D_8$, which is dihedral group of order $8$ or degree $4$) was really exciting to me. Even with some elementary knowledge about groups, if someone defines $D_4$ as $$\langle x, a\ |\ a^4 = x^2 = e, axa = x \rangle$$ it might seem meaningless or too abstract to understand. When I first saw this, I simply thought it as something to memorize. But when I saw the visual explanation, it was a "stunning" moment for me because it was really easy to understand and there was nothing to memorize at all:

First let us take a square $S$ with vertices named as $A, B,C, D$.

Then, if we let the element $a$ rotate this square $90^\circ$ clockwise direction and let the element $x$ flip the square through the first diagonal, i.e. $x=y$ line, we can have all $8$ elements of $D_4$ as the following:

$e$: Identity element. It does nothing to $S$.

$a$: Rotates $S$, $90^\circ$ clockwise direction.

$a^2$: Rotates $S$, $90^\circ$ clockwise direction twice, i.e., rotates $S$, $180^\circ$ clockwise direction.

$a^3$: Rotates $S$, $90^\circ$ clockwise direction three times, i.e. rotates $S$, $270^\circ$ clockwise direction.

$x$: Flips $S$ through its first diagonal, i.e., interchanges $A$ and $C$.

$ax$: Flips $S$ through its first diagonal first, then rotates the flipped square $90^\circ$ clockwise direction.

$a^2x$: Flips $S$ through its first diagonal first, then rotates the flipped square $180^\circ$ clockwise direction.

$a^3x$: Flips $S$ through its first diagonal first, then rotates the flipped square $270^\circ$ clockwise direction.

Here, only thing we need to be careful is that rightmost function is applied first (for example, $ax$ means: first flip, then rotate). Now, we can verify the properties that are mentioned in the definition:

First of all, we have $a^4 = e$ because $a^4$ rotates $S$, $90^\circ$ clockwise direction four times, which means $360^\circ$ clockwise direction, which doesn't change the place of any vertex. So it is as same as the identity element $e$.

Secondly, we have $x^2 = e$ because if we flip $S$ through its first diagonal (interchanging $A$ and $C$) and then flip it again, we get $S$ again, which also corresponds to identity element $e$.

Finally, we have $axa = x$, which can be verified by rotating $S$ first, then flipping it and rotating it again. In the end what we get is as same as flipping $S$, which is done by $x$.

$$\sum_{i=1}^{\infty}\frac{1}{x^n}=\frac{1}{x-1}$$

In base $$x$$, this sum equals to 0.1111111.... and if you multiply it by x-1 you get 0.(x-1)(x-1)(x-1).... which equals to 1.

An important concept im math is $$\infty$$: the Bernoulli's lemniscate is very similar to its sign. Here the GIF of its construction from an hyperbole:

The equation is very simple:$$\left ( x^2+y^2 \right )^2=2a^2(x^2-y^2)$$

One of my favourite mathematical number is $$3$$ because

$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\dots}}}}}$$

Similary we can write

$$2=\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\dots}}}}}}$$

It can be obtained by going backward as

$$3=\sqrt 9=\sqrt{1+8}=\sqrt{1+2(4)}=\sqrt{1+2\sqrt{16}} =\sqrt{1+2\sqrt{1+3(5)}} =\dots$$

And it can be written further in the same manner.

"Cantor's Diagonal Argument" is something I find amazingly beautiful. It is not geometry, it is not artistically stunning, but it visually captures an implausible mathematical truth. It allows you to prove that there are as many rational numbers as natural numbers.

The example below uses two of the simplest infinite sets, that of natural numbers, and that of positive fractions. The idea is to show that both of these sets, $${\mathbb {N} }$$ and $$\mathbb Q^{+}$$ have the same cardinality.

First, the fractions are aligned in an array, as follows: $$\begin{array}{cccccccccc}{\frac {1}{1}}&&{\tfrac {1}{2}}&&{\tfrac {1}{3}}&&{\tfrac {1}{4}}&&{\tfrac {1}{5}}&\cdots \\&&&&&&&&&\\{\tfrac {2}{1}}&&{\tfrac {2}{2}}&&{\tfrac {2}{3}}&&{\tfrac {2}{4}}&&{\tfrac {2}{5}}&\cdots \\&&&&&&&&&\\{\tfrac {3}{1}}&&{\tfrac {3}{2}}&&{\tfrac {3}{3}}&&{\tfrac {3}{4}}&&{\tfrac {3}{5}}&\cdots \\&&&&&&&&&\\{\tfrac {4}{1}}&&{\tfrac {4}{2}}&&{\tfrac {4}{3}}&&{\tfrac {4}{4}}&&{\tfrac {4}{5}}&\cdots \\&&&&&&&&&\\{\tfrac {5}{1}}&&{\tfrac {5}{2}}&&{\tfrac {5}{3}}&&{\tfrac {5}{4}}&&{\tfrac {5}{5}}&\cdots \\\vdots &&\vdots &&\vdots &&\vdots &&\vdots &\\\end{array}$$ And now the numbers in this construction can be counted as follows, leaving out the fractions which can be simplified.

