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Let f be continuous real-valued function with domain (a,6). Show that if =0 for each rational number r in (a,6) , then f(z) = 0 for all x € (a,6). (b) Let f a...

Question

Let f be continuous real-valued function with domain (a,6). Show that if =0 for each rational number r in (a,6) , then f(z) = 0 for all x € (a,6). (b) Let f and g be continuous real-valued functions on (a,b) such that f(r) = g(r) for each rational number in (a,6). Prove f() = g() for all & € (a,6).

Let f be continuous real-valued function with domain (a,6). Show that if =0 for each rational number r in (a,6) , then f(z) = 0 for all x € (a,6). (b) Let f and g be continuous real-valued functions on (a,b) such that f(r) = g(r) for each rational number in (a,6). Prove f() = g() for all & € (a,6).



Answers

Show that $f$ is a discontinuous function for all $x,$ where $f(x)$ is defined as follows: $$ f(x)=\left\{\begin{array}{ll} 1 & \text { for } x \text { rational } \\ -1 & \text { for } x \text { irrational } \end{array}\right. $$ Show that $f^{2}$ is continuous for all $x$.

We want to determine. But what values of X This piece wise function of f of X, equally zero if X is irrational and one effects is rational, could be continuous. Well, let's just maybe try to draw a sketch of what this graph is. Now. Remember, the rational numbers are all the numbers that could be written in as a fracture with Interred your top and bottom. So that's what I say over here with the P over Q with p o ver que being in Z and then all the irrational numbers. Our numbers such that they cannot be written as the ratio of to, uh, intruders. So examples of this will be like pi e square root of two Natural Log on to anything kind of like that kind of falls into the camp of Irrational. Well, let's draw this craft so it says it. Zero. When X is your rational, well, there's there's simply many. So I mean, I just have a ton here and then it also says it's one. If X is rash on again, there's employees, many of these, so we just kind of have this dotted line here. Well, let's think about what would be the limit as exp approaches a for any value of X will does his approach anything and you might notice that it doesn't. Because if this here is A as we get closer and closer, they is just going to constantly be bouncing on either side between the blue and the green line. And since there are infinitely many numbers in between, them will never actually end up getting close to anything. So what we could say is that this limit here does not exist, since it never gets close to a number because it bounces bounces between zero and one constantly and in the next chopper weaken probably do this a little bit more rigorously, as opposed to just saying that this limit here doesn't exist. And so, since limit exists doesn't exist, we wouldn't have it be continuous either. Actually, that's what we should probably say after this, uh, us. No limits exist, and it would be nowhere continuous

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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.


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