It Looks Impossible, But...

[00:00:00]We finally made it.

[00:00:02]>> The equation room. They say there's a mysterious equation inside this room.

[00:00:07]Let's go in right away.

[00:00:11]>> What? What is this?

[00:00:12]>> I see. We can't solve this equation.

[00:00:16]>> Really? It looks easy to me. Let's try solving it. Multiplying x and y gives zero. Then at least one of x and y should be zero. Huh? Wait. It says here that neither x nor y is zero. That seems like a contradiction. What is going on here?

[00:00:35]>> It is perfectly natural to feel that way. In other words, this equation says that two nonzero numbers multiply to zero. Is that even possible?

[00:00:45]>> Under ordinary circumstances, no.

[00:00:48]However, in the familiar world of numbers, many different kinds of numbers exist. And if we take one step beyond that world, we may find some strange numbers that satisfy this equation.

[00:01:02]>> H maybe. But what are we supposed to do?

[00:01:05]Take one step beyond the familiar world of numbers. I don't really understand what that step is.

[00:01:12]>> For now, let us continue deeper into this laugh dungeon. We may find a clue.

[00:01:17]>> What? We can move on without solving this.

[00:01:24]We found something already. Let's see.

[00:01:27]It only says ax. What is it trying to tell us?

[00:01:31]>> How fascinating. [music] What? What do you mean?

[00:01:34]>> In this expression, the variable x is multiplied by a. More specifically, the number [music] a has the role of multiplying the variable x by a. We can reinterpret the meaning of a number in that way.

[00:01:49]>> I have no idea what you're talking about. Look at this expression too. Oh, there was another one. Let's see. I suppose XY represents a vector. [music] It looks like the components of the vector are written vertically here. And this vector is being multiplied by some number A. Multiplying a vector by [music] A means multiplying each of its components by A. Here we can think of the number A as having the role of multiplying the vector by A.

[00:02:18]>> H I see. Yes, that does make sense.

[00:02:22]>> Watch out. Something is coming.

[00:02:24]>> What? Wa! What is this? Where are we?

[00:02:29]And what does this equation mean?

[00:02:31]>> This is most fascinating. Since Jay is wearing a hat, let's read this as J hat.

[00:02:37]>> Please explain it in detail.

[00:02:39]>> This is my first time seeing it too.

[00:02:41]When the vector xy is multiplied by something called jhat, the result is the vector yx. In other words, multiplying by jhat swaps the components of the vector. That is what this equation says.

[00:02:56][music] >> What? It swaps the components of a vector. That is not even a number anymore, is it?

[00:03:02]>> That is certainly a reasonable question.

[00:03:04]Strictly speaking, J hat is not a number. At least not in a usual sense.

[00:03:10]But for now, we will treat J hat as if it were an ordinary number and describe this as multiplying by jhat.

[00:03:19]It feels a little strange, but I'll accept it for now. This might teach us something about that equation.

[00:03:25]>> This jhat also has an aspect that closely resembles an ordinary number.

[00:03:30]>> Really, a short calculation should make it clear. Then try calculating this expression.

[00:03:37]>> Okay. The vector xy is being multiplied by jhat squared. What does that mean?

[00:03:43]Jhat is not an ordinary number, [music] right? That is true. But do not worry about the details here. For now, split jhat squar into jhat * jhat and calculate it. [music] Give me a moment.

[00:03:57]So this means we multiply the vector xy by jhat twice, [music] right?

[00:04:03]Multiplying by jhat means swapping the components of the vector. So after using jhat once, the vector becomes yx.

[00:04:12]But there is still one more jhat left.

[00:04:15]So we swap the components of the vector one more time and that brings us back to the original vector.

[00:04:21]>> Excellent sun. To summarize, multiplying the vector xy by jhat squ returns it to the original vector xy. This means that flipping it twice [music] brings it back to where it started.

[00:04:35]Now let's imagine that there is a hidden one on the right hand side. [music] Then do you notice anything? Well, it looks like jhat squared corresponds to 1.

[00:04:47]>> Exactly. In other words, multiplying by jhat squared can be regarded as the same as multiplying by 1. Let's write this as jhat squar equals 1.

[00:04:59]>> jhat^ 2 = 1. That means flipping it twice brings it back to where it started.

[00:05:05]>> At first glance, an equation saying that something squared equals 1 may seem ordinary. However, j hat is clearly neither one nor negative 1. After all, it is not an ordinary number.

[00:05:19]>> It is neither one nor negative 1, but its square is 1. That is strange. We encountered something called jhat that swats the components of a vector. Jhat squared equals 1. But jhat is neither one nor negative 1. That is strange.

[00:05:38]>> This jhat is quite troublesome. If we treat it like an ordinary [music] number, something troublesome happens.

