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Chapter 5: Problem 17

If \(f:[0,1] \rightarrow \mathbb{R}\) is continuous and has only rational [respectively, irrational] values, must \(f\) be constant? Prove your assertion.

Short Answer

Expert verified
Yes, a function \(f:[0,1] \rightarrow \mathbb{R}\) that is continuous and only takes rational or only irrational values must be a constant function.

Step by step solution

01

Statement of the Problem

We are given a continuous function \(f:[0,1] \rightarrow \mathbb{R}\) that only takes rational or irrational values. We want to determine if the function must be constant and provide a proof for our assertion.
02

Understanding Continuity

A function is continuous at a point if the limit at that point exists and is equal to the function value at that point. A function is continuous on an interval if it is continuous at every point in the interval.
03

Understanding the Density of the Rational and Irrational Numbers

The rational numbers and irrational numbers are dense in the real numbers. This means that between any two real numbers, no matter how close they are, there is always a rational number and an irrational number.
04

Case for Rational Values

Assume that the function \(f\) only takes rational values. Let's denote two arbitrary points in [0,1] as \(x\) and \(y\), with \(x < y\). From the density of irrational numbers between \(x\) and \(y\), we can find an irrational number \(z\). However, \(f(x)\) and \(f(y)\) are both rational numbers. By the Intermediate Value Theorem (IVT), since \(f\) is continuous, for every number \(N\) between \(f(x)\) and \(f(y)\), there must exist some \(c\) in [\(x,y\)] such that \(f(c) = N\). Hence, \(f(z)\) must be rational, which contradicts the assumption that \(z\) is irrational.
05

Case for Irrational Values

Assume now that \(f\) only takes irrational values. If we repeat the argument from Step 4 but now use the density of rational numbers, we reach a similar contradiction.
06

Conclusion

Given the contradictions in Steps 4 and 5, we conclude that our initial assumption that \(f\) only takes rational or irrational values must be false. Therefore, a continuous function \(f:[0,1] \rightarrow \mathbb{R}\) that only takes rational or only irrational values must be constant.

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