Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 47 / 4 / b / ii / Solution

For every binary vector $x$, $S_1(x,x)=1$, and henceThus $d_2$ is not a metric, because it fails identity of indiscernibles. It is nonnegative and symmetric. In fact it also satisfies the triangle inequality trivially: every value lies in $[1/2,1]$, so $d_2(x,z)\le 1\le d_2(x,y)+d_2(y,z)$. That does not repair its nonzero self-distance.