Editing Surreal number (section)

=== Division ===
The definition of division is done in terms of the reciprocal and multiplication:

<math display=block>\frac xy = x \cdot \frac 1y</math>

where<ref name="Con01" />{{rp|21}}

<math display=block>\frac 1y = \left\{\left.0, \frac{1+(y_R-y)\left(\frac1y\right)_L}{y_R}, \frac{1+\left(y_L-y\right)\left(\frac1y\right)_R}{y_L} \,\,\right|\,\, \frac{1+(y_L-y)\left(\frac1y\right)_L}{y_L}, \frac{1+(y_R-y)\left(\frac1y\right)_R}{y_R} \right\}</math>

for positive {{mvar|y}}. Only positive {{math|''y''{{sub|''L''}}}} are permitted in the formula, with any nonpositive terms being ignored (and {{math|''y''{{sub|''R''}}}} are always positive). This formula involves not only recursion in terms of being able to divide by numbers from the left and right sets of {{mvar|y}}, but also recursion in that the members of the left and right sets of {{math|{{sfrac|1|''y''}}}} itself. 0 is always a member of the left set of {{math|{{sfrac|1|''y''}}}}, and that can be used to find more terms in a recursive fashion. For example, if {{math|1=''y'' = 3 = { 2 {{!}} }}}, then we know a left term of {{sfrac|1|3}} will be 0. This in turn means {{math|1={{sfrac|1 + (2 − 3)0|2}} = {{sfrac|1|2}}}} is a right term. This means
<math display=block>\frac{1+(2-3)\left(\frac12\right)}2=\frac14</math>
is a left term. This means
<math display=block>\frac{1+(2-3)\left(\frac14\right)}2 = \frac 38</math>
will be a right term. Continuing, this gives
<math display=block>\frac13 = \left\{\left. 0, \frac14, \frac5{16}, \ldots \,\right|\, \frac12, \frac38, \ldots\right\}</math>

For negative {{mvar|y}}, {{math|{{sfrac|1|''y''}}}} is given by
<math display=block>\frac1y=-\left(\frac1{-y}\right)</math>

If {{math|1=''y'' = 0}}, then {{math|{{sfrac|1|''y''}}}} is undefined.