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T.F.J There exists function from real numbers to real numbers such that the function is Continuous on rational numbers...

Question

T.F.J There exists function from real numbers to real numbers such that the function is Continuous on rational numbers

T.F.J There exists function from real numbers to real numbers such that the function is Continuous on rational numbers



Answers

A rational function can have infinitely many $x$ -values at which it is not continuous.

What we want to thio is Thio Show that the function f of X that is equal to one if X is rational and zero if X is irrational, um, is discontinuous at every point. Okay, um and so we know that, um, a couple things we, though, that a, um a non empty interval of real numbers, um, contains both rational and irrational numbers. And so we're gonna go ahead and use that, um, to help us show that at every point, um, this function is discontinuous. Now, there's several ways we could do that. We could use the Epsilon Delta definition, or we can just kind use some logic on. We're gonna go ahead and just use, um, logic. So, um, we're gonna let see be any real number. And we know that if a function is continuous at sea, then the limit as X approaches see of that function must exist. Okay. And so since there are both rational an irrational clips, irrational numbers arbitrarily close to see, um, the limit is not going to exist. And it's really because as approaching see, my limit of f of X is going to balance between one and zero. depending on, um what value? Whether it's rationally rational, I land on. And since it's not going to stabilize at a single value, the limit as a whole does not exist. Now, we can also use probably wth e epsilon, um, Delta definition or the definition of continuity. Thio prove that. But just like logic, we can actually show that, um, that because, um, we're letting see, just be any real number. And so it could be any real number could be irrational, irrational. As I am approaching that real number, my function is not going to settle on one value because as I approach that really number, that I'm gonna be landing on both rational, irrational numbers at any given time, which is going to about set function value from 1 to 0. And so the limit as a whole does not exist. Now, Um, let's, um, talk about, um um and let's talk about is f of X right, continuous or left continuous. And what that means is, is does the limit. As I'm approaching a real number from the right, does it land on a single value? Or, as I'm approaching the exile, you or that see value from the left. Does my function land on a single value, or does it approaches single value And by the same logic previously both of these do not exist. And so the answer is no. The function is not right continuous or left continuous or, for that matter, um, continuous at any, um, real number for that function.

Okay. What we want, Thio, what we're starting with is, um f of X is equal to one. If x is rational and zero if X is irrational, okay. And so we want to show that f of X is discontinuous at every point. Okay? And so if we want to show continuity the definition Ah, continuity, um says that, um f of sea. So your function evaluated at some X value, um, has to be defined. And the limit as X approaches see of your function has to exist And those to the function evaluated at that see value, Hasi, with the limit as X approaches, see of the function. So that is the definition of continuity. And so what we need to do is to show that if the functions discontinuous at every point that it does, this definition of continuity does not hold true. And so let's go ahead and get started. Um, we're going to use, um the fact that non empty um interval, um, of real numbers, um, contain, um, both rational and irrational numbers. Okay. And so we're gonna let, um, we're gonna let, um, first of all, if we're gonna let see, um B um, a rational number. And I'm gonna do it for the rational number case. But it can be still shown We can do it again for irrational numbers is gonna hold true. And so, um, if I have my function evaluated at some rational, um, number, then that is going to equal wine. Now, if I let the limit as X supposed approaches, see of that rational number, this does not exist. And it is because when we're approaching this rational number, we are having to, um, go through an interval that is comprised of rational and e rational numbers. And so, when I am approaching the sea value, um, I can go from a function value of 101010 so it will never settle on any value. So this limit does not exist. And so therefore, because those two do not equal each other and more importantly, this too thought exist, then my, um, function is discontinuous. Um, for that see value or more importantly, for every, um, point. And so we can also kind of show this for, um, you Russell numbers as well. Now, another question poses itself because really, what? We're doing here? We were looking at, um, ex approaching. See, we're really in our mind's eye looking at it as we're approaching it from the left and from the right. And so if I'm approaching it from the left, I could be approaching. When I get arbitrarily close to see, I could be landing on an irrational number, which means I'm approaching zero. Whereas if I'm looking at it from the right, I could be arbitrarily approaching a rational number, which then my wife values approaching one, so I will never get connected at the same value. And so this can be, um this can be, um Also look at, um the question is is f of X, um, right, continuous or left continuous. And the answer is no by the same justification. Um, because if, um, the limit as X approaches see, from the right of the function as well does not exist because we are arbitrarily approaching C and so see bounces ASM approaching. See from the right, I am. I am bouncing from rational to irrational numbers so it will never approach a single value. And the same can be true for the limit as X approaches see from the left as well does not exist, and since those right and left limits do not exist, then the function is not continuous, either right, continuous or left continuous.

