Isn't it just obvious that $\frac {m_j} {j} < x < \frac {m_j + 1} {j}$? Isn't it just obvious that $\frac {m_j} {j} < x < \frac {m_j + 1} {j}$ in the solution below?
I can't understand how this solution concludes the above statement from contradiction (for example, it wrote "if there were three rationals having j as denominator, ..... , contadicintg...")
Could you help me understand it?
Do I need to prove $\frac {m_j} {j} < x < \frac {m_j + 1} {j}$?

 A: The fact that
$$\frac{m_j}j < x < \frac{m_j+1}j$$
is not being proven. That inequality is part of a sentence, the end of which is as important as the beginning. The red box gives it unfortunate emphasis.
A: The person who wrote this 'proof' starts with a neighborhood centered around x with width 2/j.  He states that in this interval there can be at most two rational numbers (in lowest terms) with denominator j.  So far he is fine and he is right. It hardly needs further proof, since the interval is specifically defined that way.  
However, he seems to think he does need more proof, so he proceeds to get himself into trouble and confuse his reader.  He says that if there were 3 rational numbers in that interval, all with denominator j, x would have to be one of them and thus rational.  I simply do not follow that logic. There are lots of numbers between m/j and (m+1)/j, some rational some not.  The fact that x is one of them does not imply it is rational.
I'm also a little suspicious of his choosing j < n.  If he wants to close in on x, he had better let those j's get large.  Maybe he meant 1/j < 1/n, which would make more sense.
The result is not that hard to prove, but he is doing a poor job of it.  Why don't you look at another book.  
