Having trouble debunking my friend's "proof" that "There is no real number greater than 0"

Question

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"?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.