real numbers that cannot be written without i

Added: 2026-06-05

[00:00:00]Today we're going to look at a really interesting result involving cubic polomial equations and the fact that they give rise to real numbers that cannot naturally be expressed without i the square root of minus1. In other words, we've got real numbers that kind of have to be written in terms of complex numbers with nonzero imaginary part. So, all of this is going to start with the following quadratic or cubic equation. x cubed plus px plus q. That should be a px. And you might say, well, could we have a more general cubic like one where there's a squared term? Well, in fact, we could start with one where there's a squared term, but you can do a pretty easy linear substitution to turn that into one where there is no square term. I'll let you look that up if you need to. So what we're going to start with is we're going to find a closed form for one of the solutions of this equation. In other words, we're going to derive the cubic formula for this type of cubic polomial equation. Okay. So it all starts with this. Let's perform a substitution. Let's set x equal to u + v.

[00:01:19]But then we can use a binomial expansion to pretty quickly see that x cubed which is clearly equal to u + v cubed will be equal to u cub + 3 u ^2 v + 3 u v^2 + v cubed. Again you can do that with binomial coefficients or just by multiplying it out. But now if we put this version of x and this version of x cubed into our equation, we'll see that we get the following thing. And I'm going to maybe skip some of the simplification steps. But this is going to turn into u cubed plus v cubed. And I'm going to group those two terms together. and then plus 3 u v plus p times u + v plus q. So again, that's just some symbolic manipulation to get from here to here using those expansions that we've used. But now if we want this to be equal to zero, let's see. Well, we could have 3 uv + p equal to zero. That's like using the idea of wishful thinking, if you will. So, the simplest way to make this thing equal to zero would be this coefficient. Maybe I'll underline that in orange equal to zero. And then, well, we could have this stuff that I'm underlining in red also equal to zero.

[00:02:56]And so of course a priori there could be a lot of other ways that this thing could be equal to zero but we could wishfully think that we have these two conditions and that will definitely give us zero. So let's maybe put that up here as a hope. So let's hope that we can have u cubed plus v cubed equal to minus q.

[00:03:27]So it's got to be equal to minus q because when we add it to q it has to be equal to zero. And then also u v = minus p over 3. That's from setting this orange thing equal to zero.

[00:03:42]And maybe I'll color code this here. So this orange dot comes from that orange underline and then this red dot comes from that red underline. But the next leap is going to be look at looking at the following quadratic function. So let's consider this. So we'll have t minus u cubed * t minus v cub. You might say well why would we look at that quadratic function? Well, notice if we multiply this thing out, we're going to get t^2 - u cub + v cub * t and then plus u cubed * v cub. In other words, u * v cubed. So in other words, what we have here is we've got this quadratic function which we can turn into a quadratic formula if you will by setting this equal to zero and solving that is in terms of these things that we know. U cub + v cub is minus q and then u * v is minus p over 3. We know those and remember our q and our p are given numbers. These 's and v's are uh maybe like parametric tools or tool parameters that we're using along the way. Okay. So anyway, if we rewrite this as t ^2 and then this is going to be plus q * t and then after that it's going to be minus p cub over 27. We have like a more reduced quadratic.

[00:05:22]Okay. So now let's solve this quadratic equation. And notice that on the one hand if we set this equal to zero and solve our solutions will be u cubed and v cubed. We'll have we'll have those two solutions. But on the other hand we get two solutions by using the quadratic formula on that quadratic equation. So let's observe that using the quadratic formula on that quadratic equation will give us well minus b plus or minus so on and so forth all over 2 * a but that's all over two. So I'm just going to leave that two in the denominator of the first term. Then I'm going to bring it under the radical of the second term. So here we're going to have minus q over 2 plus minus the square<unk> of so now we're going to have q^2 [snorts] over 4 and then plus p cubed over 27. So we've got something like that. But of course what we can do at this point is take the cube root of both of these parts and we'll have u and v. So notice that u and v will come from minus q / 2 and then plus minus the square<unk> of q ^2 over 4 + p cubed over 27 and all of that is under a cube root. So u is going to be one of those one of those I mean one of the ones with a plus or minus and v is going to be the other one with a plus or minus. But that means that u plus v will be the sum of these two things over here on the right hand side.