[00:05:45]>> Something troublesome.

[00:05:47]>> Calculating this expression should explain everything. [music] Here, let us assume that expressions containing J can be calculated in the same way as ordinary expressions. Are we really allowed to do that? Huh? The hat on J [music] has disappeared.

[00:06:04]>> So, you noticed, but ignore it for now and keep going. I don't really understand but okay let's see we are multiplying 1 + j and 1 minus j. Let's calculate it. Oh before I forget I'll write down the equation saying that j ^2 = 1. Now back to the calculation. If we expand the expression normally we get this. I simply use the distributive law to remove the parenthesis. Now since minus j and + j cancel each other out only 1 and minus j^ 2 remain j² right j ^² = 1. So this expression becomes 1 - 1 which is zero. Thank you zundaman. Let's summarize. When we multiply 1 + j [music] and 1 minus j the result is zero. This is actually quite a strange result. Um why?

[00:07:03]>> When we see an equation [music] of this form, we are tempted to conclude that J must be either 1 or negative 1.

[00:07:10]>> Yes, that does seem natural. If multiplying 1 + j and 1 - j gives zero, then either 1 + j or 1 - j must be zero.

[00:07:22]So j equals either 1 or -1.

[00:07:26]>> Wait a moment, Zundan. J is neither 1 nor1. It is not an ordinary number. In other words, it is not a real number.

[00:07:35]>> Oh, that's right. What does this mean?

[00:07:38]>> If we accept J as a number, then even if two numbers multiply to zero, it doesn't necessarily mean that one of them is zero.

[00:07:47]>> H, that certainly is a strange result.

[00:07:50]>> Let's change our viewpoint slightly and consider this equation. Suppose Z1 * Z2 equals Z. While neither Z1 nor Z2 is zero. This is the same equation we saw at the beginning. We have only changed the variable names. This equation has no solution in the real or complex number systems. [music] But if we accept J as a number, this equation has the solution Z1 = 1 + J and Z2 = 1 minus J.

[00:08:19]>> Well, what is this? This equation was supposed to have no solution. But by extending the number system, we created a solution.

[00:08:28]>> Yes, that is essentially what happened.

[00:08:31]>> Madam, look at this.

[00:08:33]>> But what is this?

[00:08:35]>> This I hat looks similar to the Jhat from before. Jhat had the role of swapping the components of a vector. But this I hat does more than simply swap the components. It also puts a minus sign on one of them. This is interesting as well. What happens if we calculate it in the same way as before?

[00:08:54]>> Let's try it. Um, what should I do?

[00:08:57]>> Previously, we multiplied the vector xy by jhat squ. [music] So this time, let's multiply it by i squ.

[00:09:05]>> I see that makes sense. Let's do it.

[00:09:09]First, we split ihat squar into ihat * ihat. This means [music] we multiply the vector xy by ihat twice.

[00:09:18]First, when we multiply by ihat once, a vector's components are swapped and only the first component gets a minus sign.

[00:09:26]Then, when we multiply by ihat one more time, the components are swapped again and only the first component gets a minus sign. So, this is - xy. That looks correct. To summarize, multiplying the vector xy by i^ squ gives the negative of the original vector.

[00:09:48]Now let's imagine that there is a hidden one from the right hand side. Then I see it now. I squar corresponds to -1.

[00:09:58]>> Exactly. In other words, multiplying by i^ 2 can be regarded as the same as multiplying by -1. Let's write this as i^ 2 =1.

[00:10:10]H. This is strange. An ordinary number cannot become negative when squared. But I feel like I've seen this somewhere before.

[00:10:18]>> What do you mean?

[00:10:20]>> Multiplying the vector x y by ihat gives the vector - yx.

[00:10:26]This ihat has the property that I squared equals -1. Could this be the imaginary unit I?

[00:10:34]>> Very good. Continue.

[00:10:36]>> Yes. Now let's associate the vector xy with the complex number x + y i. Of course x and y are real numbers. When we multiply this by the imaginary unit i, the result is - y + [music] x i. This is because when we multiply i by y i, i^ 2 is -1. So the result is - y. And when we multiply i by x, the result is x i.

[00:11:09]Therefore, the vector - yx corresponds to the complex number - y + x i. I see that is an interesting interpretation.

[00:11:21]Originally, this ihat is not a number.

[00:11:24]Strictly speaking, it can be regarded as a type of transformation that maps one vector to another. However, when we associate the vector XY with the complex number X + Y I, we can indeed say that this Ihat represents the imaginary unit I. So this was a story about complex numbers. However, this is a somewhat simplified approach. So we have omitted the part that reproduces the algebraic structure of complex numbers. I see. For those who would like to study this properly, please look up the matrix representation of complex numbers.

