If you're anything like me, you've probably wondered what the quadratic formula would look like in an alternate universe where triangling numbers was the default quadratic function instead of squaring.

Okay, maybe not.

But that's what this blog post is about! I've named this alternate universe Triangland.

*Welcome to Triangland*

Triangland (technically, the Triunited Kingdom) is a place much like our own, but with one major difference: instead of squaring numbers, people triangle numbers. If you ask the average Trianglish elementary school student what "three squared" is, they won't know the answer, but ask them for "three triangled" and they'll instantly respond "six." This blog post will cover some of the cool features of triangling numbers that every schoolchild in Triangland learns, culminating in the Trianglish Quadratic Formula.

Triangland (technically, the Triunited Kingdom) is a place much like our own, but with one major difference: instead of squaring numbers, people triangle numbers.

But first, let's step back.

*What Does It Mean to Triangle a Number?*

To square 4, we can imagine creating a square of dots with side length 4. This square must contain 16 total dots, so 4 squared equals 16.

Triangling 4 works in the analagous way: we imagine creating a triangle of dots with side length 4. (The triangle doesn't have to be right, but it's easier to depict that way.) This triangle must contain 10 total dots, so 4 triangled equals 10.

In Triangland, this would be written 4△ = 10. (In our world, it's typically written T_4, but I prefer the Trianglish way.) For a bit of flavor, I'll also be writing the previous result as 4☐ = 16.

For the rest of the pictures in this post, I'll be using little squares instead of dots, as shown below, as they make some properties easier to see. (Speaking of the pictures in the post, they were made by me using __Manim__, which is used in __3Blue1Brown__ videos!)

If you've heard of the triangular numbers, they were probably defined as a sequence where you add 1, and then add 2, and then add 3, and so on to get the next term. This is the exact same thing. If you think about rows, it should be pretty clear why n△ = (n–1)△ + n for all n.

In any case, we can see several cool facts about squaring numbers from this picture that do not hold for triangling numbers. For instance, a square with side length 4 has an area of 4 squared, or 16. But a triangle with side length 4 does not have an area of 4 triangled—it has an area of half 4 squared. In fact, we can see that triangling doesn't actually make a triangle—it makes a staircase. This is a bit annoying. (It's worth noting that if n△ *was *defined to be the area of a triangle with side length n, even an equilateral triangle, it would just be n☐ times a constant and therefore would be boring.)

Triangling doesn't actually make a triangle—it makes a staircase. This is a bit annoying.

This picture also provides a visual proof of the beautiful Square Identity: that n☐=n*n. While we often take this fact for granted, in Triangland proving it is a canonical problem given to advanced high school and college students if they choose to learn about squaring numbers. (It's slightly less trivial to prove algebraically, since in Triangland squaring is typically defined in terms of triangling, but that's just an aside.)

*The Triangle Identity*

You may wonder if there is a similar identity for triangle numbers, and it turns out the answer is yes! We can arrange two copies of n△ to get an n-by-n+1 rectangle, so n△ = n(n+1)/2. This is the Triangle Identity.

In fact, this will be more than an identity. Since triangling is currently not defined on numbers besides positive integers, we simply declare that n△ = n(n+1)/2 (or (n☐+n)/2 for *all* n: negative integers, fractions, irrational numbers, even complex numbers. This will make things easier, even if it doesn't* mean* anything.

(There's a long tradition in math of defining things in this way. People often ask why any number to the zeroth power equals 1. The __"mwahaha, I'm right"__-style proof goes something like "When we raise something to the zeroth power, we're not multiplying anything, and 1 is the multiplicative identity" which is true but has probably never convinced anyone of anything. A more accurate reason is that it *isn't* inherently anything. It's just that (3^4)/(3^3)=3, and (3^3)/(3^2)=3, and (3^2)/(3^1)=3, so it sure would be nice if (3^1)/(3^0) equaled 3. To make this work, mathematicians just *define *n^0 to equal 1 for all n. (__Except, mayyyyyybe, n=0__.))

Just as we can define triangling in terms of squaring, in Trianglish squaring is defined in terms of triangling, as in the picture below. This is also how the idea of "squaring" a number is generalized.

*The Trianglish Quadratic*

Right, it's time for the fun part. In our world, a typical quadratic equation looks like x☐ – 5x + 6 = 0. There are multiple ways to solve this. The easiest is to factor it as (x – 2)(x – 3) = 0, which you can check is equivalent. Now, we have two terms multiplying to 0, so one of them has to be 0. Therefore, either x – 2 = 0 (so x = 2) or x – 3 = 0 (so x = 3). Those are our two solutions.

