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Suppose $$m\le n < m+1$$

$n$ rational and $m$ a positive integer. How do we prove $m$ is unique?

I realize this is true if $m$ is an integer, so $m$ must be a unique number that is an integer that is less than $n$ and the next integer that is greater than $n$. but how would you prove this fact?

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2 Answers 2

up vote 3 down vote accepted

$\textbf{Hint:}$ Suppose $k$ is an integer such that $k\neq m$ and $k\leq n <k+1$. Since $k\neq m$, we have $k<m$ or $m<k$. In either case find a contradiction.

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Ok, i got it. seems simple enough. BTW, are there any good books to read up on abstract math? Coming from the logistical side of math, I don't get this proof stuff. I just like to crunch numbers. –  shosh Feb 1 '13 at 22:59
@shosh As an initiation to mathematical proofs, I like this one a lot: amazon.com/How-Prove-It-Structured-Approach/dp/0521446635. Is it something like this you're looking for? –  Git Gud Feb 1 '13 at 23:07
I'd have to see, thanks. I hope this book includes not only the right way to write proofs, but some of the technical details to come up with the solution in solving proofs. –  shosh Feb 1 '13 at 23:17

Let $S = \{ m \in \mathbb{Z} \mid m \leq n < m+1 \}$. By the Well-Ordering Principle, $S$ has a smallest element; call it $m_0$. Now suppose that $m_1$ is an integer satisfying $m_1 \leq n < m_1 + 1$. Then $m_1 \in S$, so $m_0 \leq m_1$. Also, $$m_1 - m_0 \leq n - m_0 \leq n - n = 0,$$ so $m_1 \leq m_0$. Since $m_0 \leq m_1$ and $m_1 \leq m_0$, we must have $m_0 = m_1$.

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