If all you know is rational numbers, what is the square root of #2# and how can you do arithmetic with it?
1 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