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

[michonamona]

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

 

[Hurkyl]

Because then f would be discontinuous! (assuming the statement is true, which it is)

 

[boboYO]

Remember that between any 2 rational points (in R) there is an irrational point.

 

[michonamona]

You guys are life savers.

Thank you!

M