 An important strategy for doing mathematics, generalize. In the 1840s, William R. Hamilton considered the problem of generalizing the complex numbers. So just to refresh your under-complex numbers themselves, if a and b are real numbers, then a plus bi is a complex number where i squared is equal to negative 1. Now we add complex numbers component-wise where we treat i as a variable like x. So for example, if we add 3 plus 5x and 2 plus 3x, we get 5 plus 8x. And similarly, if we add 3 plus 5i and 2 plus 3i, we get 5 plus 8i. We can also multiply complex numbers with the added simplification that i squared is equal to negative 1. And so if we multiply 3 plus 5x by 2 plus 3x, we get. But if we multiply the complex numbers, 3 plus 5i times 2 plus 3i, we get. And since i squared is equal to negative 1, we can go one step further and get. Now let's talk about William R. Hamilton. By the age of 9, the Irish mathematician William Rowan Hamilton could speak French, Italian, Latin, Greek, Hebrew, Persian, Arabic, Sanskrit, Caldian, Syriac, Hindustani, Malay, Maratha, Bengali, and others. Hamilton was on track to study linguistics, but then he encountered another prodigy. Sarah Colburn of Vermont was a calculating prodigy whose father took him on tour. People paid to see him answer questions like what is the cube root of 268,336,125? Or factor 247,483? Or how many black beans does it take to make three white beans? After touring the eastern states, the Colburns made their way to Europe, and in September 1813 they arrived in Dublin. Now Colburn and Hamilton were both around 9 at the time and both had a reputation for a computational facility, and so in September 1813 they fought a computational duel, which Hamilton lost, and decided to put more effort into studying mathematics. Colburn on the other hand eventually returned to the states and would become a professor of languages at Dartmouth College. In the 1840s, Hamilton tackled the problem of generalizing complex numbers. That useful property of complex numbers involves modulus. The modulus of the complex number a plus bi, written this way, is the square root of a squared plus b squared. Now remember a and b are both real numbers, so this is always going to be the square root of a non-negative numbers. Now complex numbers satisfy what's known as the modulus rule, and we'll express it this way, if I find the product of the moduli of two complex numbers, what I get is the modulus of the product of the two complex numbers. And so the question is, could we make a hypercomplex number a plus bi plus cj that satisfies the modulus rule as well? So let's consider a hypercomplex number a plus bi plus cj, where a, b, and c are real numbers, and i squared and j squared are both equal to negative one, and we'll assume that plus or minus one, plus or minus i, plus or minus j are all different. Now if we want to make sure the modulus rule holds for these hypercomplex numbers, then we have to extend our definition of the modulus, and the natural way to do that is to define the modulus of a plus bi plus cj to be the square root of a squared plus b squared plus c squared. So we want to preserve the modulus rule. So let's consider the square of a hypercomplex number. Preserving the modulus rule is going to require that the square of the modulus be the modulus of the square. Now since the left-hand side is the square of a modulus, we can calculate the modulus and square it, and we get, meanwhile the right-hand side requires us to perform the multiplication first. So let's expand this in the way that we might do so, and so we get, and the thing to notice here is that if I take the modulus of everything except for the last term, I get, and we simplify, we find that the modulus is, and what this means is that if we ignore this last term, the modulus rule holds. But how can we ignore that last term? This seems to require that this product ij has to be equal to zero. Hamilton wrestled with this problem for a long time, and it became something of an obsession with him. It occupied all of his waking hours, and when they came down in the morning, his sons would ask him, have you learned how to multiply triples yet? And Hamilton replied, sadly, I can only add or subtract them. And so he struggled with finding a solution until one day on October 16, 1843, Hamilton took a walk with his wife along the Royal Canal in Dublin. As they approached the Broom Bridge, this is how it's spelled, Hamilton realized he could save the modulus rule if he made an unintuitive choice. If ij was equal to negative ji, in that case, if ij was negative ji, then when we expanded a plus bi plus ej squared, then the bcij and the bcji terms would cancel. Our last term would drop out, and the modulus rule would be saved. Hamilton was so impressed with his discovery that he carved the fundamental rule of multiplication into the Broom Bridge. Now while to save the modulus rule, it did raise a new question, what was ij? To determine that, we note that if we look at the product one plus i times one plus j, we get one plus i plus j plus ij. So if ij is equal to one, then the product one plus i times one plus j will be two plus i plus j. And since we want to preserve the modulus rule, we must have the product of the moduli equal the modulus of the product. But if we actually calculate these moduli we get, and that says ij can't be equal to one. And by a similar argument, ij can't be negative one or i or negative i or j or negative j. In fact, ij can't be any number of the form a plus bi plus cj for any real numbers a, b, and c. ij must be a new imaginary k. And the result of all of this is a new type of mathematical object. The only way to form hyper complex numbers that satisfied the modulus rule was to take three imaginaries, i, j, and k, where i squared, j squared, k squared, and the product ijk are all equal to negative one. Hamilton referred to a hyper complex number of this form as a quaternion after a Roman infantry unit of four soldiers.