Section one of four Problem 1 19 They give us a function this defined piece wise, so this function is F of X equal to zero. Um, may make this little bit clear here. F of X is equal to zero when X is a rational number and f of X is equal to K X when X is an irrational number, so wants to prove that this is only continuous when X is equal to zero and just show it's only continues there. So I need to show that it's continuous zero and improve that it's discontinuous everywhere else. So easy part is to prove that it's continuous at zero, so you can figure out that f of zero is equal to zero because that's a rational number. And then what is the limit As, uh, hurt me the limit as X approaches zero of f of X, okay. And so we see that as X approaches zero um, for the irrational numbers. Ah, it's the same as what happens to Cave X when X approaches zero. So his ex approaches zero. This is going to approach zero, and then for very small, rational numbers, we know that is already zero. So what I know here I know that the limit of f of X as X approaches zero exists and is equal to F zero. So by definition, this is continuity. So I have It is continuous at zero. So I know that f f of X is continuous at X equal zero. Okay, Now I want to show that it's not continuous at any other point. Okay, so what I have to do here is I know that for every in order to prove this the definition of continuity for every Epsilon greater than zero there exists a delta such that if if x minus a absolute value that is less than delta, then f of X minus f of a absolute value is less than epsilon. So what? It says if I can pick up any for any neighborhood and why that I can pick There is a neighborhood and ex that I can pick. So my point falls within the ex neighborhood. It also falls within the wine neighborhood. So let's go ahead and take a look at this. So let's let let's let my Absalon equal one. I'm gonna prove this by contradiction. Okay? And that means that there has to be a number X that is less than a plus. Delta a minus Delta. Okay, so there's a delta so that I can find a X somewhere in this interval and then come up with my continuity. So, um, let's start out with, um let's let a be rational. So let a be rational. So if a is a rational number, um, then what I come up with is that the absolute value of f of X minus f of a has to be less than one for this Delta. If a is rational, then we know that by definition, for any rational number, F of that number is zero. So f of a is going to be zero. So I have f of X less than one now, what you can do, you confined by the density. So by the density of the irrational numbers, yeah, I can find an ex that's in this interval a minus Delta a plus. Delta, find that ex That's an irrational number in this interval. Okay? And so if I do that, then what this turns into. Let's just go to a new page f of X absolute value. Vex, Listen, one if X is an irrational number, I'm going to get just what k x less than one K is any non zero, um, real number. So let's just say let Kay equal one over X. You pity many choices you could do there. This just turns into one over x times X less than one one less than one. So this is not true. So I've got a contradiction here. Okay, so this tells me that for the rational numbers, I've proven that it's not continuous. Now, let's just go do the same thing for the irrational. Let a be an irrational number. Then when you look at f of X minus F of a same same epsilon of one we know now that f of A has to be equal to what k? A. That right. So if I look at, um and if a is a rational number than F of a is equal to K A. Okay, so what this gives me is that f of X minus K A is less than one. And then, by the density of the rational numbers I confined in X, that's in that interval a minus Delta A plus Delta. That is a rational number. There's an infinite number of rational numbers between any two numbers. So let XB find a rational number in that interval. And because you know that a rational number you're going to get zero when you apply this function, she got zero minus K. A is less than one. Que is any non zero number. So let's just let Kay equal one over A. You get zero minus one over a a lesson one one lesson one. So prove my contradiction. Okay, so what I did was I started out with the original definition of the function. Easy part was to prove that F zero exists. The limit exists and they're the same. It's continuous at zero, and then by contradiction, I was able to fight

So you see the at any I in green number here on now. Consider acting coach you got. So consider eliminar f X. What excuse do I? And this one? Because the rational and irrational is Dan in real number on this limit, we would not exist. And, uh, that's on at X. Go to buy the limit, Aisha. Oscillating. That's everything. Uh, from when Jews? Oh, that's why the f thanks a not continue us everywhere.


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