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In my textbook "Elementary Number Theory with Applications" by Thomas Koshy on pg. 16, there is an example given just after the well ordering principle:

Prove that there is no positive integer between $0$ and $1$.

My question is how can you even start this proof? I checked the book and google for a formal definition of integers, but they are pretty vague. The book just says $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$, and some of the websites say that it's a number without a fractional component. Then what is a fractional component? I took a course on Algebra and we defined them as equivalence classes... but to obtain that, it would require the knowledge of basic things like $1$ is the least positive integer. So I'm getting stuck in this loop. Here's what the textbook says.

Proof. (as in textbook)
Suppose there is a positive integer a between $0$ and $1$. Let $S = \{n \in \mathbb{Z}^+ | 0 < n < 1\}$. Since $0 < a < 1$, $a \in S$, so $S$ is nonempty. Therefore, by the well-ordering principle, $S$ has a least element $l$, where $0 < l < 1$. Then $0 < l^2 < l$, so $l^2 \in S$. But $l^2$ < $l$, which contradicts our assumption that $l$ is a least element of $S$. Thus, there are no positive integers between $0$ and $1$.

However, we have no definition of $l^2$ (and we can't say anything about its order).
I'm just simply not getting convinced by this proof. Can someone help?

B2VSi
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    I learned in (basic) number theory theorems that are what we would consider "so obvious", that the method of proof is mostly using Contradiction. Same in your case, Assume there is......you reach a contradiction, hence there isn't... – imranfat Aug 20 '17 at 17:38
  • why do you need the definition of $l^2$, don't you already know that? – Vidyanshu Mishra Aug 20 '17 at 17:39
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    You need to be given that $\ell^2=\ell \times \ell$, and integers are closed under multiplication. – Mark Viola Aug 20 '17 at 17:50
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    Lemma: For a real number $0<a<1$, $a^2<1$. Proof: Suppose $a^2 \ge 1$, then $a^2-1\ge 0\implies (a+1)(a-1)\ge 0$ which is false as $(a+1)>0$ while $(a-1)<0$, hence $a^2<1$ – Vidyanshu Mishra Aug 20 '17 at 17:57
  • For a less trivial example of this sort of descent see this proof on the irrationality of square roots. – Bill Dubuque Dec 16 '21 at 09:38

2 Answers2

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I think this is a weird example chosen to illustrate the version of induction that uses the well ordering of the positive integers. That means you can assume whatever you need about arithmetic, including squaring, and the fact that for real numbers $a < 1$ implies $a^2 < a$.

Then, as the argument says, the square of some (hypothetical) integer between $0$ and $1$ will be less than $a$.

Ethan Bolker
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  • The fact that the implication "for real numbers $a<1$ implies $a^2<1$ is used in this proof, isn't that perhaps some form of circular reasoning if you want to proof that 1 is the very next integer after zero? Your opinion is appreciated. – imranfat Aug 20 '17 at 17:45
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    @imranfat No. $a^2 - a = a(a-1) < 0$ since you can prove directly from the order axioms that a positive times a negative is negative. – Ethan Bolker Aug 20 '17 at 20:54
  • ok then, thanks! – imranfat Aug 21 '17 at 01:32
  • @imranfat You might gain better intuition on this type of descent from a less trivial example, e.g. see the proof of irrationality of square roots given here – Bill Dubuque Dec 16 '21 at 09:39
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Your concerns are valid - I looked over the first 16 pages and it certainly does not show a linear progression. It jumps around from the real numbers to integers and you are expected to keep pace. The author wants to expose you to ideas.

But that is not to say you can't learn from it. Your question shows that you want to dig deeper, and you can fill in the gaps via independent investigations.

It was a great idea googling for a "formal definition of integer", but on page 4 of the book you will find

The German mathematician Leopold Kronecker wrote, “God created the natural numbers and all else is the work of man.”

So try googling "formal definition of the natural numbers".

Both Example 1.9 and Example 1.10 are trying to get ideas into your head, but besides reading the proofs, try thinking about the author's claims for an hour or two and see what you can come up with.

My take is that Example 1.10 looks silly. I mean come on, 'extending the principle to the whole numbers', i.e. adding in $0$. lol. It looks like the author has a sense of humor, so don't be awestruck with some of the Examples. Also, usually the well-ordering principle is stated as

$\quad$Every nonempty set of natural numbers has a least element.

(the natural numbers include $0$).

See if you can say something about a nonempty set $A$ of integers that is bounded from below (first understand what the phrase "bounded from below" means). Again, turn it around and see if you can make a claim about a set of integers "bounded from above". Draw some number line pictures with some points on it and think about it.

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