Question

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

Answer 1

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

Explanation:

Suppose you only know the rational numbers `QQ`

This is a set of numbers of the form `p/q` where `p, q` are integers and `q != 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 `QQ xx QQ`. We can define some arithmetic operations on this set as follows:

`(a, b) + (c, d) = (a+c, b+d)`

`(a, b) * (c, d) = (ac+2bd, ad+bc)`

The set `QQ xx QQ` is closed under these operations and they obey all of the properties required of addition and multiplication in a field. For example:

`((a, b) * (c, d)) * (e, f) = (ac+2bd, ad+bc) * (e, f)`

`color(white)(((a, b) * (c, d)) * (e, f)) = (ac+2bd, ad+bc) * (e, f)`

`color(white)(((a, b) * (c, d)) * (e, f)) = (ace+2bde+2adf+2bcf, acf+ade+bce+2bdf)`

`color(white)(((a, b) * (c, d)) * (e, f)) = (a, b) * (ce+2df, cf+de)`

`color(white)(((a, b) * (c, d)) * (e, f)) = (a, b) * ((c, d) * (e, f))`

Then the rational numbers correspond to pairs of the form `(a, 0)` and `(0, 1) * (0, 1) = (2, 0)`. That is: `(0, 1)` is a square root of `2`.

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

`P(color(black)()(a, b)) = { (a > 0 " if " 2b^2 < a^2), (b > 0 " if " 2b^2 > a^2), ("false" " " "otherwise") :}`

Then we can define:

`(a, b) < (c, d) " " <=> " " P(color(black)()(c-a, d-b))`

Then this is an extension of the definition of the natural order of `QQ` to `QQ xx QQ`. With this ordering, `(0, 1)` is positive.

We can use the notation `sqrt(2)` to stand for `(0, 1)` and write `(a, b)` as `a+bsqrt(2)`.

We have:

`(a+bsqrt(2)) + (c+dsqrt(2)) = (a+c)+(b+d)sqrt(2)`

`(a+bsqrt(2))(c+dsqrt(2)) = (ac+2bd)+(ad+bc)sqrt(2)`

`color(white)()`
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 `{ a+bsqrt(2) : a, b in QQ }` forms an ordered field extending the rational numbers in a way consistent with their arithmetic and order.