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I have to prove the following:

$$\text{Prove that there is no smallest positive real number}$$

Argument by contradiction

Suppose there is a smallest positive real number. Let $x$ be the smallest positive real number:

$$x : x \gt 0, x \in \mathbb{R}$$

Let $y$ be $\frac{x}{10}$. Contradiction. This implies that $y < x$ which implies that you can always construct a number that is less than the "smallest positive real number". QED.

Can someone please verify the write up of the proof and the proof itself?

Thanks for your time!

P.S. I have seen this and this but I'm not looking for a way to approach the problem but rather verification and write up help.

P.P.S If there is another novel way of approaching this problem, I would like to know!

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    $\begingroup$ what you are doing is just fine. Just wait with announcing "contradiction" until you actually have one. To be very accurate, you want to say why $y<x$ and whey $y$ is a positive real number. Then you have your contradiction. $\endgroup$ – Ittay Weiss Apr 23 '15 at 3:14
  • $\begingroup$ Ahh okay. So contradiction should come after $y < x$ etc? $\endgroup$ – Jeel Shah Apr 23 '15 at 3:15
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    $\begingroup$ (To be pedantic, there was a contradiction as soon as we posited a smallest positive real, we just didn't know it yet :-) ) $\endgroup$ – Kyle Miller Apr 23 '15 at 3:17
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    $\begingroup$ @JeelShah A proper proof would go: "Assume, for contradiction, that $x$ is the least positive real number. Let $y$ be $\frac{x}{10}$. We have $y<x$ (because [fill in]), $y>0$ (because [fill in]), $y\in\mathbb R$ (because [fill in]). So $y$ is a positive real number less than $x$, which contradicts our assumption that $x$ is the least positive real number. Therefore, there is no least positive real number.". $\endgroup$ – user26486 Apr 23 '15 at 3:22
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    $\begingroup$ If you want to see how write it less gappy (as @user31415 indicated) books.google.com/books?id=6cMSAAAAQBAJ&pg=PA109 and probably two dozen other books have it. $\endgroup$ – Fizz Apr 23 '15 at 13:53
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Here is a slightly different way to organize the proof. What we will do is split it into two parts:

  1. For every positive real number there is another positive real number less than it. Proof: Let $x>0$. Then since $0<\frac{1}{2}<1$, we have $x>\frac{1}{2}x>0$, and so $\frac{1}{2}x$ is such a number.

  2. There is no smallest positive real number. Proof: Assume for sake of contradiction that $x$ is the smallest such. Then by 1 there is a smaller such number, contradicting minimality.

The idea with splitting the proof into two statements is that we have isolated the proof by contradiction into a very small part. The risk with proof by contradiction is that, since you are in fact assuming something which is false from the beginning, any mistaken reasoning after that will look like a valid completion to the contradiction proof.

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    $\begingroup$ If you're simply aware that you can only either use the false assumption or correct reasoning, no problems will arise. $\endgroup$ – user26486 Apr 23 '15 at 4:00
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    $\begingroup$ The point is that in a more complex argument it is much more likely to accidentally introduce incorrect reasoning. This structural organization is meant to limit the possibility of a mistake, and as this question is about proof technique it seems useful to consider this aspect for future proofs. While it's true as you say that you won't have problems if you make no mistakes, mistakes do happen. $\endgroup$ – Kyle Miller Apr 23 '15 at 4:06
  • $\begingroup$ I don't see any advantage of such a separation as a way of proving this particular statement. One should not fear proofs by contradiction as a means of proving a negative statement like this (non-existence) result; it is the easiest and most direct thing to do. From the point of view of abstraction it might be useful to prove once and for all "a totally ordered set $S$ does not have a smallest element" (or better "a partially ordered set $S$ has no minimal element") iff "for every $s\in S$, there exists $t\in S$ with $t<s$". Which tells you where to look next time. But it is not easier. $\endgroup$ – Marc van Leeuwen Apr 23 '15 at 13:27
  • $\begingroup$ @MarcvanLeeuwen Another reason to separate it like this is that there really are two separate concerns. Usually a contradiction proof for a statement like this would have 2 with 1 as a substatement, but the fact is that 1 does not depend on 2 so one could re-order the statements. Splitting it does not require writing them in a list as I have done; this was didactic. To be clear, the purpose isn't out of a fear of contradiction proofs, but out of a general sense of architecture. $\endgroup$ – Kyle Miller Apr 23 '15 at 17:30
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Another way to show this is using the Archimedian property of natural numbers. That is, the natural numbers don't have an upper bound in the reals.

Consider any small real number, $\epsilon>0$. Since natural numbers are unbounded, there exists some $n \in \mathbb{N}$ such that $n>\frac{1}{\epsilon}$. Rearranging gives that $\epsilon>\frac{1}{n}$. Thus for any small positive real number $\epsilon$, there is a smaller positive real number $\frac{1}{n}$.

I like this proof because it shows the connection between really big numbers and really small numbers.

