Alternative Proof that $\sqrt{p}$ is Irrational when $p$ is Prime I have found various proofs that $\sqrt{p}$ is irrational on this site, but I didn't find one similar to the one that I am about to post, so I am wondering if it is free of logical problems.
Here is the definition of infimum that I am using and a lemma
Definition. $m = \mathrm{inf} \, S$ where $S$ is a set iff (i) $m \leq s$ for all $s \in S$, and (ii) if $m < m'$, then there exists $s' \in S$ such that $m < s' < m'$.
Lemma. Let $x, y \in \mathbb{R}$ such that $x \leq y+\epsilon$ for every $\epsilon > 0$. Then $x \leq y$.
Suppose to the contrary that $\sqrt{p}$ is rational where $p$ is not a perfect square. Then there are natural numbers $a$ and $b$ such that
$$\sqrt{p} = \frac{a}{b} \implies a = b\sqrt{p}$$
Define the set$\Lambda = \{k \sqrt{p} \, : \, k, k\sqrt{p} \in \mathbb{N}\}$.
By definition, $\Lambda \subseteq \mathbb{N}$ and $\Lambda \neq \varnothing$, because $a \in \Lambda$.
The well-ordering property implies that $m = \mathrm{min}(\Lambda)$ exists in $\Lambda$. Note that $m$ is also the infimum of $\Lambda$.
Let $\epsilon > 0$.
Because $m \in \mathbb{N}$, obviously $m < m+\epsilon$, so $m + \epsilon$ is not a lower bound of $\Lambda$.
Using the definition of the infimum, there exists $m' \in \Lambda$ such that
$$m < m' < m + \epsilon \implies m' < m + \epsilon$$
But the lemma allows us to deduce that $m' < m$, which is a contradiction.
Therefore, such a set cannot exist and we must conclude that $\sqrt{p}$ is in fact irrational.
Update
This update is more for my benefit. As the other users have pointed out, the problem is when the definition of infimum was used in claiming that an integer $m^{\prime} \in \Lambda$ exists such that $m \leq m^{\prime} < m+\epsilon$. However, supremum and infimum only make sense in ordered fields; and since $\mathbb{Z}_+$ is not a field, we cannot use the concept of infimum.
 A: This is a great example of a proof that should have jumped out at you as being "too easy to be correct". I like this quote from Terrence Tao's blog:

If you unexpectedly find a problem solving itself almost effortlessly,
  and you can’t quite see why, you should try to analyse your solution more sceptically. In particular, the method may also be able to prove much stronger statements which are known to be false, which would imply that there
  is a flaw in the method. [source]

To be fair, by posting the proof here, I suppose you were being sceptical of it.
After defining your set $\Lambda$, you never use any property of it ever again, other than that it's a subset of the natural numbers! So everything you said applies just as well to any such subset, in other words, you've "proven" that there are no subsets of $\mathbb N$.
The definition of the infinimum of a subset of $\mathbb R$ is the greatest real number which is less than or equal to every element of that subset. Nobody said the infinimum had to be a limit point of the subset. For instance, the infinimum of $\{1\}\cup(5, 9]$ is $1$.
A: The step where you claim the existence of $m'$ is wrong.
Clearer counterexample: The set $S = \{ 1,2,4 \}$ has an infimum $1$, but for say $\epsilon = 1/2$ there exists no $s \in S$ such that $1 < s< 3/2$. 
A: No, it is not correct, and intuitively this should be quite clear. You never use the fact that $p$ is prime!
Yes, $m$ is the infimum but this simply means it is the greatest lower bound. All you can say is that for every $m' \in \Lambda$ you have $m' \geq m$. There doesn't even have to exist an $m' \neq m$ just because $m$ is the infimum. Moreover, from $m' > m$, how do you conclude $m' < m + \epsilon$?
