If all you know is rational numbers, what is the square root of 2 and how can you do arithmetic with it?

Sep 18, 2016

We can construct the square root of $2$ using ordered pairs of rational numbers...

Explanation:

Suppose you only know the rational numbers $\mathbb{Q}$

This is a set of numbers of the form $\frac{p}{q}$ where $p , q$ are integers and $q \ne 0$.

They are closed under addition, subtraction, multiplication and division by non-zero numbers.

In technical language, they form a field.

The rational numbers contain no solution to the equation:

${x}^{2} - 2 = 0$

The set of ordered pairs of rational numbers is denoted $\mathbb{Q} \times \mathbb{Q}$. We can define some arithmetic operations on this set as follows:

$\left(a , b\right) + \left(c , d\right) = \left(a + c , b + d\right)$

$\left(a , b\right) \cdot \left(c , d\right) = \left(a c + 2 b d , a d + b c\right)$

The set $\mathbb{Q} \times \mathbb{Q}$ is closed under these operations and they obey all of the properties required of addition and multiplication in a field. For example:

$\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right) = \left(a c + 2 b d , a d + b c\right) \cdot \left(e , f\right)$

$\textcolor{w h i t e}{\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right)} = \left(a c + 2 b d , a d + b c\right) \cdot \left(e , f\right)$

$\textcolor{w h i t e}{\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right)} = \left(a c e + 2 b \mathrm{de} + 2 a \mathrm{df} + 2 b c f , a c f + a \mathrm{de} + b c e + 2 b \mathrm{df}\right)$

$\textcolor{w h i t e}{\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right)} = \left(a , b\right) \cdot \left(c e + 2 \mathrm{df} , c f + \mathrm{de}\right)$

$\textcolor{w h i t e}{\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right)} = \left(a , b\right) \cdot \left(\left(c , d\right) \cdot \left(e , f\right)\right)$

Then the rational numbers correspond to pairs of the form $\left(a , 0\right)$ and $\left(0 , 1\right) \cdot \left(0 , 1\right) = \left(2 , 0\right)$. That is: $\left(0 , 1\right)$ is a square root of $2$.

Define a predicate $P$ (for "positive") on this set of ordered pairs as follows:

$P \left(\textcolor{b l a c k}{} \left(a , b\right)\right) = \left\{\begin{matrix}a > 0 \text{ if " 2b^2 < a^2 \\ b > 0 " if " 2b^2 > a^2 \\ "false" " " "otherwise}\end{matrix}\right.$

Then we can define:

$\left(a , b\right) < \left(c , d\right) \text{ " <=> " } P \left(\textcolor{b l a c k}{} \left(c - a , d - b\right)\right)$

Then this is an extension of the definition of the natural order of $\mathbb{Q}$ to $\mathbb{Q} \times \mathbb{Q}$. With this ordering, $\left(0 , 1\right)$ is positive.

We can use the notation $\sqrt{2}$ to stand for $\left(0 , 1\right)$ and write $\left(a , b\right)$ as $a + b \sqrt{2}$.

We have:

$\left(a + b \sqrt{2}\right) + \left(c + \mathrm{ds} q r t \left(2\right)\right) = \left(a + c\right) + \left(b + d\right) \sqrt{2}$

$\left(a + b \sqrt{2}\right) \left(c + \mathrm{ds} q r t \left(2\right)\right) = \left(a c + 2 b d\right) + \left(a d + b c\right) \sqrt{2}$

$\textcolor{w h i t e}{}$
So what have we done?

We have constructed an irrational number $\sqrt{2}$ from rational numbers using ordered pairs and some algebra. The resulting set $\left\{a + b \sqrt{2} : a , b \in \mathbb{Q}\right\}$ forms an ordered field extending the rational numbers in a way consistent with their arithmetic and order.