1. ## real analysis

$\displaystyle Let \ x \in R \ with \ x>0 \ . show \ that \ there \ is \ n \in N such \ that \ n-1 \leq x < n .$

2. Do you know that the positive integers are well ordered?
Let $\displaystyle K=\left\{k\in \mathbb{Z}^+: x<k\right\}$.
Let $\displaystyle n$ be the first term in $\displaystyle K$.

3. Originally Posted by flower3
$\displaystyle Let \ x \in R \ with \ x>0 \ . show \ that \ there \ is \ n \in N such \ that \ n-1 \leq x < n .$
Suppose the contrary that such number does not exist, then for any $\displaystyle n\in\mathbb{N}$, we have $\displaystyle x<n-1$ and $\displaystyle x\geq n$, which implies $\displaystyle n-1>x\geq n$ or equivalently $\displaystyle -1\geq0$.

4. Originally Posted by bkarpuz
Suppose the contrary that such number does not exist, then for any $\displaystyle n\in\mathbb{N}$, we have $\displaystyle x<n-1$ and $\displaystyle x\geq n$, which implies $\displaystyle n-1>x\geq n$ or equivalently $\displaystyle -1\geq0$.
The nagation in the above is incorrect.
$\displaystyle n-1\le x < n$ means $\displaystyle n-1\le x\text{ and }x < n$.
The negation of that is $\displaystyle n-1> x\text{ or }x \ge n$.

5. Originally Posted by Plato
The nagation in the above is incorrect.
$\displaystyle n-1\le x < n$ means $\displaystyle n-1\le x\text{ and }x < n$.
The negation of that is $\displaystyle n-1> x\text{ or }x \ge n$.
Oops! How did I do that mistake
Then, let me correct.
Suppose that for every $\displaystyle n\in\mathbb{N}$, $\displaystyle x\not\in[n-1,n)$. However, since $\displaystyle x$ is fixed, we can find $\displaystyle m\in\mathbb{N}$ such that $\displaystyle x<m$, i.e., $\displaystyle x\in[0,m)$
Then, we know that $\displaystyle x\not\in[m-1,m)$, similarly $\displaystyle x\not\in[m-2,m-1)$, and by following similar arguments, we show that $\displaystyle x\not\in[m-m,m-m+1)=[0,1)$.
This shows that $\displaystyle x$ must be negative, and thus is a contradiction.

But I have to say that this is not a nice proof.

Thanks to Plato for pointing out my mistake.