Sadly, not every quadratic can be factored like this. x☐ – 5x + 5 = 0 is also a perfectly good quadratic, but it can't be factored using just whole numbers.

However, we can plug our coefficients into the mystical magical quadratic formula (negative b plus or minus the square root of b squared minus four ac, all over two a). If we laboriously calculate this (or if we just plug our quadratic into __WolframAlpha____)__, we get that x = 5/2 – (√5)/2 (~1.38) or 5/2 + (√5)/2 (~3.62).

A typical Trianglish quadratic looks like x△ – 3x + 3 = 0. How can we solve this?

Well, by the Triangle Identity, we can rewrite this as x(x+1)/2 – 3x + 3 = 0. This equals x(x+1) – 6x + 6 = 0, which expands to x☐ – 5x + 6 = 0 using the Square Identity. Now, shock surprise, we've reduced it to the previous example. We can factor or apply the quadratic formula, and x = 2 or 3.

We can actually do this for variables if we want. The solutions to ax△ + bx + c = 0 are given by (–(a+2b)±√(a+2b)^2–8ac)/2a. Unless I made a mistake. But the exact equation is unimportant.

But let's be real. No one in Triangland would solve a quadratic like this.

But let's be real. No one in Triangland would solve a quadratic like this. First of all, it involves the Square Identity, which most students don't even learn about. Second of all, it involves taking a square root! How do we, in our square-centric world, compute square roots? Guess and check. Computers are now *really good *at calculating them with guess and check, but it's still guess and check. In Triangland? No way. Trianglish cultural considerations aside, it's just not a nice way to solve an equation. Luckily, we can do better.

*The Triangular Root*

To solve a Trianglish quadratic the Trianglish way, we have to introduce the triangular root. The triangular root of a number is exactly what it sounds like: what we have to triangle to get that number. Like square roots, most numbers have two triangular roots. For example, the triangular roots of 6 are 3 and –4, since 3△ = (–4)△ = 6.

In Triangland, the √ symbol is used to denote the triangular root, but that would probably confuse this audience. Therefore, I'll be notating the triangular root of n inline as trirt(n), which is a bit ugly but at least palindromic. When I have access to full math notation, I like to use the symbol for a cube root but with a triangle instead of the 3.

Note also that while our √n is defined to be specifically the positive square root of n (therefore √16 = 4 but not –4: that's what the ± is doing in the quadratic formula), in Triangland, trirt(n) is defined to be ambiguous, so trirt(6) refers to *both *3 and –4. That's because the triangular roots are not negatives of each other, so writing one in terms of the other is awkward. (I guess you could do ±(trirt(6)+1/2)–1/2? But that's awful.)

Interestingly, triangular roots are real for all numbers greater than or equal to –1/8. This seems random at first, but consider this: our formula for x△ is x(x+1)/2. When x is positive, both x and x+1 are both positive, so x△ is positive. When x is below –1, both x and x+1 are negative, so x△ is *still positive*. It's only when x is between –1 and 0 that the output is negative. It seems sensible that x△ would be minimized at x=–1/2, and this is in fact the case. That minimum is (–1/2)(1/2)/2, or –1/8. The triangular root of a value below –1/8 would be a complex number.

Any decent Trianglish calculator will have a triangular root button, but if you want to calculate them in terms of square roots, you can basically plug x(x+1)/2 = n into the quadratic formula to get the triangular root of n.

You might be wondering: if we wanted to take the triangular root of 2, which isn't a triangular number, what does that *mean*? Unfortunately, I don't have a satisfying answer. The square root of 2 is the side length we would need for a square to have area 2. But, as I mentioned, triangular numbers really make staircases instead of triangles, so there's not really a nice geometric interpretation. At least, I haven't found one that I like.

Any decent Trianglish calculator will have a triangular root button, but if you want to calculate them in terms of square roots, you can basically plug x(x+1)/2 = n into the quadratic formula to get the triangular root of n. If you do this, you get that the triangular root of n is equal to (±√(8n+1)–1)/2, which isn't pretty, but, you know, you can compute it. If you do, you'll find that the triangular roots of 2 are approximately 1.56 and –2.56.

*Solving the Trianglish Quadratic*

So we've got the triangular root in our toolbox now. This will allow us to easily solve any Trianglish quadratic that looks like x△ = 5: we simply get trirt(5).

We can now get a formula for any quadratic of the form ax△ + c = 0. Divide everything by a, that gets us x△ + (c/a) = 0. Move the c/a to the other side, and we have x△ = –c/a. Now we're done: x = trirt(–c/a).

