 In the previous video for lecture 26, we talked about proof of existence, that to prove that something exists, all you have to do is provide a example of such a thing. Just provide one single example. And that method of proof of existence is what's usually referred to as a constructive proof. A constructive proof meaning that we have an explicit example for everyone in the world to see. The two proofs we saw in that video were examples of constructive proofs. We proved that there is an even prime number. We're like, hey, here it is, it's the number two. We proved that there's a number which can be written as a sum of two perfect cubes in two different ways. We're like, oh, here's the number, 1,729. We were explicit and here is the example. Now, it turns out there's another way to prove existence and it's a little bit more fuzzy here. This is what we refer to as a non-constructive proof. It gets the name non-constructive because we prove that something exists but we don't explicitly know what it is. Which can be a little bit concerning sometimes. Clearly when it comes to existence, people usually prefer constructive proofs because we're like, hey, here's the thing. We know at least one of these objects and maybe we have them information to play around with it, find other objects or we can utilize it somehow. But it turns out that non-constructive proofs, even if we don't know what the object is, can still be extremely valuable. And we've actually seen some examples of this in our lecture series. For example, previously when we introduced the well-ordering principle, which was a property equivalent to the mathematical induction principle, remember that the well-ordering principle guarantees that every non-empty subset of natural numbers has a minimum element. But the well-ordering principle doesn't give us any clue on what that minimum element is. We just know that it exists. And as an example, we use the well-ordering principle to prove the existence of quotients and remainders. So if we have some integers A and B, we were able to use the well-ordering principle to prove there exists, I guess I should have a then and there, then there exists integers, we'll call them Q and R, such that A equals BQ plus R, and R sits between zero, but is strictly smaller than B. So the existence of these numbers were given to us by the well-ordering principle. But after we proved they existed, we're like, oh, we don't actually know what they are. So we have to develop the division algorithm, AKA just long division we learned from grade school, we can use the long division to actually find those numbers. But we know that the algorithm will terminate because we know they exist, okay? Another example we did had to do with the greatest common divisors, right? So if we have two integers A and B again, then there exist, again, integers, we'll call them R and S this time, there exist integers such that the GCD of A and B is equal to A R plus B S, that the greatest common divisor of any two integers can be written as a linear combination of those two integers. This is extremely useful when the numbers A and B are relatively prime, and hence their greatest common divisor is equal to one. Now, again, we proved this in this lecture series using the well-ordering principle, but the well-ordering principle did not tell us what these numbers were, it just showed that they did in fact exist. We then had to introduce the Euclidean algorithm, which then constructs these numbers, but the reason that the Euclidean algorithm can construct them is because they exist. We were able to prove that the Euclidean algorithm gives us these numbers, we proved it by induction, we proved it gives us these numbers R and S because they exist. If we didn't know they existed, that our proof would actually fall apart. As strange as it is, sometimes a non-constructive argument is sufficient that if we just know the thing exists, that actually is good enough for us. And in those two examples I provided with the division algorithm and the Euclidean algorithm, the fact that we know they exist is actually what allows us to prove that the algorithm works. So non-constructive proofs can be very, very valuable in proving existential statements. So what I wanna do right now is actually prove the same theorem twice. Our theorem here is gonna be there exist irrational numbers called an X and Y, such that X to the Y is a rational number, okay? I'm gonna first prove it non-constructively, and then I'm gonna prove it constructively, which notice that this is an existential proof. There exist irrational numbers that have this property. Let's prove that they exist. And to illustrate, we weren't gonna use the well-ordered principle this time. Let's show you another way. We're gonna show they exist, but we're not gonna know what they are. That's kind of the fun little trick here, so let's consider it. First, consider the number of the square root of two to the square root of two power, okay? We've already proven using the technique of proof by contradiction that the square root of two is an irrational number. We know that. Now, is the square root of two to the square root of two power rational or irrational? I don't actually know that off the top of my head. There's an argument that can be made there, but here's a nice little trick. If the square root of two to the square root of two is rational, then we're done because we can be like, oh, x equals the square root of two, which is irrational. y equals the square root of two, which is irrational. And then x to the y would equal the square root of two to the square root of two. If it's rational, then, oh, that's exactly what we have. We found our numbers x and y, okay? But we don't actually know if it's rational. If it's rational, then we're done. If on the other hand, the square root of two to the square root of two power is irrational, then take the following situation. I'll take x to be the square root of two to the square root of two because we're assuming now it's irrational, okay? And then you'll take y to be the square root of two, which again, we know is irrational. We're gonna take the square root of two to the square root of two to the square root of two power, okay? Now by exponential properties, if you have this repeated chain of exponents, you multiply the exponents together. So this is equal to the square root of two to the square root of two times square root of two power. But that's just the same number. You're gonna end up with the square root of two squared as your exponent, which by virtue of being the square root of two, if you square the square root of two, you get back two. So this is equal to the square root of two, which again, by virtue of these exponential properties, the square root of two squared is equal to two and we know that's a rational number. So okay, look what we have here. If this number is rational, then we have an irrational to an irrational, which is rational. But if it's irrational, then this number, well, I mean, this number is always gonna be rational because it's equal to two, but if this is irrational and this is rational, we got it. So we've exhausted our two cases. Either this number is rational or irrational. If it's rational, we get the existential statement we want. If it's irrational, we get the existential statement we want. They're different