[00:07:01]But let's look over here and observe that our original substitution was x= u + v. So now we have x= u + v can be written as the sum of these big cube roots. And one of those has the plus that was inherited from the quadratic formula and the other one has a minus that was inherited from the quadratic formula. Okay, so let's summarize this over here and see how this actually is in line with our goal of finding a real number that can't be expressed without complex numbers. Thanks for sticking around this long of the video. If you're enjoying it, make sure and give it a thumbs up. And if you're not yet subscribed, consider subscribing. We're really close to 350,000 subscribers and we'd like to get to that number pretty quickly. Okay, so anyway, what have we done so far? We took the following like simple, if you will, cubic equation and we found a cubic formula for one of the solutions. But you might say, well, what about the other two solutions? Well, here's how it kind of goes. Regardless of what you have, the other two solutions can be given by well then this one right here. So, I'll just write that as box plus box. But then you could also have omega time the first one plus omega squar time the second one or vice versa.

[00:08:20]Omega squar time the first one plus omega* the second one. And you might say well what is omega? Well it's a so-called primitive third root of unity.

[00:08:30]And you can write it in complex form as min - one2 and then plus i * the<unk> of 3 over 2. So that's like the solution that is not equal to one of the equation x cubed minus one.

[00:08:51]But if you recall from quadratic equations, there's this thing called the discriminant. And there is here as well.

[00:08:57]And it's the thing underneath this square root which is inside of the cube root. So it'll be equal to q ^2 + p cubed over 27. Now since we have a p cub term notice that this discriminant which I'm calling delta can be bigger than or equal to 0 or less than zero. And now let's notice that if this thing is bigger than or equal to zero then our expression right here is real. So in other words this is a real number which makes this whole thing a real number and the same thing over here. So our entire expression is real. But strangely enough, that corresponds to having one real solution and two non-real solutions. But that makes sense because we've got two real numbers here that we're adding. And then we've got a non-real number times a real number and then the similar thing over here. So you could kind of guess that that's how it would go. But now if delta is less than zero, this expression right here contains nonreal numbers or nonreal expressions because we've got square root of a negative number inside of here. But that in fact corresponds to all three of the solutions being real numbers. They are just real numbers that perhaps cannot be expressed without this number i. Now sometimes they can, but sometimes they can't. And the way that I want to show you that sometimes they can is with the following example. And this is like a famous example that was first seen by Cardano who developed this cubic formula. and it was like one of his very first studies of this formula to see this weird thing that we're about to see. Okay. So anyway, let's consider x cubed minus 15x - 4 = 0. So we're going to consider that. Now I'm going to maybe just skip most of this, but you can plug all of these parts into this formula right here. And what you'll see is the following pops out. You'll see that one of the solutions of this is x = the cube root of 2 + 11 * i plus the cube root of 2 - 11 * i. So again, that's just basic arithmetic.

[00:11:20]But then also we have this strange factoring behavior. I guess it's not super strange, but it's strange in light of this equation right here. So, x cub - 15x - 4 will factor as x - 4 and then x^2arus or + 4x + 1. And you can use the quadratic formula on that second part. And what you'll see is that you have solutions four and then min -2 + the square<unk> of 3 and then -2 minus the<unk> of 3. So we in fact as promised over here have three real solutions even though our cubic formula expresses well at least one of them in terms of these complex numbers.

[00:12:22]Now well 4 - 2 + roo<unk>3 and -2 -<unk>3 one of those is this expression right here and the other two are this expression right here maybe flanked by omega. So the idea is to determine well which one is this? Is this strange number right here a weird way of writing four or one of the other two things? So let's see. So here's kind of the idea.

