Dirichlet's Function: Why is it Difficult to Draw the Graph?

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SUMMARY

Dirichlet's function, defined as f(x) = 1 for rational x and f(x) = 0 for irrational x, presents significant challenges in graphing due to the dense interspersion of rational and irrational numbers. The graph cannot be drawn as a continuous line because any interval contains both types of numbers, leading to constant switching between values. The lower Riemann sum L(x_0,...,x_n) is always 0, while the upper Riemann sum U(x_0,...,x_n) is always 1, reflecting the density of rationals and irrationals in any chosen interval.

PREREQUISITES
  • Understanding of Dirichlet's function and its definition
  • Familiarity with concepts of rational and irrational numbers
  • Knowledge of Riemann sums and their calculations
  • Basic graphing skills in mathematics
NEXT STEPS
  • Study the properties of dense sets in real numbers
  • Learn about Riemann integrability and conditions for functions
  • Explore the implications of discontinuous functions in calculus
  • Investigate other examples of functions with similar properties, such as the Thomae's function
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Mathematics students, educators, and anyone interested in real analysis or the properties of functions involving rational and irrational numbers.

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Homework Statement


Define Dirichlet's function f by putting f(x) = 1 if x is rational and f(x) = 0 if x is irrational. Explain why it is difficult to draw the graph of f. Prove that the lower Riemann sum L(x_0,...,x_n) is always equal to 0 and the upper Riemann sum U(x_0,...x_n) is always equal to 1.


Homework Equations


Equations for upper and lower Riemann sums.

The Attempt at a Solution


Hi everyone,
Here's what I've done so far:

The graph is difficult to draw because there are infinitely many rational numbers and infinitely many irrational numbers, all interspersed among one another, so you will continuously be switching between f(x) = 0 and f(x) = 1.

For every rational number, there is an irrational number, so any chosen interval will contain both a rational [f(x) = 1] and irrational [f(x) = 0] number.
m = 0 and M = 1
So the lower Riemann sum will be zero, as two points side-by-side (i.e. a rational and an irrational with only 'vertical' area between them) will have minimum vertical area 0.
And, for the upper Riemann sum, two points side-by-side will have maximum vertical area 1.



Is this correct?

Thanks for any help
 
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Yep, graph difficulty is correct. The graph would look like two solid lines, y=1 and y=0.

The second part is close. Between every two rational numbers lies an irrational number. In fact, the irrationals are dense in the reals. Furthermore, between every two irrationals is a rational, and generally between any two real numbers there are both irrational and rational numbers. And you got the rest.
 

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