Set of irrationals between two reals is uncountable I know that between any two reals, there is an irrational number.
See: Proving that there exists an irrational number in between any given real numbers
Now let a, b $\in$ $R$ such that a < b. And let M be the set of irrationals between a and b. 
I want to show that M is uncountable. To do this, I think I need to show that there does not exist a bijection from M to the natural numbers.


*

*Can someone give me a hint about how to start to show this?

*What's the best way to approach "non-existence" proofs in general?
Thanks
 A: Instead you could show that there is a bijection between this set and interval (0,1)
A: First,
show that there is a
rational number
between any two reals.
This can be done using the
Archimedean axiom
(or theorem depending on your
axioms)
that,
for any two positive reals
$x$ and $y$,
there is an integer $n$
such that
$nx > y$.
Then,
using the same reasoning,
show that there is another rational
between the two reals.
Let these rationals be
$r$ and $s$.
Finally,
construct a linear mapping
from $[0, 1]$
to $[r, s]$.
This mapping maps rationals to rationals
and irrationals to irrationals.
Since there an uncountable number of irrationals
in $[0, 1]$,
there are an uncountable number of irrationals
in $[r, s]$.
Another proof,
this time by contradiction,
 can be modeled on
Cantor's original proof that
the reals are uncountable,
but I'll leave that as an exercise.
A: I assume that you know that the interval $(a, b)$ is uncountable and that the set of rational numbers in that interval is countable. So the problem reduces to the following:
If $Y \subseteq X$ are sets such that $X$ is uncountable and $Y$ is countable, then $Z = X \mathrel{\backslash} Y$ is uncountable.
To see this, note that $Z$ is either (a) finite, (b) countably infinite or (c) uncountable. But the union of a finite set (or a countably infinite set) and a countable set is countable so (a) and (b) can't hold. So (c) holds, which was what we wanted. 
