Editing Dedekind cut (section)

==Construction of the real numbers==
{{See also|Construction of the real numbers#Construction by Dedekind cuts}}
A typical Dedekind cut of the [[rational number]]s <math>\Q</math> is given by the partition <math>(A,B)</math> with

:<math>A = \{ a\in\mathbb{Q} : a^2 < 2 \text{ or } a < 0 \},</math>
:<math>B = \{ b\in\mathbb{Q} : b^2 \ge 2 \text{ and } b \ge 0 \}.</math><ref>In the second line, <math>\ge</math> may be replaced by <math>></math> without any difference as there is no solution for <math>x^2 = 2</math> in <math>\Q</math> and <math>b=0</math> is already forbidden by the first condition. This results in the equivalent expression
:<math>B = \{ b\in\mathbb{Q} : b^2 > 2 \text{ and } b > 0 \}.</math></ref>

This cut represents the [[irrational number]] <math>\sqrt{2}</math> in Dedekind's construction. The essential idea is that we use a set <math>A</math>, which is the set of all rational numbers whose squares are less than 2, to "represent" number <math>\sqrt{2}</math>, and further, by defining properly arithmetic operators over these sets (addition, [[subtraction]], multiplication, and division), these sets (together with these arithmetic operations) form the familiar real numbers.

To establish this, one must show that <math>A</math> really is a cut (according to the definition) and the square of <math>A</math>, that is <math>A \times A</math> (please refer to the link above for the precise definition of how the multiplication of cuts is defined), is <math>2</math> (note that rigorously speaking this number 2 is represented by a cut <math>\{x\ |\ x \in \mathbb{Q}, x < 2\}</math>). To show the first part, we show that for any positive rational <math>x</math> with <math>x^2 < 2</math>, there is a rational <math>y</math> with <math>x < y</math> and <math>y^2 < 2</math>. The choice <math>y=\frac{2x+2}{x+2}</math> works, thus <math>A</math> is indeed a cut. Now armed with the multiplication between cuts, it is easy to check that <math>A \times A \le 2</math> (essentially, this is because <math>x \times y \le 2, \forall x, y \in A, x, y \ge 0</math>). Therefore to show that <math>A \times A = 2</math>, we show that <math>A \times A \ge 2</math>, and it suffices to show that for any <math>r < 2</math>, there exists <math>x \in A</math>, <math>x^2 > r</math>. For this we notice that if <math>x > 0, 2-x^2=\epsilon > 0</math>, then <math>2-y^2 \le \frac{\epsilon}{2}</math> for the <math>y</math> constructed above, this means that we have a sequence in <math>A</math> whose square can become arbitrarily close to <math>2</math>, which finishes the proof.

Note that the equality {{math|1=''b''<sup>2</sup>&nbsp;=&nbsp;2}} cannot hold since [[Square root of 2#Proofs of irrationality|<math>\sqrt{2}</math> is not rational]].