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In real analysis, or calculus, the very first and most common example of a subset of rational numbers with no least upper bound within $\mathbb{Q}$ is given by $\{x\in\mathbb{Q}\,\,|\,\, x^2<2\}$. Its proof is not too difficult, but I didn't find that the proof can be easily, by elementary computations, explained to students of say B.Tech (or students, that are not undergraduate science students). This raised following natural question:

Is there any other example of a subset [EDIT: non-empty and bounded above] of rational numbers with no least upper bound in $\mathbb{Q}$, whose verification can be done easily by first learners of real analysis (or of supremum, infinimum things in real numbers)?

(One will object about what is meant by easy? etc, but if we try to see the proof of no least upper within rationals bound for $\{x\in\mathbb{Q} \,\,|\,\, x^2<2\}$, then the arguments are not so obvious ones.

On this platform of stackexchange, a question of similar nature has been posted earlier, about asking from $x$ with $x^2<2$, how can we construct little larger number with same inequality? or what is motivation for proof of this fact given in the Baby Rudin? And the answers in post were containing some connections with numberical methods.

(Sorry: I did editing since, while writing question, I was completely in flow of only one question - about set $\{ x\in\mathbb{Q} : x^2<2\}$.

Beginner
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  • I don't see why this could not be explained by elementary computations. The fact that $\sqrt 2$ is irrational is discussed in school already. Furthermore you only need that that there are infinitely many rationals arbitrary close to $\sqrt 2$. This also isn't hard. – sampleuser Aug 28 '20 at 12:08
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    What is B.Tech? – Calvin Khor Aug 28 '20 at 12:42
  • Also, $\mathbb N$ has no least upper bound – Calvin Khor Aug 28 '20 at 12:45
  • @CalvinKhor Maybe a misprint for BTEC? – Angina Seng Aug 28 '20 at 12:50
  • Instead of $\sqrt 2$ you could use $\alpha=.1010010001\cdots $ or any other decimal which you can easily show to be irrational. Now just argue that there are rationals arbitrarily close to $\alpha$ which are less than $\alpha$. – lulu Aug 28 '20 at 13:01
  • Sorry; I forgot to mention that subset to be bounded above! (Can I edit now; because one answer has been posted.) – Beginner Aug 28 '20 at 13:55
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    @CalvinKhor: In India it means Bachelor of Technology which is an undergrad degree for many engineering courses (most notably offered by IITs). Some other institutes in India offer BE (bachelor of engineering) which is at same level as BTech. – Paramanand Singh Aug 29 '20 at 02:06
  • Consider this answer. It does not use anything more than operations $+, -, \times, /, <, >$ on rationals and can be understood by anyone who knows how to operate on rationals and has some idea of using symbols instead of literal numbers. However I think many BTech students would not find the topic interesting enough which may come in way of understanding. – Paramanand Singh Aug 29 '20 at 02:13
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    @ParamanandSingh thank you for the background. Given the OP said he is located in India, that sounds right. There are apparently many things called B Tech or BTEC. Also, to OP: the link above is also good for one of your sub-questions "or what is motivation for proof of this fact given in the Baby Rudin?" I highly recommend reading everything there – Calvin Khor Aug 29 '20 at 03:47

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Two very easily verified examples of subsets of $\Bbb{Q}$ with no least upper bound are $\varnothing$ and $\Bbb{Q}$.

Servaes
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It won’t be easy...

For an above bounded set $S$ of rationals, having a (real) upper bound $A$ which is not a rational means that $A$ is irrational.

In an example like the one you request, you’re likely to have to prove that $A$ is irrational. And $\sqrt 2$ is an irrational number for which the irrationality proof is quite easy.

mathcounterexamples.net
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