Countability of the irrational numbers

One of the excersises in my book says "Is the set of all irrational numbers countable? Justify your response"

Here is what I said (For the sake of writing let $\displaystyle \mathbb{I}$ the set of irrationals):"

1. First let us establish that $\displaystyle \mathbb{Z}$ is an infintely countable set. This is readily seen since there is a 1-1 mapping of $\displaystyle \mathbb{N}$ onto $\displaystyle \mathbb{Z}$ so $\displaystyle \mathbb{N}\sim\mathbb{Z}$, thus the integers are countable.

2.** Lemma**: If $\displaystyle A$ is a countable set and if $\displaystyle B_n$ the set of all n-tuples $\displaystyle (a_1,a_2,\cdots,a_n)$ where $\displaystyle a_1,a_2,\cdots,a_n\in{A}$, then $\displaystyle B_n$ is countable.

3. So realizing that the rationals may be expressed as $\displaystyle \frac{m\in\mathbb{Z}}{n\in\mathbb{Z}}$. So now since if we express every rational number as $\displaystyle (m,n)$ we can show that the rationals may be expressed as a set of ordered pairs (2-tuples). We have that the rationals may be expressed as an analogue of #2 with $\displaystyle \mathbb{Q}$ expressed as $\displaystyle B_2$, and since $\displaystyle A=\mathbb{Z}$, a countable set, it follows that the rationals are countable.

4. **Lemma**: The set of all Real numbers is uncountable

5.** Lemma**: The union of any number of infinitely countable sets is countable.

6. Assume that $\displaystyle \mathbb{I}$ is countable, That would imply that $\displaystyle \mathbb{R}=\mathbb{Q}\cup\mathbb{I}$ is countable. A contradiction.

Therefore $\displaystyle \mathbb{I}$ is uncountable $\displaystyle \blacksquare$"

Does that look alright?