The Effect of Functions on Addresses

Recall addresses on the inch scale ruler can be expressed by using two functions

f0(x) = x/2 and f1(x) = x/2 + 1/2

Suppose the point x has address a1a2a3... . Note that starting the address with a1 emphasizes the relation between the address and the number:

x = a1/2 + a2/22 + a3/23 + ... + an/2n + ...

What are the addresses of f0(x) and of f1(x)?

This picture illustrates the effect of f0 on addresses. Note that the position of f0(x) within [0, 1/2] is identical to that of x within [0, 1].

This picture illustrates the effect of f1 on addresses. Note that the position of f1(x) within [1/2, 1] is identical to that of x within [0, 1].

Studying these pictures, we see f0 shifts the address one place to the right (this is what "the position of f0(x) within [0, 1/2] is identical to that of x within [0, 1]" means) and inserts 0 in the left-most position (because f0(x) lies in [0, 1/2]).

Similarly, f1 shifts the address one place to the right (this is what "the position of f1(x) within [1/2, 1] is identical to that of x within [0, 1]" means) and inserts 1 in the left-most position (because f1(x) lies in [1/2, 1]).

That is, we see

f0(a1a2a3... ) = 0a1a2a3...

and

f1(a1a2a3... ) = 1a1a2a3...

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