Question

A number is dyadic if it can be expressed as the quotient of two integers \(\mathrm{m} / \mathrm{n}\), where \(n\) is a power of 2 . (These are the fractions into which an inch is usually divided.) Let \(f(x)= \begin{cases}0 & \text { if } x \text { is dyadic } \\ 3 & \text { if } x \text { is not dyadic }\end{cases}\) Why does fnot have a definite integral over the interval \([0,1] ?\)

Short answer

Answer: No, the function \(f(x)\) does not have a definite integral over the interval \([0,1]\) because it lacks continuity in this interval due to the presence of infinite dyadic numbers.

Step-by-step solution

Definition of Dyadic Numbers

A dyadic number is a rational number that can be expressed as the quotient of two integers \(m/n\), where \(n\) is a power of 2. These are fractions into which an inch is usually divided. For example, the numbers \(1/2\), \(3/4\), \(5/8\) are all dyadic since they can be expressed as a quotient where the denominator is a power of 2.

Given Function

The given function \(f(x)\) is defined as follows: $$ f(x)= \begin{cases} 0 & \text{ if } x \text{ is dyadic}, \\ 3 & \text{ if } x \text{ is not dyadic}. \end{cases} $$

Integrability of f(x)

A function is considered integrable over an interval if each point within the interval has a well-defined value, and the function is continuous over that interval. We will examine the continuity of function \(f(x)\) using its piecewise definition.

Setting up the Integral

We want to find the definite integral over the interval \([0,1]\). So, let's set up the integral: $$ \int_{0}^{1} f(x)\, dx $$

Continuity of f(x)

Since there are infinitely many dyadic numbers within the interval \([0,1]\), there is a "jump" or discontinuity between the values of 0 and 3 at every dyadic number. This means that there is no continuous path between non-dyadic and dyadic numbers within the interval \([0,1]\). As a consequence, the function is not continuous over the interval \([0,1]\).

Conclusion

The function \(f(x)\) lacks continuity within the interval \([0,1]\) due to the presence of infinite dyadic numbers. Therefore, function \(f(x)\) does not have a definite integral over this interval.