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

##### 1 Answer

#### Answer:

We can construct the square root of

#### Explanation:

Suppose you only know the rational numbers

This is a set of numbers of the form

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

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

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

The set

#((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

Define a predicate

#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

We can use the notation

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)#

So what have we done?

We have constructed an irrational number