The Machine That Repeats Forever

At a glance

• Part 1: After the first pass, the output is fixed. Therefore every positive number of passes—including 11 and 111—agrees with one pass.
• Part 2: The hidden multiplier is 𝑐 =1. Repeat-stability leaves only 0 and 1, and multiplying by 0 is not reversible.

Key idea: repeat without losing information

The repeat rule alone does not identify the machine. Multiplication by 0 and multiplication by 1 both satisfy “two passes equal one pass.” Reversibility separates them: multiplying by 0 sends different inputs to the same output, while multiplying by 1 preserves every input.

1. Follow every positive number of passes

Begin with any input 𝑥, and call the first output 𝑦:

The given repeat rule says

Thus the first output 𝑦 is fixed by the machine. After one pass the output is 𝑦. If the output after any positive number 𝑟 of passes is 𝑦, then one more pass gives

Beginning with the first pass and repeating this argument proves that every positive number of passes has output 𝑀(𝑥). In particular, 11 passes and 111 passes both have exactly the same output as one pass.

Notice that this proof never uses the value of 𝑐.

2. Find the hidden multiplier

Use the input 1. One pass produces

A second pass multiplies by 𝑐 again:

One pass and two passes agree, so

Because 𝑐 is a nonnegative whole number, there are three possibilities to consider: 𝑐 =0, 𝑐 =1, or 𝑐 ≥2. If 𝑐 ≥2, then 𝑐 copies of 𝑐 have a sum greater than just one copy of 𝑐. Equivalently,

That contradicts 𝑐 ⋅𝑐 =𝑐. Therefore 𝑐 is either 0 or 1.

Now suppose 𝑐 =0. The different inputs 1 and 2 collide:

The undo rule would then require both

and

One undo machine cannot send the same input 0 to two different outputs. Thus 𝑐 ≠0, so the only remaining possibility is

Check

For 𝑐 =1,

for every nonnegative whole-number input. Two passes return 𝑥, every longer chain returns 𝑥, and the same machine serves as its own undo. All conditions are satisfied.

Why the answer is unique

Every allowed multiplication machine must satisfy 𝑐 ⋅𝑐 =𝑐. Whole-number comparison leaves only 𝑐 =0 and 𝑐 =1. The explicit collision rules out 𝑐 =0, and 𝑐 =1 works, so there is exactly one allowed multiplier.

Another facet

The same jewel has a more general face. Suppose 𝑇 is any machine whose outputs may be fed back through it. Suppose an undo machine 𝑉 satisfies

for every input, and suppose

for every input. Apply 𝑉 to both sides of the repeat equation:

The undo law turns the left side into 𝑇(𝑥) and the right side into 𝑥. Therefore

for every input. In more advanced language, any left-invertible idempotent transformation is the identity.

Teaching note

A complete response to the worksheet needs the fixed-output argument in Part 1 and all three moves in Part 2: derive 𝑐 ⋅𝑐 =𝑐, eliminate every whole number at least 2, and exhibit the 1/2 collision when 𝑐 =0. “Another facet” is optional enrichment, not a graded part.