In an attempt to prove Theorem 1.2 below, I shall first establish Theorem 1.0 and Theorem 1.1, the Archimedean Property of the Real Number System.

Let Theorem 1.0 be that the set of positive integers, $\mathbb Z^+$, is unbounded above. This may be proved via the Completeness Axiom, the details of which are left to the reader.

As a corollary to Theorem 1.0, we may establish Theorem 1.1 as follows.

Theorem 1.1: $\forall x, y$ where $ x > 0$ and $ y \in \mathbb R$, $\exists n \in \mathbb Z^+$ such that $nx > y$.

Proof. Algebraic manipulation of $nx > y$ yields $n > \frac{y}{x}$, so if Theorem 1.1 was false, $\frac{y}{x}$ would be an upper bound for the set $\mathbb Z^+$ of positive integers, contradicting Theorem 1.0.

Theorem 1.2 If three real numbers $a$, $x$, and $y$ satisfy the inequalites $$a \le x \le a + \frac{y}{n}$$ $\forall n \in \mathbb Z^+$, then $x = a$.

Proof. We shall break this proof into two sections. First, assume, for the purposes of contradiction, that $x > a$. This implies that $x - a > 0$, so by Theorem 1.1, we may say that that there exists some positive integer $n$ that satisfies the inequality $n(x - a) > y$. Conversely, we may assume, for the purposes of contradiction, that $x < a$, in which case $a - x > 0$, and we can again say that there exists some positive integer $n$ satisfying the inequality $n(a - x) > y$. However, simple algebraic manipulation of the original inequality of the theorem in question yields $$n(x - a) \le y.$$ It's clear to see, therefore, that neither the result $n(x - a) > y$ nor $n(a - x) > y$ from our previous assumptions was correct, leading to a contradiction. If $x \ngtr a$ and $x \nless a$, by the Law of Trichotomy, $x$ must equal $a$. $\square$

I am wondering if my proof is correct, and whether or not there would be any better ways to write it. Thank you.

  • $\begingroup$ @StefanMesken In Theorem 1.1, I explicitly stated that $x > 0$. Also, if we can't find some $n > \frac{y}{x}$, then that means $\frac{y}{x}$ would be an upper bound for $\mathbb Z^+$ (since $n$ must be smaller if it's not greater). That would contradict 1.0 proved via the Completeness Axiom. $\endgroup$ – Jamie Corkhill Jun 9 '18 at 20:53
  • $\begingroup$ Yeah, that's right. I didn't catch the implied 'suppose not' in your proof. Sorry about that. $\endgroup$ – Stefan Mesken Jun 9 '18 at 20:58

Yes your proof is correct. You can make it lighter and simpler by not treating the case $x<a$ since the hypothesis is the contrary.

  • $\begingroup$ Thank you. Is Stefan Meskens comment on my answer correct? $\endgroup$ – Jamie Corkhill Jun 9 '18 at 20:45
  • $\begingroup$ I do not understand the first part of his comment. For the other parts, you explicitly mentioned that x>0 in the hypothesis of 1.1 so they do not apply. $\endgroup$ – brunoh Jun 9 '18 at 20:47
  • $\begingroup$ Thank you. I mentioned it to him so I will see what he says. $\endgroup$ – Jamie Corkhill Jun 9 '18 at 20:54
  • $\begingroup$ Yep. Your comment to him is clear. Good work, Jamie. $\endgroup$ – brunoh Jun 9 '18 at 20:56
  • $\begingroup$ @JamieCorkhill And do not forget to accept my answer if you find it acceptable! $\endgroup$ – brunoh Jun 9 '18 at 21:10

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