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

1 Answer
Sep 18, 2016

Answer:

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.