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My friend and I were talking about math stuff as usual, when he brought up a fake proof for the statement:

There is no real number greater than $0$.

Now obviously this isn't true, because any positive number is $>0$. But I could not argue convincingly enough that his "proof" was wrong, because I can't pinpoint the specific step at which it went wrong.

The following is his "proof":

Suppose that there is at least one real number greater than $0$. Then it follows that if we list out all the real numbers greater than $0$ and sort them in ascending order, then there should be a first number greater than $0$.
Now let's suppose that $\varepsilon$ is the first number in this list. But, $\varepsilon/2$ is smaller than $\varepsilon$ yet still greater than $0$, because $\varepsilon/2$ is $\varepsilon/2$ above $0$, which indicates that $\varepsilon/2$ should be the first number in the list.
This contradicts our original assumption that $\varepsilon$ is the first number, so therefore, there are no real numbers greater than $0$.

I tried to argue that the entire premise of the proof is questionable, because it's impossible to define "first" numbers for infinite lists (like $\Bbb Z$), but he counters by saying that many infinite lists actually do have first numbers, like $\Bbb N$.

What is wrong with my friend's "proof"?

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    $\begingroup$ "...then there should be a first number greater than 0." : false $\endgroup$
    – user2661923
    Mar 17 at 2:55
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    $\begingroup$ This "argument" is doubly wrong. Not only do the positive reals not have a first element when listed (which would also cause, say, the positive rationals to fail too), but the positive reals are uncountable, which means that you can't even put them into a list! $\endgroup$
    – Sammy Black
    Mar 17 at 3:04
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    $\begingroup$ Many human beings have swum across the English Channel. Your friend is also a human being (I assume), but that fact alone doesn't guarantee that your friend can swim the English Channel. (Though it would be amusing if it turns out he already has.) $\endgroup$
    – David K
    Mar 17 at 3:06
  • $\begingroup$ Is your friend educated in this? It sounds like they're giving you a hint with the mention of naturals. $\endgroup$
    – Džuris
    Mar 17 at 12:13
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    $\begingroup$ What he has there looks to a be a pretty good proof that there is no such thing as a smallest real number in a set open on the left. Just that the final step jumps to the wrong conclusion when it goes to "no such thing as a positive number" when the actual contradiction is with the assumption of a smallest positive number existing. $\endgroup$
    – ilkkachu
    Mar 17 at 12:21

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Your friend is assuming that because one fact is true in a certain circumstance, it would mean that it is true in all circumstances. Just because the usual notion of ordering of numbers yields a well order of $\mathbb N$, that does not mean it will yield a well order of $\mathbb R$ too.

Basically (s)he is using inductive logic, proceeding from a special case to conclude the general case; whereas math propositions require deductive logic, proceeding from the general case to a special case.

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  • $\begingroup$ Sorry, my math isn't all that great. What does "well order" mean? $\endgroup$
    – Aiden Chow
    Mar 17 at 3:06
  • $\begingroup$ I have added the wp link. Basically it means an ordering of elements such that any nonempty subset has a least element. Our usual notion of $\le$ yields such an ordering on the natural numbers but not on the real numbers. $\endgroup$
    – Shahab
    Mar 17 at 3:09
  • $\begingroup$ Is there some kind of proof that there isn't a well order of $\Bbb R$, or is it just universally accepted by definition? $\endgroup$
    – Aiden Chow
    Mar 17 at 3:16
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    $\begingroup$ According to our usual axioms of set theory, there does exists a well order on R, its just that our known $\le$ isn't it. So theoretically there does exists a notion of 'less then equal to' among numbers where R becomes well ordered - (possibly where 1 is less then 0 and our notions of positive/negative are quite different). $\endgroup$
    – Shahab
    Mar 17 at 3:21
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    $\begingroup$ @AidenChow: As Stilez has pointed out in their answer, your friend has given you a proof that $\mathbb{R}$ isn't well-ordered (and nor is $\mathbb{Q}$). :D $\endgroup$
    – Chris Bouchard
    Mar 17 at 12:24
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To directly answer your question, the step at which the proof goes wrong is the very first step: "if we list out all the real numbers greater than 0." This is impossible. (The most well-known proof is known as Cantor's diagonal argument).

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    $\begingroup$ This isn't the core of falsity in the fake proof, though. They could have argued about Q instead of R. $\endgroup$
    – Damian
    Mar 17 at 11:39
  • $\begingroup$ I disagree @Damian. The core is exactly that you can't list out the real numbers. You can list the rationals, just the sense of "ascending" would be a bit different. $\endgroup$
    – Džuris
    Mar 17 at 12:10
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    $\begingroup$ @Džuris Listing is relevant to the exact wording of the proof in the question, but is only there to suggest that there is a "first number greater than 0". Ultimately "listing" is just a distraction from the key argument about the existence of a smallest positive value in whatever set we're talking about. $\endgroup$
    – TheRubberDuck
    Mar 17 at 13:07
  • $\begingroup$ @Damian Perhaps not, but if someone makes this sort of error in the second sentence of their proof, it is a good indicator that they do not really understand what they are talking about (and thus it is not really worth the effort to read the rest of their proof). And it is indeed the first place the proof goes wrong. $\endgroup$
    – manthanomen
    Mar 24 at 7:41
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Suppose that there is at least one real number greater than $0$. Then it follows that if we list out all the real numbers greater than $0$ and sort them in ascending order, then there should be a first number greater than $0$.

