...and wiggles decimal

Consider the following wiggle expression :

`\qquad\qquad\qquad 5_{10}3_{10}5;_{10}4_{10}9`

Telescoping from the left and right,

`\qquad\qquad\qquad 5 xx 10  =  50,`

`\qquad\qquad\qquad 50 + 3  =  53,`

`\qquad\qquad\qquad 53 xx 10  =  530,`

`\qquad\qquad\qquad 530 + 5;  =  535;`

and

`\qquad\qquad\qquad 9 -: 10 = \frac{9}{10}  (=  0.9),`

`\qquad\qquad\qquad \frac{9}{10} + 4  =   4\frac{9}{10}\qquad(=  0.9 + 4  =  4.9),`

`\qquad\qquad\qquad 4\frac{9}{10}  -: 10  =  \frac{49}{100}   =  4.9 -: 10  =  0.49,`


so that, adding, and using the decimal form of the fraction, we get

`\qquad\qquad\qquad 5_{10}3_{10}5;_{10}4_{10}9  =  535.49.`

The wiggle form on the left closely corresponds to the decimal form on the right.  The superdigits on the left are exactly the digits on the right.  The face value marker on the left corresponds to the decimal point on the right.  The interbases on the left are all explicitly `10`.  The number on the right is implicitly in our usual decimal or base ten notation.

Wiggles general...

Consider the wiggle expression `2_{4}3_{5}4;_{6}5_{7}6`.  One way to calculate its value is to telescope from left and right towards the face value superdigit.  Telescoping from the left :

`2 xx 4  =  8,`

`8 + 3  = 11,`

`11 xx 5  =  55,`

`55 + 4;  =  59;`

Telescoping from the right :

`6 -: 7  =  \frac{6}{7},`

` \frac{6}{7} + 5  =  5 \frac{6}{7}  =  \frac{41}{7},`

` \frac{41}{7} -: 6 =  \frac{41}{42}.`

Combining these, we get

`2_{4}3_{5}4;_{6}5_{7}6  =  59 \frac{41}{42}.`

We shall eventually see a more efficient process for the fractional part of such problems than telescoping from the right, but not just yet.

More boxed wiggle

Now, as a second exercise, let's using the counting board again to convert  5 yd., 2 ft., 4 in. completely into yards.  This time, the relevant wiggle notation is `5;_{3}2_{12}4`, with the semicolon now marking the leftmost superdigit.  What we do with the counting board will help us understand what to do with this wiggle notation.

First, move the 4 left across the border.  Since we are moving left, we divide by the 12 at the border.  The 4 becomes `\frac{1}{3}`.  This should not be a surprise---4 inches is a `\frac{1}{3}` of a foot.  Add this to the 2 feet already there, to get `2\frac{1}{3}` feet, which can also be written `\frac{7}{3}` feet.  Now move this `\frac{7}{3}` left across the border again.  This time, we need to divide by 3, giving `\frac{7}{9}`.  Adding to what is already there, we get `5\frac{7}{9}`.  Finally, then, we have no inches, no feet, and `5\frac{7}{9}` written in the yards rectangle.  So 5 yd., 2 ft., 4 in. is the same length as `5\frac{7}{9}` yards.

As a third exercise, we convert 5 yd., 2 ft., 4 in. completely into feet.  Starting as before, we move the 5 rightward, multiplying by 3, and we move the 4 leftward, dividing by 12.  Adding to the 2 already there, we get

`\qquad\qquad\qquad 15 + 2 + \frac{1}{3}  =  17\frac{1}{3},`


so that 5 yd., 2 ft., 4 in. is `17\frac{1}{3}` feet.


Summarizing, then,


`\qquad\qquad\qquad 5_{3}2_{12}4  =  5_{3}2_{12}4; =  (((5 xx 3)  +  2) xx 12)  +  4`
`\qquad\qquad\qquad\qquad   =  208,`

`\qquad\qquad\qquad 5_{3}2;_{12}4  =  (5 xx 3)  +  2  +  (12 \\ 4)`
`\qquad\qquad\qquad\qquad   =  17\frac{1}{3},`

`\qquad\qquad\qquad 5;_{3}2_{12}4  =  5  +  (3 \\ (2  +  (12 \\ 4)))  =  5  +  \frac{2  +  \frac{4}{12}}{3}`
`\qquad\qquad\qquad\qquad   =  5\frac{7}{9}.`

From these arithmetical calculations, we see that we have arranged things so that

`\qquad\qquad\qquad 5" yd." + 2" ft." + 4" in." `
`\qquad\qquad\qquad\qquad  =  5_{3}2_{12}4;" in."  =  5_{3}2;_{12}4" ft."  =  5;_{3}2_{12}4" yd."

The unit maker, ";", behaves something like a decimal point.  As we shall soon see, we can make this analogy much stronger.