Exercise 4.4.1 (Integer part)

Question

Prove Proposition 4.4.1. (Hint: use Proposition 2.3.9.)

Richard Ganaye · Accepted Answer

ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} {\bf Proposition 4.4.1.} (Interspersing of integers by rationals). {\it Let $x$ be a rational number. Then there exists an integer $n$ such that $n\leq x < n+1$. In fact, this integer is unique (i.e., for each $x$ there is only one $n$ for which $n\leq x < n+1$). In particular, there exists a natural number $N$ such that $N > x$ (i.e., there is no such thing as a rational number which is larger that all the natural numbers). } \begin{proof} {\bf Existence.} Let $x \in \Q$. then $x = \frac{p}{q}$, where $p \in \Z, q \in \N^*$. By Proposition 2.3.9. (Euclidean division) there exist a quotient $n$ and a remainder $r$ such that $$p = q n + r,\qquad 0\leq r < q.$$ Then $x = \frac{p}{q} = n + \frac{r}{q}$, where $0 \leq \frac{r}{q} < 1$. Therefore $n \leq x < n+1$. This proves the part ``existence" of the Proposition. \bigskip {\bf Unicity.} Assume that $n \leq x < n+1$, and $m \leq x < m+1$, where $m,n \in \N$. By transitivity, $n \leq x < m+1$, thus $n < m+1$. Since $n,m$ are integers, $n \leq m$. Similarly, $m \leq x < n+1$, therefore $m < n+1$ and $m \leq n$. This shows that $m=n$. If $N = n+1$, then $N>x$, and $N \in \N$: this is the Archimedean property in $\Q$. \bigskip The unique $n \in \N$ such that $n \leq x < n+1$ is denoted $n = \lfloor x floor$. \bigskip Note : If $p = q n + r,\ 0\leq r < q$, then $n = \lfloor \frac{p}{q} floor$. This prove the unicity of Euclidean division, not exposed in Proposition 2.3.9., which can also be proved directly, without rational numbers. \end{proof}

Exercise 4.4.1 (Integer part)

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Exercise 4.4.1 (Integer part)

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