3
$\begingroup$

I'm going through Apostol's Calculus I introduction, and I'm trying to prove this, but I'm having a little trouble doing it. It's proposed as an exercise in section I 3.5: order axioms.

So, what I get this theorem says is that, if a real number is smaller than all positive reals and greater or equal to zero, then it must be zero. I think it has something to do with real numbers' density; With my limited knowledge in math, I intuit there's no successor to 0 in ℝ, because, if I said it was 0.1, then there would be 0.01 which is smaller, but then I could find 0.001, and then 0.0001, and I could find infinitely many smaller numbers 0.00000000000000000000000000... the thing is that I remembered about infinitesimals, and remembered that the definition of a positive infinitesimal is a number α, such that ∀n|n∈ℕ : [0<α<$\frac{1}{n}$]. So if 0 is a real number and α is too, then 0+α is real number too, and it would be the next number to 0 in real numbers (I'm not sure about this, this is one of the voids I have). so, coming back to the theorem, if x is smaller than all positive reals, then it's smaller than 0+α, the number next to 0, so if it's smaller than that, it couldn't be a member of ℝ+, because the first positive real would be 0+α (again, another assumption based on intuition, maybe this would make mathematicians pull their hair in despair) and x≠0+α. So, since x∉ℝ+, but the sufficient conditions also states that 0⩽x, too, then x would have to be 0, because it's not 0+α, its successor.

The problem is that I have proven formally the other theorems proposed in the book, using the order and body axioms given there, and not using, I don't know... weird stuff like I do in here, also assuming things I intuit, instead of formalism. I put the aforementioned reasoning, just so you know I have tried to understand what I have to prove, instead of just putting it on the internet for other people to solve it for me, without putting in any effort. So, is an approach using infinitesimals valid for proving this, if so, is the way I'm trying to see it correct? if not, then, how should I prove this?

Thanks in advance.

$\endgroup$
  • $\begingroup$ You are thinking too hard about this trivial problem. It follows from the density of reals. $\endgroup$ – Paramanand Singh Nov 24 '17 at 5:38
5
$\begingroup$

This is about real numbers; infinitesimals do not belong.

The proof of this theorem depends on what you have available, but basically it goes like this:

Suppose that $x\ne0$. Then $x>0$ by the trichotomy axiom. Now you can take $h=x/2$ and you have a contradiction, because $0<h<x$. So $x=0$.

$\endgroup$
  • $\begingroup$ It does not seem like a correct proof. One can take $h = x/2$, but then there exists another $x_2$, such that $0 < x_2 < \frac{x}{2}$, so no assumption is violated. The condition $0 < x < h$ states that there exists SOME $x$, not a particular $x$, which can be divided by 2. In Apostol, the task sounds like "If x has the PROPERTY that ...". Property is the quality of a set, thus $x$ is a set of numbers, not one number. $\endgroup$ – John Apr 15 at 14:35
  • 1
    $\begingroup$ What assumption is "not violated"? The "assumption" here is that $x\leq h$ for all $h>0$. If $x>0$, the number $x/2$ violates that a lot. $\endgroup$ – Martin Argerami Apr 15 at 14:42
  • $\begingroup$ Assumption that there is some $x$ in between $0$ and $h$. $x$ is just a symbol (which denotes a set as i understand, or why would Apostol use the term property?). For any $h = \frac{x}{2}$, there is at least $x = \frac{x}{4}$, which is clearly $0 < x = x/4 < h=x/2$. $\endgroup$ – John Apr 15 at 14:48
  • $\begingroup$ Ok, I think i figured this out, it is a notation issue. So $x$ is just some given number. Then it makes sense, and your proof is clearly correct then. Thx $\endgroup$ – John Apr 15 at 14:57
  • $\begingroup$ Yes. This exercise is about the property that one uses in every single "$\varepsilon$" argument in analysis: if a nonnegative number is less than all positive numbers, then it is zero. $\endgroup$ – Martin Argerami Apr 15 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.