The Pythagorean Theorem shows how strange our concept of distance is. Using the rule a^{2} + b^{2} = c^{2}, we can trade some "a" to get more "b".

Starting with

means "A 13-inch pizza equals a 13-inch pizza". Sure. But we can trade an inch and get:

Huh? A 12-inch pizza and a 5-inch pizza equal a 13-inch pizza?

The math works (144 + 25 = 169) but, but... we gave up an inch and got a five-inch pizza!

Let's understand why the tradeoff happens, and how to use it.

Contents

- Explanation 1: Shaving the Square
- Explanation 2: Sliding the Chopstick
- So What's the Tradeoff?
- Optimizing The Tradeoff
- Maximizing Even Payouts
- General Technique: Finding the Best Direction
- Maximizing Uneven Returns
- The Gradient Principle
- Questions I Had That You Might Have Too
- Final Thoughts
- Other Posts In This Series

## Explanation 1: Shaving the Square

A key insight: **Bigger numbers are harder to square**.

Imagine laying tiles on a porch -- as your porch grows, the outer layer needs more tiles. Trimming a 13x13 porch to 12x12 frees up 25 tiles, which is enough to make a new 5x5 porch!

I call this "shaving the square". Trimming 1 unit from the outside of a large square has more "shavings" which can contribute to a smaller one (trimming an inch from a giant fro can make a sweater for an infant). As we continue to trim, the benefit diminishes because our starting point is smaller and smaller.

## Explanation 2: Sliding the Chopstick

A second insight: **Slide a little, pivot a lot**.

Imagine a chopstick wedged in a corner: the length is fixed, and the ends of the chopstick must touch a wall. What're the options?

Well, laying on a single wall means 100% for one side (like saying 13^{2} + 0^{2} = 13^{2}). Not that interesting.

By sliding the chopstick (from 13 to 12) we can swing it *out* by 5 on the other wall!

You need to try it -- a small slide gives a giant pivot. As we keep sliding, the tradeoff (How much pivot do we get?) changes.

## So What's the Tradeoff?

Time to see how the a/b tradeoff works. First, let's use grid coordinates: x & y (horizontal and vertical). Given a fixed distance (13 units, let's say), our options lay on the circle where x^{2} + y^{2} = 13^{2}:

A few points:

- Each possibility is the same distance, but has a different ratio of x to y (100% x, 100% y, or a mix like (12,5))
- We can only move to neighboring points on the circle (options at the same distance)
- The tradeoff we face is how much "x" we get for "y" when moving to a neighbor. If we're at (0, 13) we could move to (5, 12). This trades 1 y for 5 x's.

This is the "chunky" tradeoff where we're using an entire unit at a time. What about .5 units? .01?

Enter the tangent! The **tangent line** shows the trajectory of our current path, the direction to our neighbor. We follow the tangent for a tiny, microscopic amount to get our next neighbor. The tangent is an approximation -- it's not pointing exactly at our nearest neighbor, but it's pretty close.

**The tangent shows the tradeoff you are about to make.**

What's the actual amount? Any point (x,y) has a slope of y/x, and a tangent line with slope -x/y, so the tradeoff is...getting confused yet?

Less mindless algebra, more intuition:

- Circles have a tangent line perpendicular to the current point
- If you're at (5,12) then tangent slope is some ratio of 5 and 12
- Remember "shaving the square": you get a better deal in the direction of the smaller coordinate (increasing a large square is tough).
- So, at (5, 12) you're "heavy on the y" and the trade will favor improving your x: it should be "trade 5 y's for 12 x's". And why not the other way? It doesn't make sense that the more y you have, the
*easier*it is to get y! That'd spiral off into exponential growth, not a circle. - Lastly, we can't trade an entire chunk of 5 y's! The tangent is about our nearest neighbor. We have a trade of 12/5 or 2.4 to 1. Our next, tiny movement will be at this ratio (and then we'll be at a new point, with a new tangent).

General principle: Our neighbors are on a circle, which encourages balance. You get a better deal in the direction of the smaller coordinate: at (x,y) the tradeoff is y:x.

## Optimizing The Tradeoff

Now we know the tradeoff for any point (x,y) -- let's optimize!

