Prove that if a>0, then there exists n∈N such that 1/n <a<n.

Question

Here is my solution: If we have two real numbers \(\displaystyle a\) and \(\displaystyle 1\), then by Archimedean property we have $$ a \lt 1 \times n_1 \\ a \lt n_1$$ Again, given two real numbers \(\displaystyle a\) and \(\displaystyle 1\), by Archimedean property we have $$ 1 \lt n_2 a \\ 1/n_2 \lt a $$ Let's take \(\displaystyle n = ~max~(n_1, n_2)\) we have $$ 1/n \leq 1/n_2 \lt a \lt n_1 \leq n \\ 1/n \lt a \lt n$$ Hence proved. Is my solution correct?

Answer 1

I would say that it is important to include something like "by Archimedean principle there exist natural numbers n1 and n2 such that ... That of course is just my opinion and I am not a mathematician (I have a bachelors and masters degree in math)

Answer 2

Adesh said: Here is my solution: If we have two real numbers \(\displaystyle a\) and \(\displaystyle 1\), then by Archimedean property we have $$ a \lt 1 \times n_1 \\ a \lt n_1$$ Again, given two real numbers \(\displaystyle a\) and \(\displaystyle 1\), by Archimedean property we have $$ 1 \lt n_2 a \\ 1/n_2 \lt a $$ Let's take \(\displaystyle n = ~max~(n_1, n_2)\) we have $$ 1/n \leq 1/n_2 \lt a \lt n_1 \leq n \\ 1/n \lt a \lt n$$ Hence proved. Is my solution correct? Click to expand... It's not important, but you should have included something like "by Archimedean principle there exist natural numbers $n_1$ and $n_2$ such that ..."

Answer 3

Jomo said: Define n1 and n2! Why can't n2a Click to expand... I thought if I mentioned that "by Archimedean Property" it was quite clear that all \(\displaystyle n\)s were natural numbers. But if you say so I shall from next time write it out explicitly that \(\displaystyle n\)s are natural numbers.

Answer 4

Please do not put your question in the heading. Define n1 and n2! Why can't n2a