1 files changed, 2 insertions, 2 deletions
diff --git a/bip-0032.mediawiki b/bip-0032.mediawiki
index 7a4ae84..b441658 100644
--- a/bip-0032.mediawiki
+++ b/bip-0032.mediawiki
@@ -151,7 +151,7 @@ The total number of possible extended keypairs is almost 2<sup>512</sup>, but th
 * Calculate I = HMAC-SHA512(Key = "Bitcoin seed", Data = S)
 * Split I into two 32-byte sequences, I<sub>L</sub> and I<sub>R</sub>.
 * Use parse<sub>256</sub>(I<sub>L</sub>) as master secret key, and I<sub>R</sub> as master chain code.
-In case I<sub>L</sub> is 0 or ≥n, the master key is invalid.
+In case parse<sub>256</sub>(I<sub>L</sub>) is 0 or parse<sub>256</sub>(I<sub>L</sub>) ≥ n, the master key is invalid.
 
 <img src=bip-0032/derivation.png></img>
 
@@ -201,7 +201,7 @@ In addition to the expectations from the EC public-key cryptography itself:
 the intended security properties of this standard are:
 * Given a child extended private key (k<sub>i</sub>,c<sub>i</sub>) and the integer i, an attacker cannot find the parent private key k<sub>par</sub> more efficiently than a 2<sup>256</sup> brute force of HMAC-SHA512.
 * Given any number (2 ≤ N ≤ 2<sup>32</sup>-1) of (index, extended private key) tuples (i<sub>j</sub>,(k<sub>i<sub>j</sub></sub>,c<sub>i<sub>j</sub></sub>)), with distinct i<sub>j</sub>'s, determining whether they are derived from a common parent extended private key (i.e., whether there exists a (k<sub>par</sub>,c<sub>par</sub>) such that for each j in (0..N-1) CKDpriv((k<sub>par</sub>,c<sub>par</sub>),i<sub>j</sub>)=(k<sub>i<sub>j</sub></sub>,c<sub>i<sub>j</sub></sub>)), cannot be done more efficiently than a 2<sup>256</sup> brute force of HMAC-SHA512.
-Note however that the following properties does not exist:
+Note however that the following properties do not exist:
 * Given a parent extended public key (K<sub>par</sub>,c<sub>par</sub>) and a child public key (K<sub>i</sub>), it is hard to find i.
 * Given a parent extended public key (K<sub>par</sub>,c<sub>par</sub>) and a non-hardened child private key (k<sub>i</sub>), it is hard to find k<sub>par</sub>.