To establish this, one must show that formula_2 really is a cut (according to the definition) and the square of formula_2, that is formula_22 (please refer to the link above for the precise definition of how the multiplication of cuts are defined), is formula_23 (note that rigorously speaking this is a cut formula_24). To show the first part, we show that for any positive rational formula_25 with formula_26, there is a rational formula_27 with formula_28 and formula_29. The choice formula_30 works, thus formula_2 is indeed a cut. Now armed with the multiplication between cuts, it is easy to check that formula_32 (essentially, this is because formula_33). Therefore to show that formula_34, we show that formula_35, and it suffices to show that for any formula_36, there exists formula_37, formula_38. For this we notice that if formula_39, then formula_40 for the formula_27 constructed above, this means that we have a sequence in formula_2 whose square can become arbitrarily close to formula_23, which finishes the proof.