Do we even need the real numbers??

so in beginning calculus classes you always work over the real numbers but you're never really told why the real numbers are so important they're just some kind of plug for holes that happen in the rational numbers like pi and the square root 2 and so on and so forth but I was thinking is it possible to do the normal stuff from like a calculus 1 class without the real numbers at all in other words can we define calculus on functions that have a domain of the rational numbers and a co domain of the rational numbers and I want to approach this a couple of different ways first I want to look at does continuity make sense on the rational numbers and then can we maybe restrict the type of functions we're working with so that continuity does make sense if it doesn't already and then after that maybe we'll look at some derivatives can we take derivatives over functions of the rational numbers and then finally what about the classic theorems from calculus 1 are they still true over the rational numbers ok so let's get into it so let's recall that a function is said to be continuous at a point a if the limit as X approaches a of f of X equals F of a so you know there's also this epsilon Delta definition of the limit but we're gonna go ahead and use this kind of simplified definition of the limit since we're approaching this at the level of like a calculus one class so I want to open up with a troubling example in other words some example where continuity does something weird if we're only considering functions on the rational numbers and the example I want to look at is the following with so let's take this function from Q to Q and we're gonna define it in the following way so you can get really fancy with this but we're not gonna get that fancy we're just going to define this thing as follows so it's going to be f of X and it'll be given by 1 if X is less than the square root of 2 and it'll be given by 0 if X is bigger than the square root of 2 and you might be troubled here because I'm using the square root of 2 which is known to be an irrational number but notice I haven't said what to do if you plug in the square root of 2 because the function is not defined there it's only defined for an input of rational numbers and furthermore since it's only taking on the values of zero and one it only has outputs of rational numbers as well so let's go ahead and look at a graph of this function so it's a pretty easy step function so if we go over here to the square root of two which I can put there on the real number line although since we're considering these as rational axes this actually occurs as a whole on this rational axis and then this function is doing kind of the following thing so it is one up until we get to the square root of two and it is zero after the square root of two and notice I've made this really fine dotted line instead of a solid line because we're over the rational nodes here and the rational numbers have these infinitely many holes everywhere just like this hole at the square root of two okay great now what I want to look at is the continuity of this thing so if we were over the real numbers we would say that this thing is continuous everywhere except the square root of two but the square root of two is not even in the domain of this thing notice the square root of two is not in the rational number so we don't have to consider about consider it at all so let's maybe just notice this for all a in the rational numbers we have the limit as X approaches a of f of X equals F of a great and so we can really see that because this boils down to two cases one is if a is less than the square root of two great but that means we have the limit as X approaches a of well f of X if X is less than the square root of two is equal to one and that's equal to one which is f of a and you might be troubled here but let's recall that if we're at a point a let's just put it here it's less than the square root of two regardless of how close it gets the square root of two we can always find a little interval made up of rational numbers around it to take a two-sided limit so in fact this thing is continuous for all elements of its domain which like I said is a little bit troubling because this is clearly some sort of step function which should not be considered as continuous okay well so maybe we should limit ourselves to functions that are not piecewise defined to some kind of nicer functions so that we don't have to worry about strange things like this so let's just go ahead and say we limit ourselves to rational functions in other words quotients of polynomials notice that exponential functions would also cause some trouble because at most inputs in the rational numbers you would get outputs outside of the rational numbers so I'll let you guys think about that more deeply but from here on we're only going to consider rational functions on the rational numbers and let's see what could maybe go wrong with that so I'll clean up the board and then we'll do that so in the last board we saw something troubling happen when we looked at piecewise type functions defined over the rational numbers they ended up being continuous when really they shouldn't have been continuous so we said well maybe we'll just consider rational functions in other words ratios of polynomials as the only functions that we can work with in our calculus from Q to Q so now let's go ahead and look at the derivative so I want to recall real quick