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  • $\begingroup$ Just to complete the proof (that there is no smallest), there should be a statement that if there were a smallest, call it $\varepsilon$, then the $n$ would give a smaller, contradicting minimality. $\endgroup$ – Kyle Miller Apr 23 '15 at 3:54
  • $\begingroup$ @KyleMiller but avid19 said "Thus for any small number $\epsilon$, there is a smaller number." Maybe avid19 edited after you started writing, not sure. And btw, $\frac{1}{n}$ would give a smaller, not $n$ itself. $\endgroup$ – user26486 Apr 23 '15 at 3:57
  • $\begingroup$ @user31415 The statements "For any number there is a smaller number" and "There is no smallest number" are not the same statement, though they are logically equivalent (assuming the law of the excluded middle). And by saying "the $n$ would give a smaller," I meant "the [same logic used to choose] $n$ would give a smaller [associated] number." Perhaps the terseness was too. $\endgroup$ – Kyle Miller Apr 23 '15 at 4:01
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    $\begingroup$ There is no need to invoke the Archimedean property. This result continues to hold in ordered fields which are not Archimedean (en.wikipedia.org/wiki/Non-Archimedean_ordered_field), such as $\mathbb{R}(x)$ with the following ordering: if $f(x)$ is a rational function, then $f(x) \ge g(x)$ iff $f(r) \ge g(r)$ for sufficiently large real numbers $r$. With respect to this ordering, $\frac{1}{x}$ is infinitesimal. It is nevertheless still true that there is no smallest positive element, and the proof is the same as the one presented by the OP. $\endgroup$ – Qiaochu Yuan Apr 23 '15 at 4:43
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There is absolutely no need to use contradiction. Just prove the statement directly: there is no smallest positive real number. What this means is that if $r$ is a positive real number, then it isn't the smallest one. And indeed it's not because $\frac{r}{10}$ (or whatever) is smaller.

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    $\begingroup$ I don't get the point of this answer at all. To prove "there is no least positive real number" directly, the only way to go is to assume there is a least positive real number and derive a contradiction. Logically equivalent but less direct is to show "for all real $r$, $r$ is not a least positive real number". What you want to show is "for all positive real $r$, $r$ is not the least positive real number", which is also equivalent, because the omitted case of real that are not positive (in the first place) is trivial to handle, but there is no way in which this is more direct than before. $\endgroup$ – Marc van Leeuwen Apr 23 '15 at 12:52
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    $\begingroup$ gowers.wordpress.com/2010/03/28/… $\endgroup$ – Fizz Apr 23 '15 at 13:40
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    $\begingroup$ I'm no mathematician, but I don't see how the method you describe in this answer isn't a proof by contradiction. $\endgroup$ – David Z Apr 23 '15 at 15:16
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    $\begingroup$ @Marc: again, there is absolutely no need to use proof by contradiction. The hypothesis that $r$ is the least positive real number is not used at all in the proof, which shows no more and no less than what I've shown above: that $r$ is not the least positive real number. The distinction between what I wrote and what the OP wrote is that at every step in my argument, every statement I've written down is a true statement about the real numbers. (I think there are other distinctions one can make about whether one argument is intuitionistically valid or not but I think this is less important.) $\endgroup$ – Qiaochu Yuan Apr 23 '15 at 19:07
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    $\begingroup$ The problem with proofs by contradiction is that they are easy to mess up. In a long proof by contradiction, if you reach a contradiction at any point you can't tell whether you've finished your proof or whether you've made a mistake that introduced a new contradiction. When you start making statements using hypotheses you believe to be contradictory, you can't check your work based on what you believe to be true. This is an easy way to end up believing you've proven the Riemann hypothesis or whatever. $\endgroup$ – Qiaochu Yuan Apr 23 '15 at 19:08
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I understand this is an old thread, but this is the first search result for this problem on google and I would like to add my own proof for the nonexistence of a smallest positive real number.

Proof.

Assume $a \in \mathbb{R}$ is the smallest positive real number. That is, for any positive real number $r \neq a$ , $0 \lt a \lt r$. Let $x \in \mathbb{R}$ be a real number greater than 1. Since $x$ is real, it has a real multiplicative inverse, $y \in \mathbb{R}$ such that $xy=1$. Therefore $y= \frac 1x$. Since we know $0 \lt 1 \lt x$, diving through by $x$, we see that $0 \lt \frac 1x \lt 1$. Multiplying by $a$ then yields $0 \lt a \cdot \frac 1x \lt a$. Thus ($a \cdot \frac 1x$) is a positive real number, as it is the product of two positive real numbers, and it is less than $a$. But there cannot be a positive real number less than $a$. This is a contradiction. Therefore there does not exist a smallest positive real number. $\blacksquare$

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If $x$ is the smallest real, then $0 < x < 1$.

But $1-x > 0$ so $0 < x(1-x) =x-x^2$ so $x^2 < x$.

(Just wanted to come up with a different answer.)

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