Suppose, as @Damian suggests in a comment, we use your friends argument over the set of rational numbers (fractions), to bypass the issue that reals are uncountable. Then surely there's a smallest fraction, your friend is saying.

(Reasoning: If there's no smallest positive fraction, there certainly can't be a smallest positive real, because all fractions are also real numbers too)

So, is there such a thing as a smallest fraction? And is it a problem if not?

Quick answer, no, there isn't. The proof is trivial.

If there was a smallest positive fraction, say A/B, then that's equivalent to claiming there's a largest positive integer, [B/A].

That's what infinite and infinitesimal mean. There isn't an end to bigness... but there also isn't an end to smallness either.

That's their error, put as simply as it gets......

In reality, your friend has tried a "proof by contradiction", and indeed has successfully proved one of their assumptions is wrong... but chosen incorrectly which the wrong assumption was.

I'd ask your friend to prove there is a specific biggest positive number, before they try to "prove" there's a specific smallest positive number.

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  • $\begingroup$ The proof with rationals is easily fixable. There are multiple mappings between naturals and rationals, you can surely have a "first" rational number $\varepsilon$ in the list of all rationals greater than 0, just the "ascending" would be in a slightly different sense. And $\varepsilon/2$ would also be in the list, but somewhere else, not before $\varepsilon$. The flaw in the OP, as I see it, is exactly that the list of reals is impossible. The "proof" is constructing a number that's not on the proposed list and you could do that with any proposed list of reals. $\endgroup$
    – Džuris
    Mar 17 at 12:17
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    $\begingroup$ Yes, in the sense that if you map them to N, of course your mapping will define an (arbitrary) "first", and if you call your mapping "size" then it will define a "smallest". But in the everyday sense this proof is trivial and valid. There is no positive rational X such that there is no smaller rational Y, and that's equivalent to not being an integer S such that there is no larger integer T. $\endgroup$
    – Stilez
    Mar 17 at 12:20
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    $\begingroup$ @Džuris, you seem to say that the existence of a well-order of the rationals makes the fake proof valid for Q. But assuming AC, there also is a well-order of the reals. OP's friend did not argue for arbitrary orders, they argued for the natural order of R. $\endgroup$
    – Damian
    Mar 18 at 15:22
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The point were the proof fails is in 'if we list out all the real numbers greater than 0 and sort them in ascending order'. It is just implied that this is something one can do and from the intuition of finite sets this seems obvious. Unfortunately for sufficiently large sets this is false. One can not take an arbitrary set of real numbers, list them and then sort them in ascending order.

What you do when listing them is creating a one-to-one map with the natural numbers, one element of your set is the first, another one is the second and so on. There are more real numbers than there are natural numbers. The mathematical term is that the real numbers are uncountable. A classic proof goes via Cantor's diagonal method. So essentially when you try to make a list of all the positive real numbers, there are so many of them that the list can't contain all of them.

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    $\begingroup$ The fact that you cannot list all the real numbers is not really the core problem of the argument, since you can do the same "proof" with the rationals instead, which are countable. $\endgroup$
    – Vincent
    Mar 23 at 9:02
  • $\begingroup$ @Vincent I quoted and 'a and b' statement and if you replace the reals with rationals the same 'a and b' statement would still be false. It is true that the issue is more in the sorting but it is a quote and I think it reads better that way. $\endgroup$
    – quarague
    Mar 25 at 11:46
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Here's where it went wrong :

His proof gives the impression that it supposes only one thing

Suppose there is at least one number greater than zero

In reality, it makes two suppositions :

  • There is at least one number greater than zero
  • We can "list" all real numbers greater than zero (and pick a first element)

When he gets to the the contradiction, he assumes that the first supposition is false. He actually just proved that the second supposition is false : you cannot list all real numbers greater than zero, and you cannot find a first element in the set of all positive real numbers ordered in ascending order.

As to why you cannot find a first element in R , it has to do with the different sizes of infinite sets. Not all infinites are equal ! N is countable and R is not.

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    $\begingroup$ As others have pointed out, the fact that the second supposition is false is not actually the core problem, because you can do the same thing with the rationals, which are countable. $\endgroup$
    – Vincent
    Mar 23 at 9:01
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I want to give a short answer that gets to the core of things.

Like others have pointed out, you cannot "list out all the real numbers greater than $0$", but that is not the real problem of the "proof", since we could be working with the rational numbers instead.

The core problem is that an infinite list does not necessarily have a smallest element. If you would try to list all real numbers greater than zero (never mind that this is impossible), surely you would include $\frac{1}{2}$? And also $\frac{1}{3}$? How about $\frac{1}{4}$? As you can see, you need to include all of $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{ 5}$, $\frac{1}{6}$, $\ldots$, so obviously, there is no smallest number in the list and the whole proof falls apart.

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