 So in this video, I want to talk some more about existence proofs Because we previously have shown how do you prove an existential statement like there exists some x inside of x Such that P of x holds right and how do you do that? You just provide an example That's all you have to do you have to that's how you have to provide How do you have you prove it? You just provide an example now when you when you prove it an existential statement like this You provide an example Assuming it's a constructive proof right if it's a non-constructive proof You know it exists, but you don't even know what it is that is a possibility here, of course But you provide an example you prove it exists this tells you that there's at least one One object that satisfies the conditions But how many are there total and honestly that question is getting much more into the common torques that we've explored earlier Common Torques is very much interested in Existential questions honestly like the whole goal of common Torques is basically threefold Right every common Toriel problem has three phases the first phase is going to be existence Does this common Toriel object that I'm interested in doesn't even exist or not and then if you prove existence It's similar to what we did before although it could be a lot harder The next step is going to be classification We want to classify our objects. Can we organize them in a meaningful way in a helpful way? And that's about putting things into categories Oftentimes it's about describing a partition on the objects We are interested in and then lastly once it's classified. We then want to count it We want to enumerate them Can we list all of them in order so that from these organized families? We can count each of the families and and put it all together This is the goal of all of these common Toriel problems We want to count them being able to count every possibility truly shows that we Understand completely the object that we are interested in and so while we spent some earlier examples on counting And when our common Toriel unit sometimes that's too hard Sometimes you need to you need a simpler question of classifying because if you can't organize them Then you can't count them and then sometimes even before classification you have to even know if they exist or not Existence is the fundamental common Toriel question, which we are exploring right now but in this video what I want to do is provide a very very simple common Toriel exercise that is to say With existence often comes a statement about uniqueness Because maybe when it comes to counting them there is only one You know kind of like the Highlander there there can only be one I mean there there we have existence We know that one exists, but is there a second one sometimes the answer is no There isn't a second one and this then leads to this idea of uniqueness. There exists only one Now when you want to prove that there is something but that something's unique that there's only one of them There's really two parts to any to that proof there The first part is going to be your existence proof that we've done previously So you prove existence and you're going to show that there is Greater than one such object that satisfies the condition Now for uniqueness, there's a second part you prove existence first The second part is then often referred to as this uniqueness part You have to show that there is not more than one of these things So the idea is that the number of objects is greater than equal to one and less than or equal to one When you put those together you have exactly one object and hence it is unique For so for existence we show there's at least one for uniqueness. We show there is at most one Let's revisit a proof that we had seen previously There exists a unique even prime number Okay, so the word unique here suggests there's two parts We have to prove that it exists and we have to prove that it's unique We've already proven it existed that even prime exists We did that in the previous lesson, but it's such an easy argument just for the sake of completeness I'm just going to include it again. Okay, how do we know there exist an even prime number? Well, two is an even prime number two is even and it is prime It's only divisors are one and two. So this is a very simple argument. This gives us the uniqueness part We now know that an even prime exists now We have to prove that there's not a second even prime number So consider generally q to be an even prime number because it's even that means two divides q But wait a second q is a prime number It's only divisors are gonna be one and q one is not two and two does divide q So the two which is a divisor has to equal q. So we've now shown that Every prime number that's even is equal to two. So there isn't a second one Every uniqueness proof has the two parts the existence parts which shows we have at least one and we so that this right here shows There's at least at least One even prime and then the uniqueness part shows there's at most one even prime number Okay now Previously in our lecture series. We had used the well ordering principle to show examples of Uniqueness we've I mentioned these before I'm not going to go into much too much detail about it right now But those arguments were in fact uniqueness arguments when we did the division algorithm When we did the Euclidean algorithm when we prove these things we showed that certain numbers Existed by the well-known the well ordering principle particularly we got a minimum element and then we argued and got a contradiction Thus giving us uniqueness Usually contradicting the minimality of that element that the well-ordered principle gave us so a general strategy to prove uniqueness is that you first prove existence and then By way of contradiction you suppose there is a second one and then you drive a contradiction because if there's a if you get a Contradiction with at least two that means you have a most one and you give uniqueness To demonstrate this technique where to prove uniqueness. We will suppose for the sake of contradiction that there's at least two I actually want to use a calculus example. So going back