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Take any positive real number $a$. Is there always some natural number $n$ such that $a\cdot n \geq 1$? And, I guess more generally, is there always some natural number $n$ such that $a\cdot n\geq b$ for any $b\in\mathbb{R}$?

Questionbeggar
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    This is known as the Archimedean property: https://en.wikipedia.org/wiki/Archimedean_property – whpowell96 Feb 20 '24 at 16:31
  • Thank you @whpowell96 What would I need to study to understand more about the archimedean property of the reals? Real analysis? Do you have any textbook recommendations? – Questionbeggar Feb 20 '24 at 16:47
  • Archimedean property of the reals, yes. I hear Understanding Analysis by Abbott is a good place to start. – whpowell96 Feb 20 '24 at 18:02

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Yes. First note our question follows from your first one: if $b\in \mathbb{R}$, as the integers are unbounded in $\mathbb{R}$, we get some $m>1$, and then $a\cdot n\ge 1\implies a\cdot m\cdot n\ge b$ as wanted.

Now for the first part, it is equivalent to finding an $n$ such that $a\ge 1/n$, which is equivalent to finding and $n$ such that $n\ge 1/a$ (well defined as $a$ positive). Again we are done as the integers are unbounded in $\mathbb{R}$.

Guillaume Berlat
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  • Sorry, this is going to be very basic. I take it that m is an integer here? Or is it any real? And why does it follow from the fact that the integers are unbounded in $\mathbb{R}$ that there is going to be $n\geq \frac{1}{a}$ (as long as $a$ is positive)? – Questionbeggar Feb 20 '24 at 16:55
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    $m$ is an integer $>b$. Since the integers are unbounded from above, there exists an $n$ greater than any upper bound, including $1/a$, for example. – whpowell96 Feb 20 '24 at 18:04
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Yes for the first question, using $n \ge \frac{1}{a}$. Similarly for the second question, using $n \ge \frac{b}{a}$. They are true because $a$ is positive and $n$ is not bounded from above.

d0SO'N
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  • Thanks. Why does it follow that there is such $n$ provided that $n$ is not bounded from above? – Questionbeggar Feb 20 '24 at 16:55
  • $\frac{1}{a}$ and $\frac{b}{a}$ are bounded from above once $a$ and $b$ are set. Say the bound is $m$. $n$ is not bounded from above, so you can choose any $n \gt m$ to satisfy the requirements. – d0SO'N Feb 20 '24 at 17:06