Editing Balanced ternary (section)

==Square roots and cube roots==
The process of extracting the [[square root]] in balanced ternary is analogous to that in decimal or binary. 
:<math>(10\cdot x+y)^{\mathrm{1T}}-100\cdot x^{\mathrm{1T}}=\mathrm{1T0}\cdot x\cdot y+y^{\mathrm{1T}}=
\begin{cases}
\mathrm{T10}\cdot x+1, & y=\mathrm{T} \\
0, & y=0 \\
\mathrm{1T0}\cdot x+1, & y=1
\end{cases}
</math>

As in division, we should check the value of half the divisor first. For example,
                              1. 1 1 T 1 T T 0 0 ... 
                            _________________________
                           √ 1T                          1<1T<11, set 1
                            − 1
                             _____
                   1×10=10    1.0T                       1.0T>0.10, set 1
                       1T0   −1.T0
                             ________
                   11×10=110    1T0T                     1T0T>110, set 1
                        10T0   −10T0
                               ________
                  111×10=1110    T1T0T                   T1T0T<TTT0, set T
                        100T0   −T0010
                                _________
                 111T×10=111T0    1TTT0T                 1TTT0T>111T0, set 1
                        10T110   −10T110
                                 __________
                111T1×10=111T10    TT1TT0T               TT1TT0T<TTT1T0, set T
                        100TTT0   −T001110
                                  ___________
               111T1T×10=111T1T0    T001TT0T             T001TT0T<TTT1T10, set T
                        10T11110   −T01TTTT0
                                   ____________
                111T1TT×10=111T1TT0    T001T0T           TTT1T110<T001T0T<111T1TT0, set 0
                                      −      T           Return 1
                                      ___________
              111T1TT0×10=111T1TT00    T001T000T         TTT1T1100<T001T000T<111T1TT00, set 0
                                      −        T         Return 1
                                      _____________
            111T1TT00*10=111T1TT000    T001T00000T
                                              ... 
Extraction of the cube root in balanced ternary is similarly analogous to extraction in decimal or binary: 
:<math>(10\cdot x+y)^{10}-1000\cdot x^{10}=1000\cdot x^{\mathrm{1T}}\cdot y+100\cdot x\cdot y^{\mathrm{1T}}+y^{10}=
\begin{cases}
\mathrm{T000}\cdot x^{\mathrm{1T}}+100\cdot x+\mathrm{T}, & y=\mathrm{T}\\
0, & y=0\\
1000\cdot x^{\mathrm{1T}}+100\cdot x+1, & y=1
\end{cases}

</math> 
Like division, we should check the value of half the divisor first too.
For example:
                               1.  1   T  1  0 ... 
                             _____________________
                           ³√ 1T
                             − 1                 1<1T<10T,set 1
                             _______
                               1.000
               1×100=100      −0.100             borrow 100×, do division
                              _______
                       1TT     1.T00             1T00>1TT, set 1
           1×1×1000+1=1001    −1.001
                              __________
                                 T0T000
             11×100            −   1100           borrow 100×, do division
                               _________
                      10T000     TT1T00           TT1T00<T01000, set T
        11×11×1000+1=1TT1001   −T11T00T
                               ____________
                                 1TTT01000
            11T×100             −    11T00        borrow 100×, do division
                                ___________
                    1T1T01TT     1TTTT0100        1TTTT0100>1T1T01TT, set 1
     11T×11T×1000+1=11111001    − 11111001
                                ______________
                                     1T10T000
           11T1×100                 −  11T100      borrow 100×, do division
                                    __________
                       10T0T01TT     1T0T0T00      T01010T11<1T0T0T00<10T0T01TT, set 0
     11T1×11T1×1000+1=1TT1T11001    −  TT1T00      return 100×
                                    _____________
                                     1T10T000000
                                         ... 
Hence {{radic|2|3}} = 1.259921<sub>dec</sub> = 1.1T1&nbsp;000&nbsp;111&nbsp;001&nbsp;T01&nbsp;00T&nbsp;1T1&nbsp;T10&nbsp;111<sub>bal3</sub>.