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

Hint:
In order to check the rationality of any number, Firstly suppose the number to be rational and then prove the assumed statement contradictory. Let's assume $\sqrt 2 $ is a rational number. Then write it \[\sqrt 2 = \dfrac{a}{b}{\text{ where }}a,{\text{ }}b\] are whole numbers, b not zero. Assume only the rational numbers known to you are $Q$ is a set of numbers in the form of \[\dfrac{p}{q}{\text{ where }}p,q\] are integers and q is not equal to $0.$ And also assume that $\dfrac{a}{b}$ is reduced to the lowest terms.

Complete step by step solution:
Let us suppose we only know rational numbers which are the set of form \[\dfrac{p}{q}{\text{ where }}p,q\] are integers and q is not equal to $0.$
Coming to the question,
The rational number has no solution for the equation ${x^2} - 2 = 0$,
Now defining the set of ordered pair of rational numbers $Q \times Q$ and also doing some algebraic operation with it, we will get
$
  (a,\;b) + (c,\;d) = (a + c,\;b + d) \\
  (a,\;b)(c,\;d) = (ac + 2bd,\;ad + bc) \\
 $
The set $Q \times Q$ is closed under algebraic operations and also obey all of the properties required for addition and multiplication.
So then the rational number corresponds to pair of the form $(0,\;a)\;{\text{and}}\;(0,\;1).(0,\;1) = (2,\;0)$
That means $(0,\;1)$ is the square root of $2$
Define a predicate $P$ (for "positive") on this set of ordered pairs as follows:
$P(a,\;b) = \left\{ {\begin{array}{*{20}{c}}
  {a > 0}&{{\text{if}}\;2{b^2} < {a^2}} \\
  {b > 0}&{{\text{if}}\;2{b^2} > {a^2}} \\
  {{\text{false}}}&{{\text{otherwise}}}
\end{array}} \right\}$
Then we can define:
$(a,\;b) < (c,\;d) \Leftrightarrow P(c - a,\,d - b)$
Then this is an extension of the definition of the natural order of $Q\;{\text{to}}\;Q \times Q.$ With this ordering, \[(0,1)\] is positive. We can use the notation $\sqrt 2 $ to stand for \[(0,1)\] and write \[(a,\;b)\;{\text{as}}\;a + b\sqrt 2 .\]
So we have
$
  \left( {a + b\sqrt 2 } \right) + \left( {c + d\sqrt 2 } \right) = (a + c) + (b + d)\sqrt 2 \\
  \left( {a + b\sqrt 2 } \right)\left( {c + d\sqrt 2 } \right) = (ac + 2bd) + (ad + bc)\sqrt 2 \\
 $

Note:
Notice that in order for $\dfrac{a}{b}$ to be in simplest terms, both of \[a{\text{ and }}b\] cannot be even. One or both must be odd. Otherwise, we could simplify $\dfrac{a}{b}$ further. We started the whole process assuming that $\dfrac{a}{b}$ was simplified to lowest terms.