Proving the set of irrational numbers in $\mathbb{R}$ has the same cardinality as $\mathbb{R}$ Not going to show my whole proof here, but this logic work?
We are showing |R-Q| = |R| => |I| = |R| 
Since I is a subset of R, |I| must be less than or equal to |R|. 
So all we have to show is |I| is not less than |R|.
So started by proving R is uncountable using diagonalization and a contradiction.
Then proved I is uncountable using a contradiction and the fact that I union Q = R. 
Then concluded that since I is uncountable and R is uncountable, then|I| cannot be < |R| => |I| = |R|. 
Does that make sense?
 A: Just show that there are injective functions from one set to another and vice versa. For one, just take the inclusion. For the other send R\Q in (0,1)\Q and Q to (1,2)\Q for example. ( I'm using the fact the the open interval has a bijection with R and that injections preserve cardinality). Then the result follows by the Cantor-Bernstein-Schroeder theorem.
A: You need a bijection from $\mathbb R$ to $\mathbb I$ so the image has to skip $\mathbb Q$. So, force the issue:  enumerate the rationals $\{r_n\}$ in some way and fix an irrational number $\alpha$. Now define
$f(x) = \begin{cases} 
          r_{2n}+\alpha& x=r_n \\
          r_{2n+1}+\alpha & x=r_n+\alpha \\
          x & \text{otherwise} 
       \end{cases}$.
$f$ is clearly injective, so if you know the Schroeder-Bernstein theorem, you're done. If not, then we can prove $f$ is onto: suppose $y=r_n+\alpha.$ If $n=0$, then $f(r_0)=r_0+\alpha.$ If $n$ is even, then $f\left(r_{\frac{n}{2}}\right)=r_n+\alpha$ and if $n$ is odd then $f\left(r_{\frac{n-1}{2}}+\alpha\right)=r_n+\alpha.$ Otherwise, $f(y)=y$ so $f$ is onto.
A: Your approach seem futile. The problem is that you prove that $\mathbb I$ and $\mathbb R$ are both uncountable or in other words that $|\mathbb I|>|\mathbb N|$ and $\mathbb R > \mathbb N$, but like ordinary order relation this does not imply any relation between $\mathbb I$ and $\mathbb R$. 
What you would like to do is to prove the existence of a bijection between $\mathbb I$ and $\mathbb R$. However constructing such a function is quite elaborate, instead one can use the Schröder-Bernstein theorem which says that it's enough to show that there exists an injection (or surjection) both ways. An injection from $\mathbb I$ to $\mathbb R$ is trivial to construct (ie the identity mapping). 
The other way you can do it by for example take the decimal expansion of the number and interleave it with the decimal expansion of a arbitrary irrational number. Ie say you interleave with $\sqrt{2}=1.4142\dots$ it would map $1.0000\dots$ to $1.04010402\dots$ (the zeroes are from $1.000\dots$ and the other decimals are from $1.4142\dots$).
