# Discrete and Continuous

Another aspect of the two paintings posted in Why Art is that they illustrate the difference between discrete and continuous, at least in terms of color.

In Keith Haring’s painting, every point is either yellow, red, green, or purple: four discrete possibilities. In Wayne Thiebaud’s painting, there are many continuous transitions from one color to another. Notably, for me, a field that changes from gray to brown to indigo. If you look at the body of work of these artists, you would see that they each stay pretty much in one color camp. Haring in the discrete camp, Thiebaud in the continuous.

In mathematics the discrete and continuous camps exist too. People typically pick one or the other. What camp they choose is an individual matter probably much as it was for Haring and Thiebaud. Is one camp “easier” or “simpler” than the other? Not necessarily. It’s tempting to think that the discrete camp is easier – all whole numbers, no messy extended decimals. No Zeno’s Paradox.

However, the continuum can be comforting because it is filled with solutions. We know, for example, that two non-parallel lines in the plane definitely intersect, which means that the system of equations defining two such lines always has a solution. Ask a similar question in the discrete world – does a given line intersect any lattice points in the plane? And that question is harder.

(By lattice points in the plane, I mean points with whole-number coordinates in the X-Y plane, such as (2,1) and (3, 56). These points form a regular grid of dots.)

Then there is the famous Diophantine equation (love that name, Diophantine, makes me think of an equation in a flowing white dress with a long trunk): In the continuum are many solutions. But if you ask about discrete solutions, you get Fermat’s Last Theorem, which stumped mathematicians for centuries.

On the other hand, the continuum is a very mysterious place. That’ll have to be another post.

# Why art?

I feel a need to say more about A Study in Scarlet (Rectangles). What place does a painting have in a math club?

I think it’s worth pointing out that the kinds of thinking people use when they think about art are similar to mathematical thinking. In fact, from my perspective, it’s all just thinking.

# Color

Before jumping in to some obvious observations about colors and numbers let’s look at two paintings. I think color is important in both of them:

When I look at paintings, I ask myself, what is this painting about?
If I think it’s about color, I ask myself, in what way is it about color?
This is similar to what happens when I look at math problems. A math problem can seem impossible, or like I’m stumped, until I start to ask myself, what is this about? Is it about numbers or geometry? etc.

Back to the paintings, Keith Haring’s painting uses four distinct colors. Two colors are allowed to drip over the other colors. I think the bold colors express strong or bold feeling, and the dripping shows us different relationships between colors; red next to yellow is different from red next to green.

Wayne Thiebaud’s painting shows he put a lot of thought and care in mixing colors. Shadows and outlines are highlighted in slightly nonrealistic colors, lots of purples and greens, like reflections in abalone shell. When I see this use of color it is both familiar and surprising to me; yes, there are purples and turquoises in the shadows, but no, I don’t always see them.

Mixing colors is well known to be challenging. Just rendering something like a green sweater is not a simple matter of green paint and maybe gray for the shadows. Maybe the lighter parts are more yellow, maybe the dark parts more brown. Like numbers, colors can be “added together”. But unlike numbers, which have one dimension, a size, colors are said to have three dimensions: hue, saturation and value.

It’s easy to visualize the set of all numbers as a line. What would the set of all colors look like? Would it be fully three dimensional like a cube? Would it extend in three directions, or have boundaries? Would parts of the color space fold in on itself? (There are many ways to make gray, for example.)

If you look at a mostly monochromatic painting like the one in A Study in Scarlet (Rectangles) you can see how the interplay of value and saturation can play a big role if the hue is mostly left alone.

Here is more about Keith Haring.

# A Study in Scarlet (Rectangles) Professor Bear’s exponent, Marjorie, saw this painting in a museum in France. It’s a beautiful example of a (mostly) monochromatic painting. Red. And it is a study in rectangles. It’s interesting to see where the artist broke lines and boundaries and symmetries, and where the artist continued them.

The painting has both representative and abstract features. It looks like a scene in traffic, and achieves light and depth. Small touches of contrasting colors, greens and blues, make the light very interesting. In places there is strong three-dimensional feeling, and in other places it’s more two-dimensional.

Enjoy.

# Proof that the square root of 2 is irrational

If you’re interested in some math recreation: the square root of 2 is a number like pi. It’s irrational, meaning that its decimal expansion does not terminate or repeat. It goes on forever with no pattern. You could never fully specify it as a decimal. Numbers that do have a terminal or repeating decimal expansion are called rational numbers.

Proof (I will be longwinded):

Claim: if a number has a terminal or repeating decimal expansion, then it can be written as A/B, where A and B are whole numbers.

Non-rigorous Proof of Claim: The case of the terminal expansion is easy. A decimal that ends, such as 0.12345 is just 12345/100,000.

In the case of the repeating decimal, the proof involves a trick. Suppose a number S = 0.abcabcabcabc… repeating abc forever.

Then, 1000*S = abc.abcabcabc… repeating abc forever.

1000*S – S = abc.abcabc… – 0.abcabcabc… = abc (if you don’t believe it, let a = 1, b = 2, c = 3)

Therefore, 1000*S – S = abc.

1000*S – S = 999*S (if S is an apple, this is like saying 1000 apples – 1 apple is 999 apples).

999*S = abc

This means that S = abc/999 and we have converted the repeating decimal S into a fraction of the form A/B. A similar trick works for any repeating decimal.

Exercise: try proving that the repeating decimal 0.1111… is a simple fraction. What is it?

The proof that the square root of 2 is irrational hinges on showing that it is impossible to write the square root of 2 as a fraction A/B. In other words, the square root of 2 cannot rational. This is a technique called proof by contradiction. It’s often used in squirrelly matters such as infinity.

Proofs by contradiction all have a similar format: suppose something outrageous. Prove that it violates logic. If every step of the argument is logically sound, then the original outrageous assumption must be the one that is false.

Outrageous assumption: the square root of 2 has a terminating or repeating decimal.

Implication: if true, then there are numbers A and B where the square root of 2 is A/B.

Implication: furthermore, it’s possible to find A and B where the fraction A/B is in lowest terms, in other words, A and B have no common factors. (You know from elementary school math that common factors can be canceled out. So we assume they are canceled out.)

(refresher: 1/3 is in lowest terms. 2/6 is not in lowest terms).

If A/B = square root of 2,

then (A/B) * (A/B) = 2

This means that A*A / B*B = 2

This means that A*A = 2*B*B.
This in turn means that 2 is a factor of A*A.

Claim: The only way this can happen is if 2 is a factor of A.
Proof of Claim: Let S*T be the product of two numbers. Then either:
S is even, which makes S*T even; or,
T is even, making S*T even; or,
S and T are both even, making S*T even, or
S and T are both odd, making S*T odd.
In the special case S = T = A, and A*A is even, then A must be even, so 2 is a factor of A.

The number A must be of the form A=2*K.

That means that A*A = (2K)*(2K) = 4 K*K.

If 4*K*K = 2 B*B,

then 2*K*K = B*B.

The only way this can happen is if 2 is a factor of B.

However, this violates the implication that A and B have no common factors.
Numbers A and B that satisfy the original assumption cannot be found.

(This is a more detailed version of the proof you find on wikipedia.)