But what about ax△ + bx + c = 0? That's not clear at all.

Before we tackle this directly, there's another form of Trianglish quadratic that's really easy to solve: one that looks like (x+2)△ = 5. But we can actually make any quadratic look like this!

*Completing the Triangle*

Before we tackle this directly, there's another form of Trianglish quadratic that's really easy to solve: one that looks like (x+2)△ = 5. We get that x+2 = trirt(5), so x = trirt(5) – 2.

But we can actually make any quadratic look like this! To do so, we will use the the fact that, for any m and n, (m+n)△ = m△ + mn + n△. I was *very *excited when I discovered this identity, because it's basically a godsend.

The picture below shows why it's valid: let m△ be the blue and n△ be the red. The purple is mn, and the entire picture is (m+n)△.

(You may know that (m+n)☐ = m☐ + 2mn + n☐, and wonder if the pattern continues with higher __polygonal numbers__, which I haven't actually defined here but are analogous. The pattern does in fact continue! (m+n)⬠ = m⬠ + 3mn + n⬠; (m+n)⬡ = m⬡ + 4mn + n⬡; etc. It's not too difficult to prove this algebraically. It's probably also pretty easy to prove it geometrically but I still don't have a great intuition for what these polygonal numbers look like.)

So let's take x△ + 4x + 7 = 0 as an example. How are we going to solve this?

We now know that x△ + 4x + 4△ = (x+4)△. So we add 4△ to both sides of the equation, completing the triangle, and watch magic happen:

Yay!

Now, we can do the exact same thing for variables! The only difference is that the first step is dividing everything by a, the coefficient of x△. After that, just to make things clearer, we let capital B equal b/a, and capital C equal c/a. Then we switch back at the very end.

We did it! We found the Trianglish Quadratic Formula!

x equals the triangular root of the expression (b/a)△–c/a, all minus –b/a. I think this is way nicer than our quadratic formula. It's especially nice when a=1: x is just the triangular root of B△ minus C, minus B. No "minus 4ac," no "all over 2a." Without the simplicity of the Square Identity and a natural correspondence between triangling and triangle area, Triangland has it harder in many ways, but their quadratic formula is certainly more elegant!

Oh, by the way, the method I described is pretty much how you derive the standard quadratic formula, just by completing the square instead of the triangle. (If I hadn't known about completing the square, I doubt I'd have been able to derive the Trianglish quadratic formula.) Take x☐ – 6x – 7 = 0, which we rearrange to x☐ – 6x = 7. Since x☐ – 6x + 9 = (x–3)☐, we get the equation x☐ – 6x + 9 = 7 + 9, implying (x–3)☐ = 16, so x–3 = √16 = 4 or –4, and therefore x equals 7 or –1. Do this for variables, and you've got yourself the quadratic formula.

Oh, by the way, the method I described is pretty much how you derive the standard quadratic formula, just by completing the square instead of the triangle.

*Additional Problems*

There are still many aspects of math in Triangland that I haven't explored. For one, I alluded to Trianglish complex numbers. If you try to define, for example, i in terms of the triangular root of –1, is it nice? I don't actually know.

__Tetrahedral numbers__ exist, so in theory you could also try to find the Trianglish cubic formula. I didn't do this because I haven't actually learned the derivation for the regular cubic formula. (Though I know it involves depressing the cubics, which I suppose is kind of sad for them.) Also have you seen the __cubic formula__? It's way worse than the quadratic one. Though, since the Trianglish quadratic formula is nicer than the normal one, there's an outside chance the cubic one isn't actually a total monstrosity. Maybe a good starting point would be to solve Te(x) + cx + d = 0, where Te(x) is the xth tetrahedral number. I haven't tried this (yet? oh no am I committing to trying by writing this down) so I don't know how hard it is.

Regardless, I hope you enjoyed learning about Triangland!

–Jacob

PS: Before you go, here are two triangular number-related proofs without words (the first one I found, the second one I was told):

*Choose: Triangular Number-Related Proof Without Words #1*

The picture below proves that n choose 2, the number of ways there are to choose 2 objects from among n, is equal to (n–1) triangled. See if you can figure out why!

This picture can be extended to tetrahedral numbers and so on if you imagine a similar layout in higher dimensions.

(By the way, sorry this picture isn't in Manim like the others—I had already made it in Google Slides, and as it's a slightly involved diagram, I didn't feel like remaking it in Manim.)

*(xy)△: Triangular Number-Related Proof Without Words #2*

There's a neat expression for (xy)△ in terms of some smaller triangular numbers. See if you can figure out what it is from this picture!