Continuity of Real Number line What is the property of real numbers that allows them to be seen as a continuous line, and how natural numbers, rational numbers lack this property?
 A: Sometimes the least upper bound property does not quell the doubts that arise in the mind regarding the so-called continuity of the real line. Like I have mentioned here, the best way to understand it is using Dedekind's definition of continuity. The continuity of the line though follows from completeness(l.u.b. property). 

This excerpt is from Essay's on the Theory of Numbers by Richard Dedekind. You can read more here.
Suppose you split the real numbers so that the entire set $\Bbb R$ is split into two subsets such that one subset is to the left of the splitting point and the other is to the right. If there is only one point that can cause this partition then the real line is continuous. This is how we equate the real numbers to a "continuous line". If you can split the line into two such classes then there is only one point which causes this split. If there were two the distance between these points would be a hole in the original line and hence the line would not have been "continuous" as in consistent or smooth or perfect. 
If you read through Dedekind's book on page 13 you will find a proof that there are ways to split the rational numbers into two such classes but there are more than one point (infinitely many in fact) that cause this split. Specifically this sort of partitioning is caused by cutting the rational number line at a point where an irrational number might have been. I urge you to read the first few pages of Dedekind's book to properly grasp all of this. 
As for the natural numbers they do not form a line. They form a series of Hansel & Gretel bread crumbs with enormous craters between them and hence do not resemble a line. 
A: In the axiomatic construction of the sets $\mathbb{N}$, $\mathbb{Q}$ and $\mathbb{R}$, the passage from $\mathbb{Q}$ to $\mathbb{R}$ is realized by assuming the Dedekin completeness of $\mathbb{R}$. 
Namely, for any subset $S \subseteq\mathbb{R}$ that has an upper bound, there should exist $\bar{s}=\sup{S}$ and it should belong to $\mathbb{R}$ itself. The same must happen for the lower bounds.
You can easily realize that this is not true for $\mathbb{Q}$, indeed $S=\{q\in\mathbb{Q} : q < \sqrt{2}\}$ has $\sqrt{2}$ as supremum, that does not belong to $\mathbb{Q}$.