In a boring scenario, we get paid based on pure distance, so every point (or direction to move) is the same.

The exciting scenario: our (x,y) position is an *input* into some other function which gives us a return! Now we want to maximize that function.

Here's a scenario: Popeye throws cars for cash. He lines up spectators on fences running North and East. The spectators must look straight ahead (they're in neck braces, due to earlier events) but will pay Popeye if they see a car pass in front of them.

## Maximizing Even Payouts

Suppose each spectator offers $1 if they see the car (Payout (x,y) = x + y). Where to throw?

First, assume Popeye has finite energy -- he can throw the car 13 meters. Now let's start somewhere: throwing the car pure North (0, 13):

P(0,13) = 0 + 13 = $13

Ok. What if he threw it slightly East? To (5, 12) let's say?

P(5,12) = 5 + 12 = $17

Clearly better. This should make sense: at (0,13) the tradeoff is *great* to get more East. We can give up 1 North and get a whopping 5 East, a "profit" of $4 if we do the trade. We should keep trading as long as it's profitable -- as long as we're out of balance, the circle will reward us for boosting the smaller side. Following a 45 degree angle for 13 units is the ideal:

P(13 * 1/sqrt(2), 13 * 1/sqrt(2)) = P(13 * .707, 13 * .707) = 9.2 + 9.2 = $18.4

Neat. A 45-degree throw hits 70.7% of the possible spectators for each side.

Psst. Confused about how a 45-degree through passes by 70.7% of the spectators on each side? No problem.

A 45-degree throw is along the diagonal of a square. A triangle with sides 1 and 1 has a hypotenuse of:

And has sides (1, 1, 1.414).

A hypotenuse of √(2) isn't convenient: it's hard to know what fraction a side is of the whole. We divide the triangle by the length of the hypotenuse (√(2)), making the hypotenuse 1 and the other sides a percentage:

Now we've discovered that a 45-degree throw, with sides (1, 1, √(2)), has the ratio .707, .707, 1. 70.7% of the distance along the hypotenuse shows up on each side.

## General Technique: Finding the Best Direction

We stumbled upon the way to find the best return:

- Pick any starting point / direction
- Tweak it: if our return improves, keep the new choice (it's profitable)
- Keep tweaking until our return is no longer profitable

In math slang, this is "finding the local maximum". In economics slang, it's finding the point of "zero marginal returns". Popeye calls it Squeezing the Spinach.

## Maximizing Uneven Returns

Now suppose the Northern spectators offer $2 (Eastern stay at $1), so P(x,y) = x + 2*y. Should we throw it 100% North?

P(0, 13) = 0 + 2*13 = $26

Not bad. But what about 45 degrees again?

P(9.2, 9.2) = 9.2 + 2*9.2 = $27.6

Interesting -- 45 degrees is still better! But... I think we went too far! Shouldn't we favor North since it pays more?

Yep. Let's remember how to Squeeze the Spinach (maximize our returns): start with North and change until it's not profitable:

- The payout function means 1 North = 2 Easts (North pays $2, so 1 unit North = 2 units East)
- Trades are profitable if we can beat 1 North for 2 Easts (1 North for 3 Easts, for example, would profit $1)

So... where are trades *better* than 1 North for 2 Easts? In the Northern section, where the circle rewards us by throwing Easts at us ("Please, please go East... I'll give you a bunch if you give up a little North").

Remember how circles are about x/y, x & y, x:y, etc.? Well, we have the numbers 1 and 2. (2,1) is in the East section. We want (1,2). Why? At (1,2) we have reached the perfect 1 North = 2 East tradeoff.

Following the direction (1,2) for 13 units is:

P(13 * 1/sqrt(5), 13 * 2/sqrt(5)) = P(5.81, 11.62) = 5.81 + 2*11.62 = $29.05

Tada! Over 29 smackeroos because we maximized our return.

## The Gradient Principle

We can supercharge this result:

**To maximize return, go in each direction proportional to its payoff.**

If North pays 2:1 compared to East, your trajectory should favor North by 2:1. In mathier terms:

- Payoff(x,y) = a
*x + b*y - Best trajectory = (a, b) [in our case, (East, North) => (1, 2)]

And this works in multiple dimensions! Given 3 dimensions, go in a direction (Payoff(x), Payoff(y), Payoff(z)). Vector calculus fans, this is why the gradient is in the direction of greatest increase.

