Can we have a one-one function from [0,1] to the set of irrational numbers? Since both of them are uncountable sets, we should be able to construct such a map. Am I correct?
If so, then what is the map?

Both sets $[0,1]$ and $[0,1]\setminus\mathbb Q$ have the same cardinality $\mathfrak c=2^{\aleph_0}$, so there is a bijection between them.

If you want write down some explicit bijection, you can use basically the standard Hilbert's hotel argument which shows that if $|A|\ge\aleph_0$ then $|A|+\aleph_0=|A|$.
So let us try to describe some bijection $f \colon [0,1] \to [0,1]\setminus\mathbb Q$.


*

*Choose some infinite sequence $x_n$, $n=0,1,2,3,\dots$ of irrational numbers in the interval $[0,1]$.

*Choose some bijection $g\colon \mathbb N\to\mathbb Q\cap[0,1]$.


Then you can define $f$ as:


*

*$f(x_{2n})=x_n$;

*$f(x_{2n+1})=g(n)$;

*$f(x)=x$ for $x\in [0,1] \setminus \{x_n; n=0,1,2,\dots\}$ 



Let me add links to some posts where a very similar ideas can be used to construct a bijection between two given sets:


*

*How to define a bijection between $(0,1)$ and $(0,1]$?

*Construct some explicit bijective functions

*Constructing a bijection

*Construct a bijection from $\mathbb{R}$ to $\mathbb{R}\setminus S$, where $S$ is countable
