# Continuity Question: Rational vs Irrational Functions

• kidsmoker
In summary, the first function given is continuous at all irrational values of x because as x approaches a, the distance between x and the closest rational number approaches 0. However, the second function given is discontinuous at all values of x other than x=0 because the limit of the function is 0 at all real values of x, but the value of the function is not equal to the limit unless x=0. This is because for every epsilon, there are only finitely many denominators q such that p/q is within epsilon of any given x, so as epsilon approaches 0, both p and q must go to infinity. Therefore, the limit is only equal to the value of the function at x=0.
kidsmoker
Hi.

In the book I'm reading it gives the function

f(x) = 0, if x is irrational
f(x) = 1/q, if x=p/q in lowest terms.

It says this is continuous at all irrational x. This i can understand i think, because you can show that f(x) tends to zero, as x tends to a, for all a. For this you just take epsilon to be the distance from a to the closest rational number. It can be seen from the graph:

http://img80.imageshack.us/img80/7246/26351357js6.jpg

Then they give the function

f(x) = 0, if x is irrational
f(x) = x, if x is rational.

But they say this is discontinuous at all values of x other than x=0. I don't really understand why this is the case. How come you can't use the same argument as above to show that f(x) tends to zero, as x tends to a, for all a again?

Thanks!

Last edited by a moderator:
For this you just take epsilon to be the distance from a to the closest rational number

This doesn't exist

To show that the first function is continuous, you have to notice that for every epsilon, there exists n such that 1/n<epsilon. Then there are only finitely many rationals that are 'near' (say within distance 1) an irrational x with denominator less than or equal to n, so if you pick some d smaller than the distance to the closest one of those, then

|x-y|<d implies f(y) = 1/q for some q larger than n (since if y is rational, we constructed d to force y to have a larger denominator) or f(y)=0 if y is irrational. Hence |f(x)-f(y)| = |f(y)| < 1/n < epsilon in either case. Hence f is continuous at every irrational point.

Actually, what you can show is that the limit is 0 at all real values of x. That is because, given any $\epsilon> 0$, there exist only a finite number of denominators q such that p/q is within $\epsilon$ of any given x, so, as $\epsilon$ goes to 0, both p and q must go to infinity. Since for x rational, f(x)= x, the limit, 0, is equal to the value of the function if and only if x= 0.

Office_Shredder said:
This doesn't exist

To show that the first function is continuous, you have to notice that for every epsilon, there exists n such that 1/n<epsilon. Then there are only finitely many rationals that are 'near' (say within distance 1) an irrational x with denominator less than or equal to n, so if you pick some d smaller than the distance to the closest one of those, then

|x-y|<d implies f(y) = 1/q for some q larger than n (since if y is rational, we constructed d to force y to have a larger denominator) or f(y)=0 if y is irrational. Hence |f(x)-f(y)| = |f(y)| < 1/n < epsilon in either case. Hence f is continuous at every irrational point.

Okay yeah I understand it better now.

## What is the difference between rational and irrational functions?

Rational functions are functions that can be expressed as a ratio of two polynomial functions, while irrational functions cannot be expressed in this form and may involve non-repeating decimal digits.

## What are some examples of rational and irrational functions?

Examples of rational functions include f(x) = (x + 1) / (x - 2) and g(x) = 3x^2 + 2x - 5. Examples of irrational functions include h(x) = √(x + 1) and k(x) = log(x).

## How do we determine if a function is rational or irrational?

A function is rational if it can be written as a ratio of two polynomial functions. If a function cannot be written in this form, it is irrational.

## What are some properties of rational and irrational functions?

Rational functions are continuous and smooth, meaning they have no breaks or sharp turns. Irrational functions, on the other hand, may have discontinuities or sharp turns due to the nature of their non-repeating decimal digits.

## How do rational and irrational functions behave as x approaches infinity or negative infinity?

Rational functions tend to a specific value as x approaches infinity or negative infinity, depending on the degree of the numerator and denominator. Irrational functions, on the other hand, may oscillate or not have a specific limit as x approaches infinity or negative infinity.

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