$0<p<1 \implies \exists n\in\mathbb{N}$ such that $p<1-\frac{1}{n+1}$

Question

Problem statement:

Let $p \in \{r\in\mathbb{Q}:0<r<1\}$. Then prove that $ \exists n\in\mathbb{N}$ such that $p<1-\frac{1}{n+1}$.

Context:

Source. Although it seems like a trivial thing to say, I don't feel comfortable just accepting it especially when I'm trying to get a firm foundation for analysis. Hence, my question.

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Proof Attempt:

Suppose for a contradiction, there is no such $n\in\mathbb{N}$. $\forall n\in\mathbb{N},\hspace{1mm}p\geq 1-\frac{1}{n+1}$.

$\underline{\text{Establish a supremum}}$:

Let $A = \{r\in\mathbb{Q}: \exists n\in\mathbb{N},\hspace{1mm} r = 1-\frac{1}{n+1}\}$. Since $a=1-\frac{1}{2}\in A$ and $a=1-\frac{1}{n+1}\leq 1$, $A$ is nonempty and bounded.

$\underline{\text{Claim lub}(A)=1}$:

Since this is with reference to rational numbers, I can't use the axiom of completeness. However, I should still be able to produce a least upper bound.

Clearly an upper bound for $A$ is $1$ since $1>1-\frac{1}{n+1}$ for all $n\in\mathbb{N}$. Suppose there is an upper bound $\beta<1$. Then $\beta<\frac{\beta+1}{2}<1$ which means $\beta$ is not an upper bound.

$\underline{\text{Contradiction}}$:

Since $\forall n\in\mathbb{N}$, $p\geq1-\frac{1}{n+1}$, $p\not\in A$. Then $p\geq\text{lub}(A)=1$ which contradicts our initial assumption that $p \in \{r\in\mathbb{Q}:0<r<1\}$.

Is this a correct proof? Is there an easier way to do this?

Answer 1

Since $p < 1$ we have $1 - p > 0$, and then by the Archemedian property, $\exists n \in \mathbb{N}$ such that $n \left( 1 - p \right) > p$. Adding 1 and subtracting 1 on the right hand side, we get

$$n \left( 1 - p \right) > 1 - \left( 1 - p \right)$$

On simpliifying, we get

$$p < 1 - \dfrac{1}{n + 1}$$

Answer 2

Since $p$ is rational, it can be written as $\frac ab$ with $a\in\Bbb Z$ and $b\in\Bbb N$. Then $p<1$ is equivalent to $a<b$, or $a\le b-1$. Let $n=b$. Then $$p+\frac 1{n+1}=\frac ab+\frac1{b+1}=\frac{a(b+1)+b}{b(b+1)} \le \frac{(b-1)(b+1)+b}{b(b+1)}=\frac{b^2+b-1}{b^2+b}<1$$

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Or in short, $p+\frac1b\le 1$, hence $p+\frac1{n+1}<1$ whenever $\frac1{n+1}<\frac1b$, i.e., whenever $n+1>b$.

Answer 3

Here is how one can find such an $n$ without using any unnecessary and complicated stuff.

Since $0<p<1$ it follows that $0<q=1-p<1$ and then $r=1/q>1$. Let $r=a/b$ where $a, b\in \mathbb {N} $. Consider the quotient $n$ when $a$ is divided by $b$. Such a quotient exists by division algorithm. Then $n\leq a/b<n+1$ and hence $q=b/a>1/(n+1)$ and therefore $p=1-q <1-\dfrac{1}{n+1}$.