[00:12:48]We're going to take the cube root of 2 + 11 * i. And we're going to say that that equals a + b i. So it's it's a complex number for sure. So we're going to write it as the complex number a + b i. Now what we'll do is cube both sides of this equation. So cubing the left hand side, we'll just annihilate the cube root and we'll have 2 + 11 * i. And then we can just do some standard complex arithmetic over there on the right hand side. And we'll get a cubed minus 3 a b^2. So that's the real part. That's from cubing a and then multiplying a * b i^ 2.

[00:13:31]Keeping in mind that we get that three different ways when all is said and done multiplying it out. And then we'll have + i and then 3 a^ 2 b minus b cub. So we've got our real part and our our imaginary part, but we also have our real part and imaginary part over here.

[00:13:53]So just to underline this color coding, we've got 2 must be equal to this a cub minus 3 a b^2 and then 11 has to be the other one. So we might as well write a cub - 3 a b^ 2 = 2. and then 3 a^2 b minus b cubed has to be equal to 11. So something like that. Now we can like kind of cheat here if you will and use the fact that we know those solutions up here to start guessing the values of a and b. Or we could do it more algorithmically. It turns out just doing it algorithmically is maybe not technically that difficult, but not that very fun for a video. So, we'll just say that we'll take an educated guess that we could set a equal to two. But then if a is equal to two, and let's observe that that kind of makes sense from the fact that we've got a four there and we've got two in both of those places.

[00:14:57]So if we set a equal to two, well what are we going to get right here? So this second equation is going to collapse to 8 minus 6 b^2. So 6 b ^2 = 2. So something like that. But notice that tells us that b is going to be uh plus or minus1.

[00:15:21]But then we can argue that in fact B really has to be equal to plus one. And that's because if you look at this number in the complex plane, it's going to be in the first quadrant. And then if you look at this number in the complex plane, well the quadrant depends on the sign of all of these parts. But the sign of the imaginary part has to be equal to a positive number if we want to also land in the first quadrant. given that we already evaluated a to two. So that means we have a is 2 and b is 1. So that means we have the cube root of 2 + 11 i plus the cube root of 2 - 11 i is in fact equal to 2 + i + 2 - i. And you might say, well, where' the 2 - i come from? Well, notice that the cube root of 2 - 11 I is simply equal to the complex conjugate of this 2 + 11 I. So that's going to turn into two pl 2 minus I.

[00:16:25]Adding these together, we get four. So [snorts] in fact, this is just a really weird way of writing the number four.

[00:16:33]Okay, so now that we've seen this example of the fact that there's a weird way to express four with only complex numbers or using complex numbers. Let's revisit our original question which was the real numbers that you can't write without imaginary numbers. Well, notice that all three of these real numbers you could write them with imaginary numbers, but it's very easy to write them without imaginary numbers as well. So that means we still need to find a cubic equation that gives us a satisfactory example of this setup. So let's see what that would be. Before looking at this last example, I'd like to tell you about my second channel, Math Major, which has full lecture courses in mostly upper division math classes. And I keep the ads turned off on that channel thanks to my support via channel memberships here and Patreon. So, if you'd like to help me keep that ad free, maybe consider joining the channel here or becoming a patron. I've been uploading or releasing videos early for those two groups of people. So, there is that little perk.

[00:17:38]But, of course, you can always wait for the videos and there's no other pressure to do this. Okay. So, anyway, let's start looking at this maybe last example, which should give us a satisfactory example of our goal over here. So we've got x cub - 3x -1.

[00:17:56]So let's first observe that there are no rational roots and that's because we can see by the rational root theorem that the possible rational roots are plus minus one. Again look up the rational root theorem if you don't recall exactly what it is. It has to do with the factors of the constant term and the factors of the leading term and taking their quotient. So I've set f ofx equal to that polomial so that we can look at these things a little bit more streamlined. So anyway let's just check if we look at f of 1 we can easily see that that's equal to -3 which is not zero. And if we look at f of -1 you can easily check that that's equal to one which is not zero. So that means that there are no rational roots.

