18
$\begingroup$

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?

$\endgroup$
2
  • 12
    $\begingroup$ The problem is that although $\mathbb{Q}$ is closed under addition of a finite number of terms, it is not closed under taking limits which is essentially what you are doing with an infinite sum. $\endgroup$
    – Old John
    Commented Dec 13, 2013 at 17:57
  • 2
    $\begingroup$ The issue comes from saying $1 + \sum_{i=1}^{\infty} \frac{f(\sqrt{2}, i)}{10^i} \in \mathbb{Q}$. It is true that each of the terms in the sum is a rational number, but their (infinite) sum is not. $\endgroup$
    – tylerc0816
    Commented Dec 13, 2013 at 17:58

4 Answers 4

20
$\begingroup$

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: $\pi = 3 + \frac{1}{10} + \frac{4}{100} + \ldots$.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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:

$$\displaystyle \bigg(1 + \sum_{i=1}^n \frac{f(\sqrt{2},i)}{10^i}\bigg), \ \forall n \text{ from } 1 \to \infty$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .