For all real numbers satisfying $a < b$ , there exists an $n ∈ N$ such that $a + 1/n < b$

Question

For all real numbers satisfying $a < b$ , there exists an $n ∈ N$ such that $a + 1/n < b$.

My try:

Contradiction: For all real numbers satisfying $a < b$ and any $n ∈ N$; $a + 1/n \ge b$.

Then $1/n \ge b - a$, then $n \le \frac{1}{b-a}$, take $n = \frac{1}{k(b-a)}$, where $k ∈ N$, then $a + \frac{1}{1/k(b-a)} < b$ => $a(1-k)<b(1-k)$ => $a>b$, since $1-k<0$ for all defined $k$, but $k = 1$ then I recieve $0<0$.

Answer 1

Your choice of $n$ is not correct. It need not even be an integer. Take $n= [\frac 1 {b-a}]+1$ where $[.]$ is the floor function. [ Note that $n >\frac 1 {b-a}$ .(I have added $1$ to $[\frac 1 {b-a}]$ to make sure that I get strict inequality). Now $a+\frac 1 n <a+(b-a)=b$ and this is what we want].

Answer 2

There is a bug in your proof. You said take $n = \frac{1}{k(b-a)}$. How did you find the $k$ such that $n$ is an integer, besides $k$ is an integer too? It is not possible at all.

For example $b = \pi$ and $a = e$ (nepper number). How did you find such $k$ that $n = \frac{1}{k(\pi - e)}$ will be integer too? You can find the problem here.

Answer 3

You May argue directly without using a proof by contradiction. By the properties of the reals there is some integer $n>\frac1{b-a}$. Thus $a+\frac1n<a+(b-a)=b$.

Answer 4

An option:

We have $b-a >0$, $a,b$ real, and let $n, n_0$ be positive integers

$\lim_{n \rightarrow \infty } (1/n) =0$:

Let $\epsilon >0$ be given:

There is a $n_0 >0$, s.t. for $n \ge n_0$

$|1/n| \lt |1/n_0| \lt \epsilon$.

Choose $\epsilon \lt (b-a)$.

Then

$1/n_0 \lt \epsilon = (b-a)$, or

$a +1/n_0 < b$.