For rational numbers, the numerator and denominator can be treated separately as integers.

This required only a slight variation of the base function I had been using, adding the line

    If d < 0 Then d = d - b: q = q + 1

    

shown below.

Also, to catch attempts at converting negative numbers in an ordinary positive base, I added

  If n < 0 And b > 0 Then base$ = "-----": Exit Function

  

but that's not essential to the negaternary algorithm.

Of course for negaternary, the base, b, will need to be -3.

Function base$(n, b)

  If n < 0 And b > 0 Then base$ = "-----": Exit Function

  v$ = ""

  n2 = n

  Do

    q = Int(n2 / b)

    d = n2 - q * b

    If d < 0 Then d = d - b: q = q + 1

    n2 = q

    v$ = Mid("0123456789abcdefghijklmnopqrstuvwxyz", d + 1, 1) + v$

  Loop Until n2 = 0

  base$ = v$

End Function

The function is called by the following driver for the integers -10 to 10, first for base +3, then for base -3:

DefDbl A-Z

Dim crlf$

Private Sub Form_Load()

 Form1.Visible = True

 

 

 Text1.Text = ""

 crlf = Chr$(13) + Chr$(10)

 

 For i = -10 To 10

   Text1.Text = Text1.Text & base$(i, 3) & crlf

 Next

 Text1.Text = Text1.Text & crlf

 For i = -10 To 10

   Text1.Text = Text1.Text & base$(i, -3) & crlf

 Next

 

 Text1.Text = Text1.Text & crlf & " done"

  

End Sub

The results are:

-----

-----

-----

-----

-----

-----

-----

-----

-----

-----

0

1

2

10

11

12

20

21

22

100

101

1212

1200

1201

1202

20

21

22

10

11

12

0

1

2

120

121

122

110

111

112

100

101

showing first the rejection of negative number representation in positive base 3, and then a set of values matching those in the linked explanation of negaternary representation.

One could of course change the function take only one parameter, the integer to be converted, and then use -3 wherever b is used.  Following that, one can make a function that takes in two integers (representing numerator and denominator) and apply the base conversion algorithm to each separately.

For negative fractions you presumably have a choice of having a negative numerator or a negative denominator, so that -3/8 could be either 10/112 or 120/1201. 

Indeed the following program produces 10/112 in response to an input of -3/8:

DECLARE FUNCTION base$ (nu#, b#)

DEFDBL A-Z

DO

  INPUT n$

  ix = INSTR(n$, "/")

  IF ix = 0 THEN EXIT DO

  numer = VAL(LEFT$(n$, ix - 1))

  denom = VAL(MID$(n$, ix + 1))

  PRINT base$(numer, -3) + "/" + base$(denom, -3)

LOOP

END

FUNCTION base$ (nu, b)

  IF nu < 0 AND b > 0 THEN base$ = "-----": EXIT FUNCTION

  v$ = ""

  n2 = nu

  DO

    q = INT(n2 / b)

    d = n2 - q * b

    IF d < 0 THEN d = d - b: q = q + 1

    n2 = q

    v$ = MID$("0123456789abcdefghijklmnopqrstuvwxyz", d + 1, 1) + v$

  LOOP UNTIL n2 = 0

  base$ = v$

END FUNCTION