to calculus one There was two theorems that you probably saw in your calculus course I will remind you what those statements are We have the intermediate value theorem and the mean value theorem the intermediate value theorem We're gonna list here first the intermediate value theorem tells us that if f is a continuous function on The closed interval a to b let's let's unwrap that for a little bit closed intervals We're familiar with that So this is gonna be some interval along the x-axis for which the bounds a and b are included in that F is a function in the calculus sense, of course, and then continuous Continuous we could be precise about the definition, but for the sake of this example Continuous means that we can draw the graph without picking up our pencil. Okay, that's what continuous is So if f is continuous on the closed interval a to b and if n is any value between f of a and f of b so let's see on our graph f of a is this y-coordinate here and F of b is this y-coordinate right here and n is just something in the middle hence Intermediate value value typically means a y-coordinate intermediate means it's between So we have a y-coordinate that lives between the y-coordinates of two established points on the graph And we also need to assume that f of a doesn't equal f of b Because if f of a equals f of b there is no intermediate values, okay The intermediate value theorem then tells us if we have a continuous function with these properties Then there exists some number c on the open interval a to b So it's not a and it's not b that obtains that y-coordinate You get some x-coordinate c like you see here in my diagram that then obtains the intermediate value. All right now We're not going to prove the intermediate value theorem. That's a proof that belongs to a Analysis class real analysis class. I'm just going to use it as gospel for the sake of this But I do want to make mention that the intermediate value theorem is an existential statement if certain conditions satisfied Then there exists a number that lives between these two that satisfies the property be aware that the intermediate value theorem is a non constructive Existential statement it doesn't tell you what what the number is it just tells you generally where it lives It's gonna be somewhere between a and b. We know it exists, but that's all so this is going to be one of these x Non-constructive existential statements. Okay, these happen all the time in mathematics. We're gonna use this one in just a moment the main story here is What you know if you hit this line and you hit this line you hit every line in between right? You can't the only way you can get from only way you can get from here to here is you have to cross the line There's no way to get past it unless you have like a wormhole what you teleport past it But if you have a wormhole, you're not continuous The only way to continuously go from from below to above is to cross this line in the middle That's the intermediate value theorem Another important theorem from calculus is the mean value theorem. Let me remind you what that one tells us So now we have a function f again. It has slightly different Assumptions this time. We do require that f is continuous on the closed interval a to b Again same meaning of continuity that we have before we also require that f is Differentiable on the open interval a to b so it doesn't have to be differentiable at the end points a and b does have to be continuous there But it'll be differentiable between everywhere between Differentiable means that the derivative exists everywhere between a and b or in other words the tangent line Exists everywhere between a and b or you can think of it as it's a smooth graph There's no sharp corners or vertical tangent lines or anything like that because the derivative the tangent line exists between all points a and b Okay, so with these assumptions There exists some c such that c lives between a and b so just like before c is on the open interval a to b and c satisfies the following relationship that the derivative of f evaluated at c is Equal to f of b minus f of a over b minus a what you want to what what how you want to interpret this is is that with these conditions satisfied Which let me mention of course that if your function is differentiable. It's continuous. So Honestly, this first condition just saying that has to be continuous on the end points and differentiable everywhere else. Anyways what this statement is telling us geometrically is that the tangent line is somewhere parallel to the The secant line that is if you take the average rate of change, which is illustrated right here It's the secant line that connects the start and end by the mean value theorem there will exist a tangent line that is parallel to This secant line right like so or in other words that somewhere the instantaneous rate of change is equal to the average rate of change, which is why it's called the mean value there It's not like it's a night. It's not mean as opposed to the nice value theorem It's an average value theorem that the average rate of change is somewhere equal to the instantaneous rate of change The secant line is parallel to a tangent line But I want to be aware that again the mean value theorem is a non constructive existential statement It's a conditional statement if these things hold then there exists a number C that has this property It doesn't tell you anything about see that other than it exists And it does give you an interval, but as there are literally infinitely many numbers between a and b here that Interval still doesn't narrow it down very much, but it does give you that that thing exists And so these are both existential statements Well, I should say conditional exponential statements if hypotheses hold then we're guaranteed the existence of these elements And we're gonna use this to prove a very nice result We can actually prove that the equation X-cube plus x minus one equals zero has exactly one real root and honestly I should just I should scratch this off, right? We don't the statement is the equation x cube plus x minus one equals zero has exactly one real root that