1
$\begingroup$

Let a, b be two positive real numbers. Then there exists a natural number n such that na >b. The above statement is called the archimedean property of real numbers. The theorem holds if a be positive and b be negative. Then why we take two positive numbers?

$\endgroup$
  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange. If $n>0$ and $a>0$ and $b<0$, it's trivial that $na>b$. $\endgroup$ – J. W. Tanner May 2 '19 at 15:53
  • $\begingroup$ Then why we take two positive numbers? I tnink in the theorem we shouldn't mention b as positive. We can say a is positive and b is a real number. $\endgroup$ – Sahil Laskar May 2 '19 at 16:12
  • 1
    $\begingroup$ The statement holds if $b<0$; it's just not remarkable or noteworthy in that situation $\endgroup$ – J. W. Tanner May 2 '19 at 16:25
  • $\begingroup$ In fact, in Rudin's Principles of Mathematical Analysis (third edition, page 9), he does not require that $b$ (which he calls $y$) be positive $\endgroup$ – J. W. Tanner May 2 '19 at 16:51
3
$\begingroup$

If $n>0$ and $a>0$ and $b<0$ then $na>0$ and $0>b$,

so it's trivial from properties of $ ">"$ that $na>b,$ in any ordered field.

The Archimedean property that there exists a natural number $n$ such that $na>b$ when $a,b>0$

is a special property that works for real numbers but not all ordered fields.

$\endgroup$
1
$\begingroup$

The significance is that no member of $\Bbb R$ is greater than every member of $\Bbb N$ nor less than every negative integer. So if $b\ne 0$ then there exist $n_1,n_2\in \Bbb N $ such that $-n_1<a/b<n_2.$ And if $x\ne 0$ then $|x|>1/n$ for some $n\in \Bbb N.$

This is a consequence of the definition of $\Bbb R.$ We can extend $\Bbb R$ to a larger arithmetic structure that has objects that are greater than every member of $\Bbb N.$

$\endgroup$
1
$\begingroup$

Some people prove it as a theorem. Others take it as axiom.

Either way, a theorem or definition or axiom doesn't (always) claim to apply to all possible objects it could apply to.

In this case the emphasis is not on the negative reals. Indeed, the ancient Greeks didn't believe in such. All their magnitudes were real and positive. Not even $0$ was considered a quantity.

Thus, the theorem tells you that given two quantities (thought of as positive moduli), say $\epsilon<<<<\cdots<<M,$ where I have used the notation $<<<<\cdots<<$ to signify that the difference $M-\epsilon$ may be as large as you please, then we can aggregate sufficiently many $\epsilon$'s in order to exceed $M,$ thus the claimed existence of a positive integer (called a number, arithmos, by these Greeks), say $N,$ such that $$N\epsilon>M.$$

This claim essentially says that we cannot have infinite quantities, or in other terms, infinitesimal quantities. However, we can discard this axiom and allow such quantities; then we have an example of what is called a non-archimedean field. Such are used in alternative constructions of standard objects, for example nonstandard analysis, or in the development of $p$-adic numbers.

In sum, a definition or theorem or mathematical statement doesn't claim to exhaust some class of objects. Its hypothesis or claims apply to precisely the types of objects it is interested in applying to. Here, we're concerned with positive magnitudes; we don't care a fig about any other thing the axiom/theorem applies to.

$\endgroup$

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.