### 16th Putnam 1956

**Problem A6**

Let R be the reals. Find f **:** R → R which preserves all rational distances but not all distances. Show that if f **:** R^{2} → R^{2} preserves all rational distances then it preserves all distances.

**Solution**

Let f(x) = x for x rational, x + 1 for x irrational. If x - y is rational, then either both x and y are rational in which case f(x) - f(y) = x - y, or neither are in which case f(x) - f(y) = (x + 1) - (y + 1) = x - y. So f preserves rational distances. But f does not preserve all distances. For example f(√2) - f(0) ≠ √2 - 0.

Given points A and B in the plane a distance k apart, and any ε > 0, we can find a point C such that AC and BC are rational and |AC - k| < ε/2, |BC| < ε/2. Let A' = f(A), B' = f(B), C' = f(C). Then |A'C'| = |AC|, |B'C'| = |BC|, and hence |A'B'| lies within ε of k. This is true for all ε > 0, so |A'B'| = k and f preserves all distances.

16th Putnam 1956

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999