that the derivative in real variable calculus is defined in the following way so we generally say f prime of X and that's the limit as H goes to 0 of f of X plus h minus f of X over H now we could make this a definition over rational valued functions just by saying that this limit is also occurring as H is only from the rational numbers but we can actually be a little bit more heavy-handed than that in order to really show that we're only over the rational numbers and we can do that in the following way let's make our new definition of our rational derivative in the following way we'll say f prime of x equals the limit and then instead of having H go to zero we'll have n be a natural number in n go to infinity and we're essentially making the change of variables H equals 1 over N so notice if n goes to infinity with that substitution H will go to zero so we'll have f of X plus 1 over n minus f of X over 1 over n great so now we can simplify that a little bit maybe if we wanted to so that's the limit as n goes to infinity of n times f of X plus 1 over n minus f of X great so that's our new rational valued derivative so since we're only working with quotients of polynomials we really only have to check that this works on a power function in other words a monomial like x squared or X cubed or so on and so forth and then after that everything will follow from a product rule in a quotient rule and so on and so forth which all of those will hold in this calculus as I'll let you guys check if you want to so let's just go ahead and work out the only example that we really need to work out and that would be the example if f of x equals x to the M and here I'll just say that M is some natural number maybe some natural number including 0 and then from there you can extend it to all integers or something if you want to okay so let's go ahead and see if this derivative works and gives us the same kind of thing as it would for the real variable calculus derivative so here we'll have the derivative of F is equal to this limit as n approaches infinity of n times now we're gonna plug X plus 1 over n into this power function X to the N so let's do that X plus 1 over N to the M power and then we're gonna subtract f of X but that's just X to the N great and now we're gonna use a binomial expansion formula so I just want to recall that really quick so I'll recall it over here so let's say we have a plus B to the M power that's going to be the sum as K goes from 0 to M of M choose K and then a to the M B to the sorry a to the K and then B to the M minus K so we want to apply that binomial expansion formula to this guy right here so let's see what we get that's going to give us this limit as n approaches infinity and n times I'll just write out the first couple of terms because it's kind of obvious what will happen so we'll have X to the M and then next we'll have plus M over N times X to the M minus 1 so that would be like M choose 1 and then we have 1 over N to the first power that would be like playing the role of our a to the K over there and then next we would have M choose 2 and then 1 over N squared X to the M minus 2 plus dot dot great and then finally we subtract X to the M from this ok great and now what I really want to notice here is that this X to the M term is gonna cancel with this X to the M term and all we're left with is this M over N X to the M minus 1 and then everything out here can be written as 1 over N squared times a bunch of other stuff because the exponent of 1 over N squared in the binomial coefficient is increasing as we move out that way until we get to this very very last term which is by the way 1 over N to the M power ok great so when all is said and done we have this limit as n goes to infinity notice this n is going to multiply onto this term and M X to the M minus one then this n is going to multiply onto this term and give us 1 over N times a bunch of other stuff but now as n goes to infinity this guy's going to go to 0 and we're left with M times X to the N minus 1 so it looks like it makes sense to define our derivative on this power function and then we can boost that to a definition of a rational derivative on rational valued functions in other words functions that are ratios of polynomials when you're just considering the domain as the rational numbers okay so now that we've kind of seen that the derivative works I want to look at some classic theorems from calculus wanted to see if they still hold so it looks like we were able to define a derivative or at least a hack of a derivative on our calculus for the rational numbers if we only considered rational functions that is ratios of polynomials now I want to look at three classic theorems from calculus one and see if they're still good to go so let's first look at the intermediate value theorem so let's recall that F is continuous F is continuous on the closed interval A to B and Y is some number between F of a and F of B so we don't know the order of F of a and F of B so it's just between them then there exists a C in the open interval a B such that F of C equals Y ok great and we're actually going to show that this does not hold on the rational numbers and we're going to do that by the following example so let's consider this function let's consider the function f of X which is equal to x squared minus 2 so notice that's a nice continuous function on the real numbers and thus it's continuous on the rational numbers as well another thing that I want to