Assume we have a computable, nonstandard model of Robinson arithmetic consisting of the set of integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial, with their usual arithmetic. Consider the formula sum_{i=0}^{n}i=(n^2+n)/(2)

figoveck38

figoveck38

Answered question

2022-11-14

Assume we have a computable, nonstandard model of Robinson arithmetic consisting of the set of integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial, with their usual arithmetic.
Consider the formula
i = 0 n i = n 2 + n 2
Is this formula valid in this model? Let |2x| be the nonstandard natural number represented by the polynomial 2x. In this model, is the sum of all natural numbers from |0| through |2x| equal to | 2 x 2 + x | ? If so, does this mean some hyper-finite sums are undefined? For example, the sum of |0| through |x| would be x 2 + x 2 , which doesn't exist in this model.

Answer & Explanation

Sean Sutton

Sean Sutton

Beginner2022-11-15Added 17 answers

Step 1
Elaborating on Andre Nicolas' comment: Try to write down the sentence
"For all  n i = 1 n i = n 2 + n 2 "
in the language of Robinson arithmetic. There isn't really any way to define an arbitrary summation. For a concrete example, how would you express
i = 1 X 2 + 17 X 3 i
in the language of Robinson arithmetic?
Now, stronger theories such as PA can talk about numbers which code sequences, and can prove the basic facts about these, so that we can legitimately use abbreviations like " i = 1 n ".
Step 2
EDIT: Specifically, we write " i = 1 n = x" as shorthand for "There is a number y which codes a sequence of length n, such that the first term of that sequence is 1, the last term is x, and for each k < n, the ( k + 1 )th term is the kth term plus k + 1." Note that proving that such a y exists for every n and x requires induction - which Robinson arithmetic lacks! Also note that even referring to sequences requires a lot of work in the absence of exponentiation, and while the Godel machinery for doing this (via the Chinese Remainder Theorem) goes through in Robinson, seemingly basic facts about sequences so represented will not.
We can make the same definitions in Robinson arithmetic (since the language is the same); however, we can't prove even very basic claims - such as that " i = 1 n i exists" in the first place

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?