[00:18:47]>> [snorts] >> Let's also look at this. We can look at the derivative of this function. So that's going to be 3x^2 minus 3. Which means we have critical points where that thing is zero. That thing is zero at -1 and positive one. So that means that we have a local maximum here. So I'll put like local max here and we're going to have a local minimum here. And we know it's a local maximum on the left hand side and a local minimum on the right hand side simply because this is a cubic function and we know the end behavior of this cubic function. So let's see that gives us a pretty easy way to sketch this thing. So let's observe we could put like -1 right here and then we could put one right here. We know at -1 this function is equal to 1 and at one this function was equal to -3. So that would be like maybe down here. [snorts] Now just using what we know about the end behavior we know that the general shape of our function has to be that it doesn't go through this point 0 0. So just pay no mind to that. But it has this kind of picture. The important thing about this picture is not exactly where the zeros are, but the fact that these local maximums and local minimums have these signs gives us three unique solutions or three nonrepeed solutions.

[00:20:32]So that's good to keep in mind. So let's also notice that the discriminant can be calculated and the discriminant is -3/4. So I'll let you plug that into the discriminant formula. It's pretty easy to see that that's -3/4. But notice that that is negative. So that means that we are in this world down here where we should have all real solutions.

[00:20:56]But we've already showed that there's all real solutions using standard techniques from calculus one. But this just backs up this claim of truth down here. Okay, so let's see what we can do with all of this information. So we just got done showing that there are three real solutions to this equation. And so let's just put those into this cubic formula to find one of those solutions.

[00:21:22]So one of those solutions is via this cubic formula going to be the following.

[00:21:26]It'll be the cube root of 12 + i * the<unk> of 3 / 2 plus the cube roo<unk> of 12 - i * the<unk> of 3 over2. And that's our example of a real number that cannot be written without using nonreal complex numbers. So we're not going to check carefully that that this cannot be written without nonreal complex number.

[00:21:56]That's a bit outside of the scope of this video, but what I would like to do is make an observation in terms of some trig functions. So, this is in fact equal to the cube root of the cosine of p<unk> over 3 + i * the s of<unk> over 3. Just given the fact that cosine of p<unk> over 3 is a half and sin p<unk> over 3 is <unk>3 /2. And then well we're gonna have the same kind of thing over here. This is going to be the cube root of the cosine of pi over 3 minus i * the s of pi over 3. But writing this in terms of cosine theta + i sin theta really ex us on to use oilers's formula.

[00:22:43]So in fact we can write this as the cube root of e to the i pi over 3 plus the cube root of e to the minus i<unk> over 3. But then we can easily take cube roots of those. That's going to be equal to e to the i<unk> over 9 plus e to the minus i<unk> over 9. But then again using Oilers's formula this time kind of in reverse this gives us 2 * the cosine of p<unk> / 9. But well we've just kind of broken the rules here and we have just written our solution without complex numbers as 2 * the cosine of pi over 9. And that's because well there's a bit of a rule underlying this. And when I say without nonreal complex numbers, I mean without non-real complex numbers using so-called basic operations of numbers of which the cosine doesn't count, if you will. Okay, so let's tie this all together into a theorem which guides exactly which real numbers can't be written without i. Okay, so here's the big theorem underlying all of this and it's from Holder from 1891.

[00:24:03]So let's let f ofx be a polomial with rational coefficients and it's going to be irreducible with three real roots.

[00:24:11][snorts] Then the result says that none of these real roots can be expressed using only basic arithmetic. So what I mean by that is addition, subtraction, multiplication, division, exponentiation, stuff like that. and real radicals in a finite number of steps. So it's important here that we have finite number of steps because you could in principle for some of these use an infinite number of steps like perhaps get some sort of series expansion of these. But I think that's pretty clearly possible based off of our example where we saw our solution was the cosine of pi over 9 or related to the cosine of p<unk> over 9 which has a pretty obvious series expansion. Okay. So anyway, there's the theorem and as promised, we produced a real number that you cannot express without I. And that's a good place to