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MATH 136: HOMEWORK 3

Due Wednesday, September 25 at 11:59pm.

Remember that this is a math class, so every assertion requires justification (i.e. a proof).

On this homework, we will show that decimal expansions of various irrationals r ∈ R are primitive recursive. We will actually show the stronger and more base-agnostic fact that the map n → ⌊nr⌋ (the Beatty sequence) is primitive recursive. This immediately implies that the decimal expansion is primitive recursive as well, since the k-th decimal digit of r is the ones digit of ⌊10kr⌋.

Problem 1. The decimal expansion of √2

Show that the function N → N defined by n 7→ ⌊n√2⌋ is primitive recursive.

Problem 2. The decimal expansion of e

Recall Euler’s constant e = 2.71828..., which satisfies

Part a. Show that for every n ∈ N, we have

(Hint: Show first that ).

Part b. Show that the function N → N defined by n 7→ ⌊ne⌋ is primitive recursive.

Problem 3. The decimal expansion of π

Recall the alternating series for π:

Part a. Using the standard approximation for alternating series, show that there is a primitive recursive function f : N → Q such that for all positive n ∈ N, we have |π − f(n)| < 1/n.

(By the standard approximation for alternating series, we mean the fact that if A = a0 − a1 + a2 − a3 + · · · , then |A − (a0 − a1 + a2 − a3 + · · · an−1)| < |an|).

Part b. It is a very nontrivial fact that the irrationality exponent of π is finite. In fact, it is known to be less than 8, which means that for cofinitely many positive integers q, the distance from π to the set  is greater than 1/q8 (i.e. for every p ∈ Z, we have |π − p/q| > 1/q8 ).

Using this, show that the function N → N defined by n 7→ ⌊nπ⌋ is primitive recursive.

(Hint: The only issue with an approximation is that it can end with a string of 9s or 0s, since for instance, you could have a very good approximation of a real that ends in a string of 9s, like 2.4999, when the actual value is something like 2.5000003. So you need to do a minimization like “take the first approximation that doesn’t end in a string of 9s or a string of 0s”. To make this primitive recursive, you need to bound the minimization, which the irrationality exponent can help with.)