The gradient for F(x,y,z) is

And each partial derivative (dF/dx) is the payoff for moving in that direction.

But does it all balance? Suppose x pays 3, y pays 4, and z pays 5 (at the current position). The 2-dimensional tradeoff trajectories are:

Now for the magic: the combined trajectory

satisfies all 3 requirements! On the x-z plane, x doesn't care about y -- as long as the ratio to z is (3 , ?, 5) you're getting the best tradeoff from the x-z perspective. The pairs are:

- (3, ?, 5)
- (?, 4, 5)
- (3, 4, ?)

You don't need a sudoku master to see (3, 4, 5) satisfies all those proportions.

Still not convinced? Imagine the payoff for y was zero. We don't want to waste energy in our trajectory (3, ?, 5) in a useless direction. But that can't happen, because the y-z tradeoff will be (?, 0, 5) and the x-y tradeoff will be (3, 0, ?). The x-z tradeoff lets y-z and x-y "figure out" what y should be, which is 0.

## Questions I Had That You Might Have Too

**Q: I still don't get why this works at all. Somehow 50% in x and 50% in y leads to .7 + .7 = 1.4?**

It's a deep question about *why* space behaves like this. I was going crazy staring at chopsticks on a wall.

Here's my answer: distance is distance. 13 units is 13 units. But in some situations we are "measuring our coordinates" (what are the values of x & y) and *not* the distance itself.

Cartesian coordinates (x-axis, y-axis) are very inefficient for diagonal motion (i.e., you are measuring the sides of the triangle, not the hypotenuse). When .707^{2} + .707^{2} = 1, it's a measure how how "inefficient" our x & y coordinates are being. We used 70% of each coordinate to represent an object that could have been 100% on one (i.e, if we used polar coordinates).

**Q: I have an offshore investment with 200% return, and an onshore one with 5% return. I have $1000 to spend -- should I split my money?**

Heavens, no! Remember, this principle is about *distance measurements on a grid* with the idea that 50% in x and 50% in y covers "more ground" than 100% in x. In investing 1) money is not on a grid and 2) there's no distance bonus. Putting half your money in each is plain old 0.5 + 0.5 = 1.0. Giving up $1 of the offshore investment gives you $1 for the onshore one.

Put all your money in the best investment.

**Q: So all this stuff is useless?**

Heavens, no! Ask yourself: am I measuring distance on a coordinate system?

Many things are measured in terms of x-y coordinates (physical phenomena, etc.) and *do* have the Pythagorean distance tradeoff.

But not every graph is the same. Graphs that aren't about distance (like "Money vs. Time") do *not* get any boost from the Pythagorean theorem. This confused me for a long time: the Pythagorean Theorem works for coordinate distance!

## Final Thoughts

The Pythagorean Theorem is so versatile -- it's not about triangles, it covers the nature of distance. I seem to find some new realization when I study it. Really grokking it will help you everywhere, from geometry to vector calculus.

Happy math.

## Other Posts In This Series

- Vector Calculus: Understanding the Dot Product
- Vector Calculus: Understanding the Cross Product
- Vector Calculus: Understanding Flux
- Vector Calculus: Understanding Divergence
- Vector Calculus: Understanding Circulation and Curl
- Vector Calculus: Understanding the Gradient
- Understanding Pythagorean Distance and the Gradient

## Leave a Reply

24 Comments on "Understanding Pythagorean Distance and the Gradient"

[…] Points in the direction of greatest increase of a function (intuition on why gradient works) […]

cool,nice….but what do u want to say?

sry kiddng he he…

Wow. These wonderful sparks of intuition just keep getting better and better each time – truly amazing!

As soon as my understanding of the intuition for this concept of pythogorean distance really sinks in, I’ll perhaps be able to add something to the distance concept, through folding-paper thought experiments… there’s really a lot of insights that can already be derived from just “folding paper” in unique ways.

Wonderful, intriguing work, Kalid!! :D

@Ashish: :)

@Stan: Thanks! Yep, definitely send along any insights, would be cool to see!