numbers, right? X and Y are different numbers, but regardless, then there do exist numbers X and Y, but with the proof currently in hand, I don't know what X is and what Y is. I guess Y in both cases is the square root of two. So okay, that was consistent, but is X gonna equal the square root of two or is X gonna equal the square root of two to the square root of two? I don't know, but I know one of those will be rational and so the existential statement exists and that's why this one's non-constructive because I had to try two different cases and they both led to the existence of such numbers. I don't actually know which case is true. What is this irrational number or is it not? That would require a different argument which I am not going to supply in this video here. I mean, this is actually how we get these non-constructive arguments by cases, prove by contradiction. We might prove, we might assume that the thing doesn't exist, get a contradiction, and therefore that proves that it does exist, although usually that argument goes the other way around. To prove that something doesn't exist, you assume that it does exist and get a contradiction, but we'll talk about that in another video. I promise you I was gonna prove this theorem again in a constructive manner. So here's our second proof, okay? It's a little bit longer this time, but that's okay. I want you to take x to be the square root of two, that seems to be our favorite number in this video, and then take y to be the log base two of nine, okay? We already know that the square root of two is irrational, again, we just used a proof by contradiction, but we have not yet established in this lecture series that the square root of the log base two of nine is irrational, so an argument has to be supplied there. So this is like we talked about earlier, that sometimes we have to put proofs inside of proofs inside of proofs, this is the situation. We need to establish that this is an irrational number, we haven't proven it yet, we don't know that, so we have to prove it. Unlike the square root of two to the square root of two, we won't worry about whether that's rational or irrational, we can prove this one, and we're gonna prove this one by contradiction, that's how we proved the square root of two was irrational, we'll do that for the logarithm as well. So log base two of nine, if it were rational, so for the sake of contradiction, if it was rational, there would be two integers, A and B, such that Y is equal to A and B, A over B. Remember Y is of course log base two of nine, okay? So using this equation right here, we have Y equals a fraction, and I can also assume that the denominator is positive. I know it's not zero because it's a rational number, and if B was negative, well you can just times the top and bottom by negative one over negative one, that makes B positive, and that gives you a different fraction representation, so I can assume the denominator is positive, no big deal. All right, so now by properties of logarithms, if we have the log base two of nine equal to A over B, by logarithmic properties, we can switch from the logarithmic form to the exponential form to get this equation right here. Nine is equal to two to the AB power. Then what I'm gonna do is I'm gonna take the power of B of both sides, so on the left-hand side you get nine to the B, on the right-hand side you're going to get two to the A over B to the B power by exponential properties, this becomes two to the A, all right? And so now this is where B being a positive number is significant. Since B is a positive number, I know that nine to any positive power, again this is an integer, so it could be nine to the first, nine squared, nine cubed, et cetera, this is gonna be an integer, okay? On the other hand, if two to the A was not an integer, like if, for example, if A was a negative number, we would get one-half, that would be a contradiction, an integer can't equal a non-integer, so we can assume that two to the A is likewise an integer, okay, but also remember B is still positive here. So this isn't zero, right? This is nine first, nine second, nine cubed, there's some positive integer there. In particular, three divides the left-hand side, because three divides nine and it'll divide any multiple of nine as well. And so this tells us because three divides the left-hand side of this equation and both sides are integers, three has to also divide the right-hand side. So three divides two to the A power. Now by Euclid's method, Euclid's lemma here, since you can factor two to the A as two to the A, factors as two times two to the A minus one, by Euclid's lemma either three divides two or three divides this, and then you can invoke induction there that this keeps on going down, down, down, down. Eventually you're gonna reach a point where three divides two, but as three and two are distinct prime numbers, this gives us a contradiction and this shows us that log base two of nine is in fact an irrational number. Okay, so we got that little lemma taken care of. Now let's get back to the heart of it. We have two irrational numbers, X, which is irrational, which is a square root of two and Y, which is irrational, which is the log base two of nine, okay? So now let's look at their exponents. It's very simple calculation here. Take X to the Y, if you take the square root of two to the log base two of nine power, things to note here is that nine is the same thing as three squared. Property of logarithms, if you have a log base, say A, and you have a number to the X to the N power, this is equal to N times log base A of X, like so, exponents inside come out as coefficients. We're gonna do the same thing here. This two comes out, so we get two times log base two of three. Again, by exponential properties here, because we have a factorization in the exponent, we can break it up. Excuse me, the square root of two squared to the log base two of three power. Like we saw before, the square root of two squared is equal to two, so this simplifies to be two to the log base two of three power. Now, by definition, the log base two of three is the exponent of two that gives you three. So if I take two and raise it to the log base two, log base two of three power, it's gonna give you three, and this is in fact a rational number. So we've now demonstrated, without a doubt, two irrational numbers whose exponential expression gives you a rational. This then gives us a constructive example where we actually know what X is. We actually know what Y is as opposed to the previous one. But there was some simplicity to the first approach, right? I don't actually have to show that the square root of two to the square root of two power is rational or irrational. I just was able to get the existence with a much simpler proof. While this one, of course, required a lengthy part to show that this log base two of nine is an irrational number. So there's given takes with these constructive versus non-constructive. We like to find the number itself, but non-constructive proofs can potentially be easier because they don't require you actually find the numbers, even if they're a little bit more disconcerting. Anyways, we'll see more examples of this later on in our lecture series, but that'll finish it for lecture 26. Thanks for watching. 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