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

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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!

Best Answer

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.

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