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Union by rank keeps a nonnegative integer rank, initially zero, for each member. For each input, obtain its root of a tree using the disjoint-set find operation, then compare the resulting ranks. Attach the lower-rank root of a tree below the higher-rank root of a tree, leaving both ranks unchanged. When the two disjoint-set representatives differ but their ranks are equal, choose either survivor and increase its rank by exactly one. A redundant disjoint-set union operation changes no ranks. Path compression also changes no ranks. The surviving root's rank bounds the height of its rooted tree, measured in edges; after path compression it need not equal that height.^[ref]

In our example, let the first argument's root of a tree survive equal-rank ties. The first two merges create roots $2,4$, each of rank $1$. The third gives $\mathrm{parent}[4]=2$ and increases the rank of $2$ to $2$. The fourth creates root $6$ of rank $1$. The fifth attaches $6$ below $2$ without increasing any rank. Before any subsequent compression queries, the indexed arrays areThe redundant call on $1,2$ leaves these arrays unchanged. In particular, the disjoint-set representative of $1$ is $2$, although $1$ is the minimum member of its block of a set partition.

Two short arguments explain the rank invariant. First, a root of rank $r$ has at least $2^r$ members in its block of a set partition: rank zero starts with one, and increasing rank combines two disjoint blocks of a set partition with at least $2^r$ members each. Unequal-rank linking only adds members. Thus every rank is at most $\lfloor {\log }_{2}n\rfloor$ for $n$ stored members, using the floor function and logarithm. Second, height is initially zero; unequal-rank linking adds at most one to the lower height, while equal-rank linking raises the available bound by one. Path compression only shortens routes. These arguments prove a height bound, not the stronger amortized analysis below.