Base-n arithmetic and wiggle

We saw above that

${\displaystyle 5_{10}{3}_{10}5{;}_{10}{4}_{10}9 = 535.49.}$

Being explicit about the base, we can write

${\displaystyle 5_{10}{3}_{10}5{;}_{10}{4}_{10}9 = {535.49}_{10}.}$

Translating numbers in other bases into wiggle is an immediate generalization.  As a first example, suppose we are trying to evaluate

${\displaystyle {10110111}_2.}$

We rewrite this, with the same digits---or bits---in the same order as the superdigits, and making each interbase be 2, so that



${\displaystyle {10110111}_2 = 1_2{0}_2{1}_2{1}_2{0}_2{1}_2{1}_{2}1.}$

Telescoping from left, we get

${\displaystyle 1\times 2 = 2,}$
${\displaystyle 2+0 = 2,}$
${\displaystyle 2\times 2 = 4,}$
${\displaystyle 4+1 = 5,}$
${\displaystyle 5\times 2 = 10,}$
${\displaystyle 10+1 = 11,}$
${\displaystyle 11\times 2 = 22,}$
${\displaystyle 22+0 = 22,}$
${\displaystyle 22\times 2 = 44,}$
${\displaystyle 44+1 = 45,}$
${\displaystyle 45\times 2 = 90,}$
${\displaystyle 90+1 = 91,}$
${\displaystyle 91\times 2 = 182,}$
${\displaystyle 182+1; = 183;}$

so that in the end

${\displaystyle {10110111}_2 = 183.}$

Let's try one with a fractional part.  What is

${\displaystyle {102.21}_3?}$


Rewriting in wiggle, we have

${\displaystyle {102.21}_3 = 1_3{0}_{3}2{;}_3{2}_{3}1.}$

Telescoping from the left and the right, we get

${\displaystyle 1_3{0}_{3}2{;}_3{2}_{3}1 = {(3+0)}_{3}2{;}_3{2}_{3}1,}$ ${\displaystyle = 3_{3}2{;}_3{2}_{3}1,}$
${\displaystyle = (9+2){;}_3{2}_{3}1,}$
${\displaystyle = 11{;}_3{2}_{3}1,}$
${\displaystyle = 11{;}_3(2+\frac{1}{3}),}$
${\displaystyle = 11{;}_3\frac{7}{3},}$
${\displaystyle = 11\frac{7}{9}.}$

Odder bases are also easy to calculate with.