Proof that every rational number is between 2 consecutive integers

Question

I need to prove that $\forall$ $p \in \mathbb{Q}$, $\exists n \in \mathbb{Z}$ such that $n \leq p < n+1$. What I have done is assume that such a $p$ does not exist. Which implies that either $p>n$ or $p<n$ , $\forall n \in \mathbb{Z}$. This would imply that p is either a least or greatest element of $\mathbb{Z}$ which is a contradiction. Hence I can conclude that $\exists m \in \mathbb{Z}$ such that $m<p$ and that $\exists n \in \mathbb{Z}$ such that $p<n$. Combining these 2 gives me $m<p<n$, however I'm stuck here, how can I come to the stronger conclusion that it is between two consecutive integers?

Answer 1

Take $p \in \mathbb{Q^+}$, so exists $n, m \in \mathbb{N}$ such that $p=\frac{n}{m}$. Now perform the euclidean division $n$ divided by $m$ (quotient and remainder are unique) so $n=q \cdot m+r$ with $0 \le r<m$ $\implies \frac{n}{m}=q+\frac{r}{m}$ with $0 \le\frac{r}{m}<1$.

We obtain that $q \le p<q+1.$

The case $p<0$ is the same.

Answer 2

For $0\le p<1$ it is trivial, let consider then wlog $p>1$

$$p=\frac q s =\frac{ks+r}{s}=k+\frac r s,\, 0\le r/s<1 \implies k\le k+\frac r s\le k+1$$

Answer 3

Let $\frac{p}{q}\in\mathbb Q$ where $p$ and $q$ are integers. By Euclid's division algorithm, $p=nq + r$ where $0\le r\lt |q|$.

Therefore, $\frac{p}{q}=\frac{nq+r}{q}=n+\frac{r}{q}$

If $\frac{r}{q}\ge 0$ then $\frac{p}{q}$ is between $n$ and $n+1$, otherwise it is between $n-1$ and $n$.