How to show that the is a $1-1$ correspondence between real numbers and the set of points of a line in the Euclidean plane? When we sketch a line in the plane , we say that we can label every point on this line by a real number , and for every real number  there exists only a point on the line which can be labeled by this number . This - as I think - is the reason which we consider the plane as $\mathbb{R}^2$.
but I have never seen a proof for this fact . so my question is , How to prove this ? 
I think that this can be proved by constructing  a bijection ( or showing that such bijection exist without giving it explicitly )  from the Reals to the set of points of line . 
So , Can we show that such bijection exist ? 
If not , How did mathematicians know that reals express lines in the plane without gaps ?! in the case of the lack of such proof , there is a possibility that there are gaps in our line line when we express it by real numbers. 
Added: 
to make the question clearer , We say the plane is $\mathbb{R}^2$ and so we we assume that there is a correspondence between $\mathbb{R}$ and the set of points of the $x$-axis and two sets have this correspondence if there is a bijective between them , so we have to find such bijective , otherwise , why not to say that there is a correspondence between the line and a proper subset $A$ such that there is not bijective from $A$ into $\mathbb{R}$ ( and so there is no correspondence between the line and $\mathbb{R}$)  ? 
 A: In mathematics, we normally describe a line as, for example, the set of points $\{(x,2x+1):x\in\mathbb R\}$: 

Then the function $\phi:x\mapsto(x,2x+1)$ is an easy bijection between the line and $\mathbb R$.  
But what if we defined it as the set of points $\{(x,2x+1):\mathbb Q\}$ instead?  Then it would be in bijection with $\mathbb Q$ instead.  So what would it look like then?  

Oh.  It's the same.  The rational numbers form a dense set (for any rational numbers $x,y$ there is a rational number between them (e.g., $\frac{x+y}2$), so they do 'fill in' space in some way.  But we would then be introducing more problems, such as: 
The function $x^2-2$ would have no zeroes.  That certainly doesn't coincide with our intuition from looking at the graph: 

In addition, the function defined by taking
$$
f(x)=\begin{cases}0&x^2<2\\1&x^2>2\end{cases}
$$

would be continuous!  In order to get round these, we introduce the real numbers, which are complete in the sense that any Cauchy sequence - i.e., any sequence that you would intuitively expect to converge to a value - does in fact converge.  
Note that we could have used the algebraic numbers or some other countable set instead of the rational numbers, but would have run into similar problems.  We could even use the definable numbers, which are a countable set: in that case, we wouldn't be able to construct these sorts of counterexamples to our intuition, as we could turn them into definitions of undefinable numbers; however, the definition of the real numbers isn't that hard, and it's what mathematicians are used to, and it allows us to prove lots of beautiful results, so that's what we use.  
For a discussion on the use of real numbers in the form of a dialogue, see here.  I'll just give the closing line.  
Mathematician: So, finally, we arrive at the following justification for real numbers. 1. We must go further than just the rationals. 2. When we do so we introduce certain procedures that give us new numbers. 3. Formalizing these, we end up with the monotone-sequences axiom, or something equivalent to it. 4. This axiom is not as precise as it seems, since the notion of an arbitrary monotone sequence, even of rationals, is not precise. 5. There is no need to make it precise, because we know how to reason in terms of arbitrary sequences. 6. That allows us to define the real numbers we have a use for, even if it gives us a lot of junk as well. 7. In fact, we don't really know what junk it does give us, and it's not even clear that it makes sense to ask.
A: Suppose you have a line $L$.
Case 1.
$L$ is defined by $y=a x + b$.
In this case, $(x,y)\leftrightarrow x$ is one to one and onto, which practically projects the line to x-axis.
Case 2.
$L$ is defined by $x=a$.
In this case, $(a, y)\leftrightarrow y$ is one to one and onto, which is just to rotate the line at $(a,0)$ by $90^\circ$ (and a shift by $a$).
In either case, you have an $1-1$ correspondence.
A: I think.  In classical geometry it is given as an axioms (and we have to accept axioms and the concepts of "points" and "lines" only exist as axiomatic definitions and could, in other contexts, mean other things and in other contexts it's not true that a line has the same number of points as the real numbers[1]) that


*

*A point is no distance from itself.

*Any two distinct points will have a measurable distance between them.

*Given a point and a distance one can create a point that distance from the first (and thus a line between the two points).


And... that's it, really, we're done.  Given any real number, a point for zero, and a direction to go in (right for positive, left for neg.) we can find a point for that number.  For every point on the line we can find its distance from the zero point.  If the points are different then the numbers will be different and vice-versa.  That's a one to one correspondence.  We're done[2].
====
[1] Axiomatically, we can have a consistent geometry where "lines" have only two points and the "universe" consists of precisely 3 distinct points and the three lines pair-wise determined by them.  Or we could have point set topology where there need not be any correspondence between points and numbers at all.
Or we could have real analysis where points and numbers are pretty much defined to be the same thing and a line just means the inequality ordering. (see [2]).
[2] Okay, we're not really done.  We need to demonstrate the completeness of the real numbers in the rational field (something beyond Euclid and Descartes) but if we accept the axioms of distances related to numbers, then defining the real numbers equivalently defines a line. 
A: Here is a proof using transformations. 
Clearly $\{(x,0)\mid x\in \Bbb R\}$ is a copy of $\Bbb R$ inside the plane.
Now, rotations are continuous linear transformations of the plane. Using a rotation, you can turn this line to be any other line through the origin. Because the transformation is continuous, the resulting line is (topologically) a copy of the real line.
Next, note that any translation of the plane is also a continuous transformation. The resulting line is again a copy (topologically) of the real line.
Since you can get to any line in the plane using one or both of these operations, that is one way to convince yourself that all these lines look like copies of $\Bbb R$.