[00:12:00]Please note that the notation I hat and jhat is not standard and is used only within this video.

[00:12:07]>> I wonder where you got that information.

[00:12:10]>> Now let's consider what the transformation ihat does on the plane.

[00:12:15]Ihat moves the point xy to the point minus yx. It looks like this when drawn in a diagram. In fact, this can be interpreted [music] as a 90° counterclockwise rotation about the origin. This is because after a 90° rotation, moving X to the right becomes moving X upward. So, the Y-coordinate becomes X. Furthermore, moving Y upward becomes moving Y to the left, so the X coordinate becomes - Y. This is only a rough explanation, though.

[00:12:49]>> So, that's what it meant. Wait, could we think about jhat in the same way as ihat? Multiplying the vector xy by jhat gives the vector yx. So jhat has the role of swapping the components of a vector. This jhat has the property that jhat squared equals 1. Now, just as before, let's introduce something called J that is not a real number and whose square is 1 and associate the vector X Y with the number X + Y J. Of course, X and Y are real numbers. When we multiply this by J, the result is Y + XJ. This is because when we multiply J by Y, J^ 2 is 1. So, the result is Y. And when we multiply J by X, the result is XJ.

[00:13:42]Therefore, the vector YX corresponds to the number Y + XJ. What a beautiful correspondence.

[00:13:50]In fact, numbers such as X + Y and Y + XJ are called split complex numbers.

[00:13:59]>> What? You already knew that?

[00:14:01]>> We explained [music] split complex numbers in a past video, but it seems you have forgotten. Hm. For some reason, I can't remember it at all.

[00:14:09]>> More precisely, a number written in the form x + y, where x and y are real numbers, is called a split complex number. Here, j is a nonreal number whose square is one. Originally, this j hat is not a number. However, we can regard j hat as representing the j of split complex numbers. This is also a somewhat simplified approach. So those who would like to study it properly should look up the matrix representation of split complex numbers. Let's consider what the transformation jhat does on a plane. Jhat moves the point xy [music] to the point yx. In other words, it is a transformation that swaps the horizontal and vertical components. This can be interpreted as reflection across the line y= x. H I [clears throat] see. This really does swap the horizontal and vertical directions. So this is what Jhat was doing.

[00:15:09]>> It seems you understand now.

[00:15:16]By the way, we carelessly overlooked it.

[00:15:19]But I wonder what lies beyond the store.

[00:15:21]>> H I have a bad feeling about this.

[00:15:24]>> Since we are here, let's go.

[00:15:27]>> Wait, this has become quite serious.

[00:15:31]>> We saw something. We should not have seen the vector xy is being multiplied by epsilon hat. Strictly speaking, I suppose xilon hat is also a type of vector transformation. The transformation [music] epsilon hat maps the vector x y to 0x. That is a very strange transformation >> indeed. The position of x has changed and y has disappeared. Instead, a zero appeared. What is going [music] on?

[00:16:03]>> It is difficult to see what is happening, but this can be regarded as a type of shift.

[00:16:08]>> Explain it.

[00:16:10]>> Very well. Let me explain. The transformation epsilon hat maps the vector xy to 0x. If we write out the components of each vector, we get this, [music] right? X moves from the first component to the second component.

[00:16:26][music] In other words, it shifts by one position. Yes, that is true. What happens to the other components?

[00:16:33]>> We can think of the other components as also shifting [music] by one position.

[00:16:38]However, y has nowhere to go. So, it disappears while the empty position [music] is filled with zero. I see. So, that is how we should think about it.

[00:16:48][music] It was hard to see because there are only two components. But this transformation really does shift everything by one position.

[00:16:55]>> You can interpret it [music] that way.

[00:16:58]Incidentally, if we transform it with exon hat once more, everything becomes zero.

[00:17:04]>> After two shifts, both components move outside and [music] disappear. Written as an equation, it works like this.

[00:17:12]Epsilon hat squared is zero.

[00:17:14]Transforming a vector twice with epsilon hat is the same as multiplying it by zero. [music] But epsilon hat itself is not zero. I feel like we saw something similar earlier. H that is strange.

[00:17:28]>> It is not zero but its square is zero.

[00:17:31]Epsilon hat seems somewhat like zero. In fact, in the number system called the dual numbers, the element corresponding to epsilon [music] hat is sometimes interpreted as an infinite decimal. We covered several different number systems this time. For those who would like to learn more about each one, we will place links in the [music] description. So, please take a look. Anyway, this means we've completed this math dungeon. Well done.

[00:18:03]>> Thanks for watching. [music] If you'd like to support this channel, consider becoming a member. In this video, we talk about the purpose of our membership. Check it out if you're interested. [music] Well then, take care everyone. See you later.