Update crates/consistent-hashing/README.md

aneubeck · look · web-flow · commit 99c69f3a059e · 2025-08-15T10:15:12.000+02:00
Co-authored-by: Luke Francl &lt;look@github.com&gt;
diff --git a/crates/consistent-hashing/README.md b/crates/consistent-hashing/README.md
@@ -106,7 +106,7 @@ Putting it together leads to `S(k+1, n+1) = {n} ∪ S(k, m)` and `S(k+1, n) = {m
 ### Property 2
 
 The final part is to prove property 2. This time we have an inducation over `k` and `n`.
-As before, induction start for `k=1` and for all `n>0` is inherited from the `consistency_hash` implementation. The case `n=k` is also trivially covered, since the only valid set are the numbers `{0, ..., k-1}` which the algorithm correctly outputs. So, we only need to care about the induction step where `k>1` and `n>k`.
+As before, the base case of the induction for `k=1` and all `n>0` is inherited from the `consistency_hash` implementation. The case `n=k` is also trivially covered, since the only valid set are the numbers `{0, ..., k-1}` which the algorithm correctly outputs. So, we only need to care about the induction step where `k>1` and `n>k`.
 
 We need to prove that `P(i ∈ S(k+1, n+1)) = (k+1)/(n+1)` for all `0 <= i <= n`. Property 3 already proves the case `i = n`. Furthermore we know that `P(n ∈ S(k+1, n+1)) = (k+1)/(n+1)` and vice versa  `P(n ∉ S(k+1, n+1)) = 1 - (k+1)/(n+1)`. Let's consider those two cases separately.
 