$$\begin{array}{lclclclclc}{\tfrac {1}{1}}\ _{\color {Blue}(1)}&{\color {MidnightBlue}\rightarrow }&{\tfrac {1}{2}}\ _{\color {Blue}(2)}&&{\tfrac {1}{3}}\ _{\color {Blue}(5)}&{\color {MidnightBlue}\rightarrow }&{\tfrac {1}{4}}\ _{\color {Blue}(6)}&&{\tfrac {1}{5}}\ _{\color {Blue}(11)}&{\color {MidnightBlue}\rightarrow }\\&{\color {MidnightBlue}\swarrow }&&{\color {MidnightBlue}\nearrow }&&{\color {MidnightBlue}\swarrow }&&{\color {MidnightBlue}\nearrow }&&\\{\tfrac {2}{1}}\ _{\color {Blue}(3)}&&{\tfrac {2}{2}}\ _{\color {Blue}(\cdot )}&&{\tfrac {2}{3}}\ _{\color {Blue}(7)}&&{\tfrac {2}{4}}\ _{\color {Blue}(\cdot )}&&{\tfrac {2}{5}}&\cdots \\{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\nearrow }&&{\color {MidnightBlue}\swarrow }&&{\color {MidnightBlue}\nearrow }&&&&\\{\tfrac {3}{1}}\ _{\color {Blue}(4)}&&{\tfrac {3}{2}}\ _{\color {Blue}(8)}&&{\tfrac {3}{3}}\ _{\color {Blue}(\cdot )}&&{\tfrac {3}{4}}&&{\tfrac {3}{5}}&\cdots \\&{\color {MidnightBlue}\swarrow }&&{\color {MidnightBlue}\nearrow }&&&&&&\\{\tfrac {4}{1}}\ _{\color {Blue}(9)}&&{\tfrac {4}{2}}\ _{\color {Blue}(\cdot )}&&{\tfrac {4}{3}}&&{\tfrac {4}{4}}&&{\tfrac {4}{5}}&\cdots \\{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\nearrow }&&&&&&&&\\{\tfrac {5}{1}}\ _{\color {Blue}(10)}&&{\tfrac {5}{2}}&&{\tfrac {5}{3}}&&{\tfrac {5}{4}}&&{\tfrac {5}{5}}&\cdots \\\vdots &&\vdots &&\vdots &&\vdots &&\vdots &\\\end{array}$$ Leading to a bijection between $$\mathbb{N}$$ and $$\mathbb{Q^+}$$ $$\begin{array}{cccccccccccccccc}{\color {Blue}1}&{\color {Blue}2}&{\color {Blue}3}&{\color {Blue}4}&{\color {Blue}5}&{\color {Blue}6}&{\color {Blue}7}&{\color {Blue}8}&{\color {Blue}9}&{\color {Blue}10}&{\color {Blue}11}&{\color {Blue}\cdots }\\[3pt]{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }&{\color {MidnightBlue}\downarrow }\\[3pt]1&{\tfrac {1}{2}}&2&3&{\tfrac {1}{3}}&{\tfrac {1}{4}}&{\tfrac {2}{3}}&{\tfrac {3}{2}}&4&5&{\tfrac {1}{5}}&\cdots \\\end{array}$$ Hence, the cardinality of $$\mathbb{Q^{+}}$$ and $$\mathbb{N}$$ is the same. This can be easily extended to all of $$\mathbb{Q^{+}}$$

We can visually encode factorizations of numbers using $$2$$-digit palindromes!

Define $$n\times n$$ matrices $$P_n(k)$$ and $$N_n$$ for $$x,y\in[0,n)$$ as: (the sum is entry-wise)

\begin{align} P_n(k)(x,y)&= \begin{cases} 1, & \text{if }x+nk\text{ is a two digit palindrome in number base }y+n\\ 0, & \text{else} \end{cases}\\ N_n&=\sum\limits_{k=0}^{\infty} P_n(k) \end{align}

For example, if we color nonzero elements of $$N_n$$ blue and the zeroes white, we get the following images:

You can notice that prime numbers will always be "complete blue squares" because they do not have nontrivial factors (divisors).

Other numbers will have patterns that represent their factors (divisors) in some way.

