Quantifier elimination makes $S=\{x:\exists y(y^2=x)\}$ basic. It contains every square and no negative element. If some $a>0$ were not a square, $S$ would have a boundary inside $(0,\infty)$; but multiplication by positive squares and the ordered-field inequalities force membership to be locally constant along positive multiplicative intervals, contradicting the finite-endpoint form of a basic set. Hence every positive element, and also zero, is a square.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 16J / e / Solution

 Solution (source code)