Fix $x$ and $y$ in $\Bbb R$. Suppose that $x < y + e$ for all $e>0$. Prove that $x\leqslant y$. I just started learning proofs in analysis class and this is my first time in this page.
I learned about the Completeness Axiom and tried to solve some problems, but none of them look easy to me
This is my proof:
Assume $x>y$ is true, then there exists a number $n$ in natural numbers such that $n>e$ thereby making $x+n>y+e$ by the Archimedean Property
$x+n>y+e$ and $x<y+e$ contradiction
Therefore, $x\leqslant y$
(I think my proof is seriously messy)
If it is possible can you please tell me how can I approach proof problems?
Should I first look at the answers and memorize them all or just keep trying?
I have no idea where to start
And can you also teach me where can I learn how to write a decent coded math equation so that I can clearly deliver my questions?
Thank you so much for your help
 A: Assume that $x>y$. So you have $0<x-y$ . so $0<\frac{x-y}{2}<x-y$. This means that for $e=\frac{x-y}{2}$ . You have $y+e=\frac{x+y}{2}\leq\frac{x+x}{2}=x $ which contradicts that $x<y+e\,\,,\forall e>0$ . Therefore it must be that $x\leq y$ as it is the negation of the statement $x>y$ .
It is usually a really bad idea to try and memorize everything. It is better to try and understand the logical flow. Ask yourself at each step that "if this did not hold, then what problem I am running into". Then obviously it comes down to intuition, knowledge and application of basic algebra.
A: You don't need completeness for this. For example the same statement holds in $\mathbb{Q}$.
By contraposition: suppose $x>y$; then we have
$$
x>y+\frac{x-y}{2}
$$
Can you finish?
A: Here is a direct proof (just a rewording of the other answers): since by hypothesis, the claim is true for any $\epsilon>0,$ we may consider specifically elements of the set $\{\epsilon_n\}_{n\in \mathbb N}=\{\frac{1}{n}\}_{n\in \mathbb N}.$ Now, suppose $z>y.$ Then, there is an integer $N$ such that $x<y+\frac{1}{N}<z$. We have shown that $x$ is less than any number $z$ which is greater than $y.$ That is, $x\le y.$