For example, notice that even semi-primes $$n=2p$$ where $$p$$ is prime, all have the following look:

Or for example, notice that squares of primes $$n=p^2$$ have the least details:

On the other hand, numbers with a lot of factors such as factorials like $$4!=2\cdot3\cdot4 = 24$$ or primorials like $$p_3\#=2\cdot 3\cdot 5 = 30$$ have much more details (factors), and numbers such as powers of primes like $$27=3^3$$ and $$32=2^5$$ have regular fractal-like patterns.

And so on.

I must admit that this is a part of my own question, more precisely, it is a part of the introduction to:

Thing get really interesting if we "look inside" the blue squares that represent the prime numbers. But, that part is no longer "easy to explain". (See the link I listed above for more information.)

This is "easy to explain" because:

Factorizations of positive integers $$n$$ are related to $$2$$-digit palindromes. That is, if $$n$$ can be factorized as $$n=p\cdot q$$ where $$p\lt q-1$$, this means that $$n$$ is a $$2$$-digit palindrome in the number base $$q-1$$, which we write as:

$$n=p q=p(q-1)+p=(p, p)_{q-1}, p

where $$(p,p)$$ are digits of $$n=pq$$ in the number base $$q-1$$.

There is a bijection between $$\mathbb R$$ and $$\mathbb R^2$$. That is: a line and the space have the same cardinality.

You can visualize it with (one of) the Peano curve:

P.S. We use the fact that there is a bijection between $$\mathbb R$$ and $$[0,1]$$

• The limit curve is surjective, but, unfortunately, it is not injective. However, since there is a simple injection $f:[0,1]\mapsto[0,1]^2$ defined by $f(x)=(x,0)$, the Schröder-Bernstein Theorem says that a bijection exists. – robjohn Sep 18 '20 at 20:32

A number spiral of primes with some prime-dense polynomials noted. Alignment of perfect squares is along the right horizontal axis. (The prime-free gap is accounted for by the squares and the squares minus 1.) The pronic numbers are aligned along the left side.

The spiral is plotted counterclockwise like this:

Source: http://numberspiral.com/

• all swares will be "on line"? and the 3,4,13 branch contains infinitely primes? – Luis Felipe Sep 25 '20 at 17:09

The surface obtained by spinning a cube on two diametrically opposite corners:

$$\hskip{4.5cm}$$

All the surfaces are ruled surfaces. The top and bottom are simply conical caps. The curved part in the middle is part of a hyperboloid of one sheet. It can be obtained by taking a cylinder of radius and height $$\sqrt{\frac23}$$ times the edge length of the cube and giving one of the ends a $$60^{\large\circ}$$ twist.

I have found it intuitively difficult to see that the sequence $$(1+\frac{1}{n})^{n}$$ is increasing. However the following picture makes it clear.

We have drawn the graph of the function $$\log(1+x)$$. It is clear that the graph passes through the origin. By the concavity of the function it is clear that the slope $$\log(1+a)/a$$ increases as $$a$$ decreases.

Therefore

$$\frac{\log(1+\frac{1}{n})}{1/n} < \frac{\log(1+\frac{1}{n+1})}{1/(n+1)}$$

From here the desired result follows.

In fact, one can also see from above that $$\lim_{n\to \infty} (1+1/n)^n = e$$ since the figure shows that $$\log(1+1/n)/(1/n)$$ converges to the slope of the tangent at $$(0, 0)$$, which is $$1$$.

I just saw this movie on youtube that gives a visual proof that every prime that is 1 modulo 4 can be written as the sum of two squares. This is a highly non-trivial theorem, and that it can be explained visually is impressive.

A literal meaning of visual math.

Almost any line-drawing can be traced on the cartesian plane using the fourier transformations using only one parametric equation.
In simple terms tracing multiple circles with varying time periods can make a closed curve. Here are a few of my creations (click on the names to open the graph):

1. Here is Pikachu

1. I am Iron Man

For a visual proof I suggest watching this piece of art by Grant Sanderson.

Also note that the jittery behaviour and random lines cutting through the figures are not a problem but a rendering issue, which anyways make it look cooler.

P.S. It works as gifts too. You can sketch a portrait using this on a graphing calculator.

This is a proof of the Pythagorean by US President J. A Garfield.

As you can see, when you line up the triangles like this, it forms a trapezoid. One way to find the area of the trapezoid is by adding up the areas of all the triangles that make it up, so we get $$A = \frac{ab}{2} + \frac{ab}{2}+ \frac{c^2}{2}$$. Also, we can find the area using the formula for a trapezoid, like this $$A = \frac{a+b}{2}*(a+b)$$. Now when we set the areas equal to each other, we get $$\frac{ab}{2} + \frac{ab}{2}+ \frac{c^2}{2} = \frac{a+b}{2}*(a+b) \implies ab + \frac{c^2}2 = \frac{(a+b)^2}2 \implies 2ab + c^2 = (a+b)^2 \implies a^2+b^2=c^2$$