Consider the algorithm "M(x): if x<0 return -x, else return M(x-M(x-1))/2". This algorithm terminates for all real x, though this is not so easy to prove. In fact, Peano Arithmetic cannot prove the statement "M(x) terminates for all natural x". Paper to come! Joint work with @jeffgerickson and @alreadydone

A Mastodon instance for maths people. The kind of people who make \(\pi z^2 \times a\) jokes.
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