Why does f([a,b]) contain only one rational point?

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Homework Help Overview

The problem involves a continuous function f defined on the interval [a,b] with the range f([a,b]) being a subset of rational numbers. The original poster seeks to understand why f([a,b]) can contain only one rational point.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions the reasoning behind the assertion that f([a,b]) must contain only one rational point and why it cannot be a collection of rational points.
  • Some participants suggest that having more than one rational point would imply the existence of irrational points between them, leading to a contradiction with the continuity of f.

Discussion Status

The discussion is exploring the implications of continuity in relation to the set of rational numbers. Participants are engaging with the original poster's questions and providing insights into the nature of rational and irrational numbers in the context of continuous functions.

Contextual Notes

There is an underlying assumption that the continuity of f plays a crucial role in the argument, and the discussion hints at the implications of having multiple rational points within the interval.

michonamona
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Homework Statement


Suppose f:[a,b]->R is continuous and that f([a,b]) is a subset of Q (rational numbers). Prove that f is constant on [a,b]


Homework Equations


N/A


The Attempt at a Solution


The solution states that:

f([a,b]) must contain only one point, because if it has more than that, then it would have to be an irrational point. Which is a contradiction.


My questions are, why does f([a,b]) contain only one rational point? Why can't it be a collection of rational points?

Thank you for your help.

M
 
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Because then f would be discontinuous! (assuming the statement is true, which it is)
 
Remember that between any 2 rational points (in R) there is an irrational point.
 
You guys are life savers.

Thank you!

M
 

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