$\sqrt{2}\notin\mathbb{Q}$ but ... Ok, it's easy to prove that prime roots are irrational (i.e.
$ \sqrt{p} \not\in \mathbb{Q}, \text{ if } p \in \mathbb{P} $)
Consider $ \sqrt{2} $. We can quickly prove that 
$ \sqrt{2} \not\in \mathbb{Q} $. 
Proof: (by contradiction)
Assume that $\sqrt{2} \in \mathbb{Q}$
\begin{align*}
  &\Rightarrow \exists a,b \in \mathbb{Z} \;\text{ such that }\; \frac{a}{b} = \sqrt{2}
    &(\text{By definition})\\
  &\Rightarrow \frac{a^2}{b^2} = 2 & (\text{By squaring both sides}) \\
  &\Rightarrow a^2 = b^2 \cdot 2 \\
  &\text{Now consider the prime factorization of both sides. The left hand} \\
  &\text{side has an even number of prime factors (this is true of any square).} \\
  &\text{Since 2 is prime, and $b^2$ has an even number of prime factors, The } \\
  &\text{right hand side has an odd number of prime factors.} \\
  &\Rightarrow a^2 \neq b^2 \cdot 2 \\
  &\Rightarrow \frac{a}{b} \neq \sqrt{2} \quad \quad \rightarrow\leftarrow \\
  &\therefore \sqrt{2} \not\in \mathbb{Q} \quad \blacksquare
\end{align*}
Easy, right? Ok, consider this:
Let $f(a,n)$ be a function that returns the $n^{th}$ digit to the right of the
decimal place of the number $a$ (in its decimal expansion). So,
$f(z,n) = 0,\quad\forall z\in\mathbb{Z},n\in\mathbb{N}$.
However, since $\sqrt{2} \approx 1.414213562$, $f(\sqrt{2}, 1) = 4,\;f(\sqrt{2}, 2) = 1$
and so on.
This implies that 
$\displaystyle\sqrt{2} = 1 + \sum_{i=1}^\infty \frac{f(\sqrt{2},i)}{10^i}$
Since $\mathbb{Q}$ is closed under addition (i.e. $\forall a,b \in \mathbb{Q}, a+b \in \mathbb{Q}$) We must conclude
that $1 + \sum_{i=1}^\infty \frac{f(\sqrt{2},i)}{10^i} \in \mathbb{Q}$
$\therefore \sqrt{2} \in \mathbb{Q}$
But we just proved that $\therefore \sqrt{2} \not\in \mathbb{Q}$
ACK!!
Clearly one of the steps is wrong. My suspicion is this one:
$\displaystyle\sqrt{2} = 1 + \sum_{i=1}^\infty \frac{f(\sqrt{2},i)}{10^i}$
If so, this is sort of alarming! This implies that irrational numbers (i.e. almost all
numbers on the real line) cannot have a decimal representation! Is this true?
 A: The problem is that while $\mathbb{Q}$ is closed under addition (that is, the sum of two elements of $\mathbb{Q}$ is again in $\mathbb{Q}$), in order to capture an infinite sum such as $\displaystyle \sum_{i=1}^\infty \frac{f(\sqrt{2}, i)}{10^i}$ you don't need just addition, you need to take a limit of the sequence of partial sums $\displaystyle \sum_{i=1}^N \frac{f(\sqrt{2}, i)}{10^i}$.  The rational numbers $\mathbb{Q}$ are not closed under taking limits.
A: To say in another way, $\mathbb{Q}$ is not it's own closure, and so not all infinite sequences converge inside itself. Here, the terms of the sequence in question are:
$$\displaystyle \bigg(1 + \sum_{i=1}^n \frac{f(\sqrt{2},i)}{10^i}\bigg), \ \forall n \text{ from } 1 \to \infty$$
A: Any real number can be approximated by closer and closer rational numbers. If you took such an approximating sequence and computed the successive differences $a_i$, then that real number is the infinite sum of the $a_i$. So any real number is an infinite sum of rational numbers. Additive closure of $\mathbb{Q}$ is only true for finite sums.
A: The problem is when you deduce that an infinite sum of rational numbers is rational, this is just simply not true. A finite sum of rational numbers is rational, but not necessarily an infinite sum, for example: $\pi = 3 + \frac{1}{10} + \frac{4}{100} + \ldots$.
