Prove consequence of Archimedean property

Question

I have to proof that for $x,y\in \mathbb R$ with $y>0$, it exists $n\in\mathbb N$ such that $x<ny$

My proof does not convince me because its too direct (when something seems too easy, something might be wrong).

Proof:

Suppose $x,y\in \mathbb R$ with $y>0$. By archimedean property, there exists $n\in \Bbb N$ such that $x<n$. Because $y>0$, we can say that $x<n<ny$. Thus $x<ny$, proving the existence of $n$.

My doubts of this demonstration base on the way $y$ depends on the value of $n$ when it should be the other way (having $x,y$ and finding $n$).

Yes, you are right that $y>0$ does not imply that $ny>n$. And your are also right in grammar, prove $\neq$ proof. — taue2pi, Dec 01 '13 at 03:21

score 4 · Accepted Answer · answered Dec 01 '13 at 02:47

4

HINT: Apply the Archimedean property to $\frac{x}y$.

answered Dec 01 '13 at 02:47

Brian M. Scott

616,228

Now I feel bad, it follows directly with this. Thank you Brian! – taue2pi Dec 01 '13 at 03:35
@Andrew: You’re welcome! – Brian M. Scott Dec 01 '13 at 03:39

score 1 · Answer 2 · answered Dec 01 '13 at 03:19

if $x \geq yn $ for all $n$, then put $S = \{ yn : n \in \mathbb{N} \} $. the set $S$ is clearly bounded by $x$, in fact. Hence, can apply the supremum axiom to obtain $\alpha = \sup S $. Notice $\alpha - y$ is not upper bound of $S$ by definition and since $y > 0$. $(\alpha > \alpha - y )$. Hence $\exists n_0$ such that $yn_0 \geq \alpha - y $

$$ \therefore yn_0 + y > \alpha \implies y(n_0 + 1) > \alpha $$

Since $n_0 + 1 \in \mathbb{N}$, then it follows that $\alpha$ cannot be supremum of $S$. A contradiction, hence the problem is solved.

Prove consequence of Archimedean property

2 Answers2