notice is the following that f of 0 equals well 0 squared minus 2 which is equal to minus 2 and F of well let's maybe use two so that's equal to two squared minus two which is equal to two great and so here's what we have just to spell it out really carefully f is continuous on and I'll write it like this the interval 0 to 2 but I'm intersecting that with the rational numbers and the number 0 is between F of 0 and F of 2 right well I mean that's pretty obvious because negative 2 is less than 0 which is less than 2 and 2 is equal to F of 2 and negative 2 is equal to F of 0 great but there does not exist a C such that F of C equals 0 great and we can actually see that because by way of contradiction let's suppose such a C exists well then we would have f of C equals 0 but that would be the same thing as C squared equals 2 by our equation up there but since we're only working over the rational numbers that would tell us that C was in the rational numbers which is not true so what's going on here is this kind of typical example where we have a whole so we're exploiting the fact that the rational numbers are full of holes so we've got this parabola so it obviously goes through like 1 over root sorry root 2 and negative root 2 so it has that kind of picture right there great but this isn't a solid curve notice it's got a hole right here at the point 0 because that point doesn't correspond to a rational number ok so it looks like the intermediate value theorem is not good to go even with the standard polynomial example so let's maybe like revisit this with the mean value theorem so we just showed that the intermediate value theorem failed on our calculus of the rational numbers now let's check the mean value theorem so let's recall what it says over the real numbers so we want to suppose that F is continuous on a closed interval a B and differentiable on the corresponding open interval then there exists a C which is on that open interval such that F prime of C equals F of B minus F F of a over B minus a in other words the slope of the secant line equals the slope of the tangent line okay so we're gonna exploit this hole in the rational numbers of the square root of two again to find a counter example for the mean value theorem on our calculus so and the function that we're gonna look at in this case is given by 1 over x squared minus 2 great so this thing is actually continuous at all rational numbers so continuous on all of Q great so that's actually easy to check because from calculus 1 we know the only place that this is not continuous is root 2 and negative root 2 and furthermore it's differentiable on all of Q as well and you can check that the only places that it's not differentiable on are are the square root of 2 and negative the square root of 2 so this definitely satisfies this set up that it is everywhere continuous and everywhere differentiable so now we can restrict to a certain closed interval and corresponding open interval and see what happens so let's maybe consider and obviously we want our closed interval and our open interval to sandwich one of these bad points so let's maybe consider the closed interval zero to two and I'm going to go ahead and intersect that with the rational numbers just so that we only have rational numbers in there good and now what I want to notice is that if we take F of zero we'll get 1 over negative 2 so that's equal to negative 1/2 and then if we take F of 2 we'll get 1 over 2 so that'll be 1/2 now we can go ahead and find that slope of the secant line in other words the average change so f of two minus F of zero over two minus zero so let's see what that'll be 1/2 minus negative 1/2 which is 1 over 2 so that's equal to 1/2 so we've got our average change of 1/2 now we're going to try to find a place where the instant change equals 1/2 as well so in order to take this derivative I'm going to think about it as x squared minus 2 to the negative 1 so I can employ the chain rule and so that's going to give me a derivative of minus 2x over x squared minus 2 quantity squared now I want to notice that all X values on the open interval corresponding to this closed interval will give us a negative slope here and that's because if we plug in a positive number to the numerator we get a negative number from this minus sign and then the denominator is always positive because we have this thing as squared so that's like negative divided by positive which is going to be negative and let's just reiterate this is for all X on this corresponding open interval 0 to intersect Q great but what that means is that it can never equal 1/2 because the 1/2 is obviously not negative so in other words the mean value theorem won't work in this rational calculus setup either okay and maybe I want to look at a little picture of what's going on here so let's maybe think about the graph of this thing so thinking about it from calculus we know that we've got these asymptotes at plus/minus square root of 2 and furthermore we can sketch in the rest of it in the following way so it's gonna look something like this good and now I want to what I want to point out is to the right of the y-axis this function is always decreasing in other words it has a negative slope everywhere but the secant line that we just calculated is something like that which clearly has a positive slope so obviously there's that asymptote at the middle but because that asked occurs at an irrational number our function is actually continuous on its whole domain