equation has one solution Okay, that's a uniqueness statement to say that has exactly one means that there is at least one and There is at most one. This is a uniqueness statement. So in order to prove this we're gonna first prove Existence we want to prove there's at least one. So we will first prove that the equation has a solution Now the we're gonna use the intermediate value theorem to accomplish that and the intermediate value theorem and the mean value theorem Are about functions. So we have to turn this equation Into a statement about functions. Well, it's pretty nice since the right hand side is just zero here so I'm gonna take the function f of x to be x cube plus x minus one and Notice any solution to this equation is just an x-intercept of this function. Okay So take f of x to be that polynomial function I want to mention that because it's a polynomial function. It's continuous This is a property of calculus that I won't justify any more than citing by calculus Now notice with this function if you take the limit as x approaches infinity f of x will likewise approach infinity And so that tells us that if f of x is going towards infinity There has to be a point some point x equals zero that the y-coordinate becomes positive. Okay now again right here This is a non-constructive existential statement I'm saying that there's somewhere a point where if you're going off towards infinity then your graph Somewhere has to be above the x-axis. You can't go towards infinity if you're not above the x-axis eventually So there's gonna be some point where you're above the x-axis. That's a non-constructive argument, but let's also Provide a constructive argument. You don't need both but for the for the point of demonstration You could also take the point take f of one here f of one will be one plus one minus one, which is one that's positive so this tells us that Somewhere the function is positive and specifically I know now at f at f of one the function is positive Okay, so we'll be above the x-axis by a similar argument If you take the limit as x approaches negative infinity f of x will approach negative infinity as well And so it's pointing down on the left-hand side of its graph Which again if you approach negative infinity there has to be some point where you're eventually below the x-axis and hence Your y-coordinate is negative That's a non-constructive argument, but I can make it into a constructive argument by finding a specific point Oh take f of zero here f of zero equals zero plus zero minus one, which is negative one Which is negative so this tells us that? The point f zero comma sorry zero comma f zero would be below the x-axis all right We have a continuous function which is somewhere negative somewhere positive if you take The intermediate value of zero this tells you by the enemy value theorem that there's somewhere that the Function has to cross the x-axis. We're gonna call that point r r for root And so we see that this polynomial equation does have a solution the intermediate value theorem gives us Existence and e and we know the number r lives somewhere between zero and one you have x equals zero And you have x equals one right here, but we don't have anything better than that We know it exists somewhere there so despite my efforts to Give constructive examples of where it's positive and negative in the end I get a non-constructed value here that exists so honestly you could gotten away with these ones as well This is true in general if you have a polynomial odd degree it points up on the right and points down on the bottom on the left It's gonna eventually cross the x-axis every polynomial of odd degree has an x intercept by this same argument It's non-constructive, but every polynomial every real polynomial of odd degree has a real root It's a consequence of the intermediate value theorem so now that we know that a root exists Let us argue that it is unique in order to prove that r is a unique solution to the equation Let us suppose there's at least two Distinct solutions. Let's call them a and b and since they're distinct I can assume that a is less less than b for the considerations here Okay now because a is a solution to the equation that means it's an x-intercept to my function f So f of a equals zero, but since b is also a solution to the equation It's likewise an x-intercept to my function and so f of b is likewise equal to zero. Okay, so Let's for them recall that we're talking about a polynomial here This polynomial x cube plus x minus one I might want to write down the screen We're going to need this in a second, but f of x equals x cube plus x minus one It's a polynomial so it's going to be continuous for all real numbers It's going to be differentiable for all real numbers again This is a this is a these are conditions that you can demonstrate in a calculus setting All that I care about is that this the assumptions of the mean value theorem are satisfied Therefore there exists some number c that sits in between a and b such that the derivative of F at c is equal to zero But on the other hand if this is my function I can calculate the derivative by the usual rules the power rule gives me 3x squared plus one That's this function right here notice though that x squared is always greater than equal to zero for any real number by times that by three that is still true if I add one to both sides you're going to get 3x squared plus one is greater than equal to one Which means it is strictly greater than zero 3x squared plus one is always Positive it's always positive. So the derivative is always positive, but The mean value theorem tells me that the derivative is somewhere equal to zero That is a contradiction and we got a contradiction which means our assumption was bad We do not have two X intercepts to the function thus we don't have two solutions to the equation Which means the one I found earlier was the unique solution? And so this thing is a nice example from calculus that demonstrates how you prove something is unique you prove that it exists and You prove that a second one does not exist oftentimes you do that by proof by contradiction as we did in this example