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

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In summary, the conversation discusses the proof that a continuous function with a domain of [a,b] and a range of rational numbers must be constant on [a,b]. The solution explains that this is because if f([a,b]) contains multiple rational points, it would also have to contain an irrational point, which would contradict the assumption that f is continuous. The conversation also addresses a question about why f cannot be a collection of rational points, and the response is that this would make f discontinuous.
  • #1
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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  • #2
Because then f would be discontinuous! (assuming the statement is true, which it is)
 
  • #3
Remember that between any 2 rational points (in R) there is an irrational point.
 
  • #4
You guys are life savers.

Thank you!

M
 

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