which is considered to be the rational numbers okay so I'm gonna go ahead and clamp the board and then we're gonna look at an example of the extreme value theorem so we just showed that the mean value theorem really required the real numbers it would not work on our calculus of the rational numbers now we're gonna repeat that process for the extreme value theorem so let's recall what that says we want to suppose that F is continuous on a closed interval a B then it achieves its maximum and minimum minimum on this set now we're gonna do this the same kind of way that we did before in other words build a very simple example out of the fact that the square root of 2 is not rational so let's maybe consider this function so it's going to be x squared minus 2 quantity squared so now notice this thing is continuous everywhere and differentiable everywhere on the real numbers and thus it is also on the rational numbers so what I want to do is use the fact from calculus one that we know that these extreme values have to happen at critical points in order to kind of hack together what is missing here so we'll take the derivative of this thing again we'll have to use the chain rule so that's going to give us 4x times x squared minus 2 so that means we've got critical points at x equals 0 and x equals plus minus the square root of 2 but since we're outside of the rational numbers we really only have these critical points at 0 good now the next thing that I want to do is notice that we could graph this thing and if you graph this thing it has the following picture so it's gonna look like this so it's gonna look like that but the problem here is that these minimums occur at plus minus the square root of two which means they're holes so all we have to do is consider some sort of rational interval around one of these holes and we'll see that we have a problem so let's maybe consider the rational interval from 1 to 2 in other words let's consider this interval 1 to 2 intersect rationals and what we want to claim is that this function does not achieve its minimum on this interval and I want to point out that this interval is something like from here to here good so let's maybe make that claim so claim F does not achieve its minimum here great and let's let's maybe do that so let's see how this proof goes so the first thing that we want to notice is that f of X is bigger than 0 strictly bigger than 0 for X in the interval 1 2 if we intersect that with the rational numbers and that's clear because the 0 that it has is at the square root of 2 but that's outside that rationalised interval so the next thing that we'll do is work via a contradiction so in other words we're working by way of contradiction we'll suppose that we actually do achieve a minimum and so let's by way of contradiction suppose that f of X naught equals M and that is the minimum on this interval great and then this is going to boil down into two cases and here's where this turns from like something a little bit more careful to something a little hand wavy in this first case is if X naught is less than the square root of 2 so in other words X naught is here great then what we need to do is find some N which is a natural number such that X naught plus n lives in between X naught and the square root of 2 so let's maybe see how we could do that we could have X naught which is strictly less than X naught plus 1 over head which is strictly less than the square root of 2 so we can always find a one over in which is less than the square root of two minus X naught by the Archimedean principle okay good so but now what we want to notice is that by the behavior of the function if we evaluate the function at X naught plus 1 over N we'll get a number which is smaller than X naught so let's write that down so f of X naught plus 1 by n is smaller than f of X naught but that's equal to the minimum so we found a value of the function which is smaller than its minimum which is a contradiction and then the other case is essentially the same just reflect it across this point right here okay so it looks like the extreme value theorem is not good either over our rationalized calculus now I'll clean up the board and really point towards what is going on here ok so now we've seen that a lot of things go wrong when we work in calculus over the rational numbers in other words this rationalized calculus and at the heart of it is this axiom of completeness of the real numbers so that says the following so every non-empty subset of the real numbers which is bounded above has a least upper bound so we're not going to get into the technicalities of that but I have a real analysis playlist that's kind of being built right now that has some of this stuff in there and what happens is that this axiom of completeness is one of the important seeds in the proof of the intermediate value theorem and the mean value theorem and the extreme value theorem and furthermore if you assume any of these theorems in other words if you assume the intermediate value theorem is true the mean value theorem is true or the extreme value is true with extreme value theorem is true without assuming the axiom of completeness these will also imply the axiom of completeness so in fact these are kind of like equivalent formulations ok so I think that's a good place to stop

Channel: Michael Penn

Views: 41,588

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: vV7ZuouUSfs

Length: 26min 13sec (1573 seconds)

Published: Sat Jun 27 2020