Existence of nth Roots

Real Analysis · Axiom Academy

How the completeness axiom turns "√2 should exist" into a rigorous proof that it does — and that every nth root does too. 1. The Gap the Rationals Leave Squaring rationals lets us crowd in on 2 from below ( , all with square ) and from above ( , all with square ). The two crowds press toward a single point — but in that point is missing . No rational squares to exactly 2 . crowding in from below — squares stay under 2 no rational lands exactly on the boundary 2. Completeness Hands Us a Number Collect every positive number whose square stays under 2 . This set S is non-empty ( since ) and bounded above (if then , so ). Completeness now does the heavy lifting: Every non-empty set of reals that is bounded above has a least upper bound (a supremum) in . exists — the exact right edge of S By trichotomy, is less than, greater than, or equal to 2 . We rule out the first two — each contradicts what means — so only can survive. There's room to nudge up: some still has , so . But it's bigger than — so wasn't an upper bound. Contradiction. There's room to shrink: some has , so it's also an upper bound — yet smaller than . So wasn't the least upper bound. Contradiction. both escapes blocked — the boundary squares to 2 4. The Same Move Builds Every nth Root Nothing in the argument cared that the exponent was 2 or the target was 2 . Fix any and any , swap " " for " ", and replay it line for line. Show S is non-empty and bounded above. Let — it exists by completeness.

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