Question

PLEASE USE INDUCTION, NO PROGRAMS ALLOWED

Consider the following program, where a and n are positive integers. Input: a, n x = a; m = n; y = 1; while (m > 1) {if m is even x = x*x; m = m/2; if m is odd y = x*y; x = x*x; m = (m - 1)/2;} Output x*y Show by induction that the above loop has the following invariant: a^n = x^m times y. What does the above program compute?

Answer 1

Take M={P=anxn+…+a0∈Z[x]/an>0}, with the usual addition and multiplication on polynomials. Interprete S(P) as P(X)+1.

Then M is a model of Robinson arithmetic, but there are long strings of "not-even" numbers, such as X,X+1,X+2,…. So in this model, it is false that if n is not even then n+1 is even.

If you define "n is odd" as "∃k/n=k+k+1", then it is false that every number is even or odd.

However, it is still true in M that addition is associative and commutative, and so if n is odd then n+1 is even, and if n is even, then n+1 is odd. (you will need a stranger model for this to fail)

If you want a model in which a2−2b2=0 has a solution, you can pick M={P∈Z[X,Y]/(X2−2Y2)/limy→∞P(√2y,y)=+∞ or P∈N}