Question

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Answer 1

any more info?

Answer 2

I tend to think that the proof of this would rely on the fact that for any rational number, there is an irrational number in its neighborhood.

Answer 3

nope...or similar question Let f:[a,b]→Q be a continuous function. Prove that f is a constant function.

Answer 4

If it doesn't rely on what I said previously, what does it rely on?

Following my previous line of thought: Suppose it wasn't constant so that \(f(a)\neq f(b)\) for \(a, b \in \mathbb{R}\). Then, since there is an irrational number in the neighborhood, it must hit that number so that there is some \(c\in\mathbb{R}\) such that \(f(c) \notin \mathbb{Q}\). This is a contradiction of the definition, so it must be constant.

Answer 5

ooh i get it anything from p to q is just a distance which is a constant

Answer 6

magnitude is sqrt(p^2+q^2)

Answer 7

i think it is rely on rational and irrational number

Answer 8

so if it rely on what are the answer

Answer 9

dude all line segments are just distances, constant distances.