Given any real numbers a and b such that a < b, prove that for any natural number n, there are real numbers x1, x2, x3, ... , xn such that a < x1 < x2 < x3 < ... < xn < b.(adsbygoogle = window.adsbygoogle || []).push({});

The hint I was given says : Define xi recursively by x1 = (a+b)/2 and x(i+1) = (xi +b)/2. Prove that xi < xi + 1 < b, and use this result to prove by induction that a < x1 < x2 < x3 < ... < xn < b for any n in the natural numbers.

Okay so the hint doesnt really help me at all but I think I have an idea of what the question is asking. Lets suppose a = 1 and b= 10. Then there are natural numbers between these such that 1 < .... < 10. So going back to the original proposition I could say that there are real numbers x1, x2, x3, ..., xn such that a < x1 < x2 < x3 < .. < xn < x(n+1) < b. Not really sure where to go from their.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Given any real numbers a and b such that a<b, prove that for any natural number n

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