0 like 0 dislike
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?

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 ..."
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.