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 $0<0$.
Best Answer
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 $a+\frac1n<a+(b-a)=b$.