Great new look.

@mark: Thanks!

[…] education, Without Geometry, Life Is Pointless started a new series on habits of the mind, and at Better Explained there was a nice post on how to enable students to understand (and not just learn) the Pythagorean […]

I like how you explain concepts intuitively. Pythagorean theorem is quite simple to many, but for young students who still have a vague notion of proofs, we need intuitive explanation.

[…] I’ve gained immense value from upgrading the bow that holds the Pythagorean Theorem. That “arrow” (a2 + b2 = c2) can be launched in so many ways — each year I find a new personal discovery (it’s not about distance; it can apply to any shape; it explains the gradient). […]

Wow… Extraordinary!

I didn’t know the Pythagorean Theorem could be used up to such extents, and with so many scenarios! Thanks for the elaborations. :)

@Joey: Glad you liked it — you’re welcome!

How about proving the pythagorean theorem in your hands with a new shape-making ruler. This thing is too cool!

http://www.kickstarter.com/projects/koalatools/rule-like-never-before-new-shape-making-versa-rule

kalid: I’m with you up to “We can only move to neighboring points on the circle (options at the same distance)”. To what circle are you referring? The circle that includes points (13,0), (12,5), (5,12) and (0,13)? Shouldn’t the slopes be negative? Thanks.

Hi Pat, yep, I’m referring to the circle of radius 13 (which includes the points the points you mentioned). Every point on this circle is the same amount of “effort” (distance) to get to, so it’s a question of which gives us the best payoff for that identical effort.

Some of the slopes (rise/run) are indeed negative, and you can see this as “you give up some North (rise) and gain some East (run)”.

Hi Khalid,

I did not understand

“At (1,2) we have reached the perfect 1 North = 2 East tradeoff.”

(1,2) = one unit towards East and 2 units towards North.

But for perfect trade off (equate X and Y or east and North) we would need 2 units east and 1 unit North i.e. (2,1).

Am I really missing a basic point. Would appreciate your help! Thanks!

What a wonderful post, and a truly great blog! I have a very literal mind, and I just *love* the way you describe abstract concepts in terms of physical things (chopsticks, a porch, pizza…). Thank you SO much for the time and effort you have put into this site! It has helped me so much.

@ Rohan

Kalid writes: Payoff(x,y) = ax + by

Best trajectory = (a, b) [in our case, (East, North) => (1, 2)]

As you go “up” along to the y axis, you go to North, not East (it is a convention to show the North direction as “up”). You can see in the equation of Payoff, b is associated with y axis, and thus North.

Similarly, you can see that a is associated with x, and thus East.

“Psst. Confused about how we got .707? No problem. Taking sides of 1 and 1 means the hypotenuse is 2:”

Surely this is an error? a^2 + b^2 = c^2, yeah? so 1^2 + 1^2 = c^2, then c = sqrt(2).

@Jacob: Whoops! Thanks for the correction, I cleaned up that section.

wow! just wow!!

I am viewing and reviewing your sequence on vector calculus as it is quite good and I need to improve this area.

So overall: excellent job. I have one grave concern though.

Your statement “Coordinates with perpendicular axes are very inefficient, especially for diagonal motion (i.e., you are measuring the sides of the triangle, not the hypotenuse). “

I think this is extremely misleading.

One of the key lessons from Linear Algebra is that forming orthogonal basis is incredibly useful. And that extends to diagonal motion (or any ‘line like’ motion). I think what you meant to say is that cartesian coordinates can be inefficient at times.

Indeed you could choose a different orthogonal basis like [1,-1] and [1,1] (sorry I can’t seem to get LaTeX to render in the comments) which you could much more ‘efficiently’ use to describe your diagonal line. And of course, if you wanted said basis to be orthonormal, you would scale it down a touch to be approx: [0.707,-0.707] and [0.707, 0.707], and your diagonal line in the picture would described simply by [0, 1].

Hi Derek, great catch! Yep, better to clarify that Cartesian coordinates are generally inefficient. Thanks!

i am seeing this type of maths for just about thre weeks now so i’m still trying to understand it.

This is a Godsend!

Perfect compliment to the best math teacher on YouTube, Krista King.