I think the prerequisite of this question relies on assuming that we haven't give irrational numbers a rigorous definition .

Rational numbers and the points of a horizontal straight line becomes a real correspondence when we select upon the straight line a definite origin or zero-point and a definite unit of length for the measurement of segments. Of the greatest importance, however, is the fact that in the straight line there are infinitely many points which correspond to no rational number. If we name these points non-rational points, and say every of these points corresponds to one and only one non-rational number. Similarly , we define the point corresponding to a rational number a rational point.

If a non-rational point lies to the right(or left) of a rational point, we say the non-rational number corresponding to the non-rational point greater( or less) than the rational number corresponding to the rational point. If we denote the union of all these non-rational numbers and all rational numbers $K$, can we show that between two different numbers $a_1$ and $a_2$ ($a_1<a_2$) in $K$ such that there is a rational number $a\in K$ satisfying $a_1<a<a_2$?

I've read a proof on this result within real numbers , the proof relies on the Archimedean property for real numbers , so I think my question may lead to set up the Archimedean property for $K$, so how to set up Archimedean property for $K$?

  • $\begingroup$ Using geometric intuition to define real numbers is fine, but not in the way you indicate. We have to take clues from the geometric intuition rather than base our definitions on them. There is no other way to prove your point apart from mentioning that the straight line also possesses Archimedean property. $\endgroup$ – Paramanand Singh Oct 28 '17 at 15:32
  • $\begingroup$ Thus Hardy mentions in his A Course of Pure Mathematics that if $BC$ is a line segment then it is evident that there is a positive integer $k $ such that $k\cdot BC>1$ and then remarks that the assumption that this is possible is equivalent to the assumption of what is known as the axiom of Archimedes. $\endgroup$ – Paramanand Singh Oct 28 '17 at 15:42
  • $\begingroup$ @ParamanandSingh Thanks, nice to see you again! (1) Why "We have to take clues from the geometric intuition rather than base our definitions on them"? why we have to divorce the concept of the real numbers from its geometric origins? (2) I think it is possible to prove the conclusion provided treating the least upper bound property to be an axiom. $\endgroup$ – iMath Oct 28 '17 at 16:08
  • $\begingroup$ @ParamanandSingh As Dedekind observed ,“If all points of the straight line fall into two classes such that every point of the first class lies to the left of every point of the second class, then there exists one and only one point which produces this division of all points into two classes, this severing of the straight line into two portions.” $\endgroup$ – iMath Oct 28 '17 at 16:09
  • $\begingroup$ @ParamanandSingh Utilizing an analogy, one could figure out that the number continuum should have the property that "If the system R of all numbers breaks up into two classes A1 , A2 such that every number α1 of the class A1 is less than every number α2 of the class A2 then there exists one and only one number α by which this separation is produced." This indicate that every bounded A1 must have a least upper bound , I treat it as an axiom, with the least upper bound axiom, the Archimedean property is not hard to prove . $\endgroup$ – iMath Oct 28 '17 at 16:09

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