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?
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