For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$ From Stephen Abbott's Understanding Analysis 1.2.11:

For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$

My try:
$$\forall a\in \Bbb R, \forall b\in \Bbb R, \exists n \in \Bbb N \space \text{that satisfies}: \space a<b \Rightarrow a+\frac 1n<b$$
$\because a<b $
$\therefore\exists m\in\Bbb R, m>0$ such that $b=a+m$.
So the statement we are proving becomes (substitute $b$ by $a+m$):
$$\forall a\in \Bbb R, m \in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}: a<a+m \Rightarrow a+\frac 1n < a+m$$
Which is (take $a$ from both sides):
$$\forall m\in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}:0<m \Rightarrow \frac 1 n<m$$
We write this as: $$\forall m\in \Bbb R, 0<m, \exists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$
We define set $M = \{m\mid M=\Bbb R \cap m>0\}$
And set $N = \{\frac 1n \mid n \in \Bbb N\}$
$\because\forall q \in \Bbb Z, p\in\Bbb Z, \frac p q\in\Bbb Q$
$\space\space\space n\in\Bbb N\subset\Bbb Z$
$\therefore\frac 1n\in\Bbb Q$
Mention that $m\in\Bbb R$
$\because\Bbb Q\subsetneq\Bbb R$
$\therefore\forall n\in\Bbb N, \exists m \in \Bbb R \Rightarrow \frac 1n \geqslant m$
This statement is equivalent to:
$$\forall m\in \Bbb R, 0<m, \nexists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$
Which disproves the statement, which is wrong - What is wrong with my disproof?
Also, I am learning how to use mathematical notations properly (because I am only in year 11 but anyway). Please tell me if there is any error in my expressions.
Thanks a lot!
 A: Part of an answer, too long for a comment.
Your very last statement is true: for every positive real number $m$  there is some integer such that $1/n < m$.
How you know that statement is true depends on what axioms for the real numbers you start with. One common axiom is that for every real number $r$ there is some integer $M > r$.
I think you are creating problems for yourself by thinking that proper mathematics uses lots of symbols. Words are almost always easier to understand and to write. The idea behind this assertion is that no matter how small $b-a$ is there will be a unit fraction smaller still, since the fractions $1/n$ are smaller and smaller the larger $n$. Just turn that idea into words:

Suppose $a < b$. Then $b-a > 0$ so there is an integer $n$ such that
$1/n < b-a$. Then the result follows from simple algebra.

If you need to go all the way back to the Archimedean axiom

Suppose $a < b$. Then $\frac{1}{b-a} > 0$ so there is an integer $n > \frac{1}{b-a} $  Then the result follows from simple algebra.

A: In your eagerness to make things formal (very formal), you've ended up juggling a bit too many entities. :)
So if I understand correctly, you're saying that the statement from Abbott's textbook should be true, whereas you found a proof that it isn't. Informally, your argument seems to boil down to something like this:

I have to show that, given $m\in\mathbb{R}$, $m>0$, there exists $n\in\mathbb{N}$, such that $\frac{1}{n}< m$. But it also holds that, for any $n\in\mathbb{N}$, I can always find an $m\in\mathbb{R}$ such that $\frac{1}{n}\geq m$. So Abbott's statement is false.

I guess the misunderstanding comes from using the same symbol '$m$' to refer to what are two distinct numbers.
First off, let's check that the original statement is indeed true. As Ethan mentions, your chain of reasoning (with the $m$ and the sets $M$ and $N$) is a bit overkill. More simply, just notice that the inequality you start with can be rewritten as $n>\frac{1}{b-a}$. In other words, you have to show that for any $a,b\in\mathbb{R}$ such that $a<b$, there exists an $n\in \mathbb{N}$ such that $n>\frac{1}{b-a}$.
Since $b-a>0$, it is not difficult to see that this, in general, true. As an example, suppose $a=0.02$ and $b = 0.34$ (I chose these numbers at random). Then $\frac{1}{b-a} = 3.125$. Can we find a natural number greater than $3.125$? Yes: $4$ works; $5$ works; anything greater than $5$ also works. Broadly speaking, no matter what $a$ and $b$ you start with, the bag of natural numbers is large enough that you can always find some $n$ that is large enough and that fits your needs.
This is just to give some intuition for why the original statement is true.
Now coming back to your proposed answer, I would say the last inference is a mistake. Indeed, for any $n\in\mathbb{N}$ there exists $m\in\mathbb{R}$ such that $\frac{1}{n}\geq m$. That is to say, given $n\in\mathbb{N}$ you can always find some very small $m$, smaller than $\frac{1}{n}$. But that doesn't mean that you've disproven the original statement, because this $m$ is not the same as the $m$ you've started with! It's a different $m$, which, to avoid confusion, should probably not even be denoted $m$ in the first place. Perhaps if you give it a different name, say $r$, it becomes clearer that from the last statement you introduce it does not follow that you've disproven the original claim.
A: I think your proof is OK up to the point I write below. Here's a typical trick in doing these things. To prove that:
$\forall m \in \mathbb{R}$ where $m > 0$, then $\exists n \in \mathbb{N}$ so that $\frac{1}{n}<m$.
You want to fix an arbitrary $m > 0$ and show there exhibit some $n \in \mathbb{N}$  so that $\frac{1}{n} < m$, and think about why that is sufficient! Proving this actually requires the Archimedian property, which says for the reals:
$$
\forall x \in \mathbb{R}, \text{ there is a } n \in \mathbb{N} \text{ so that } n > x
$$
and proving this requires the least upper bound property of reals.
Also, with the archimedian property, you can rather shortly conclude the proof.
