diff 56bc728..6fbd3ac pi_irrationality_exponent_v2.txt
diff --git a/pi_irrationality_exponent_v2.txt b/pi_irrationality_exponent_v2.txt
index 763c91e..383a60b 100644
--- a/pi_irrationality_exponent_v2.txt
+++ b/pi_irrationality_exponent_v2.txt
@@ -16,7 +16,7 @@ Always 0 <= ||x|| <= 1/2. For example ||2.3|| = 0.3, ||2.7|| = 0.3,
Two facts we use without proof, both immediate from the definition:
- ||x + y|| <= ||x|| + ||y|| (triangle inequality)
+ ||x + y|| <= ||x|| + ||y|| (triangle ineuality)
if ||x|| < 1/N then some integer m has |x - m| < 1/N.
@@ -129,7 +129,7 @@ Step 2, compare 1/N with q^(1-nu). Since q <= N and 1 - nu > -1,
1/N <= N^(1-nu) <= q^(1-nu).
-The middle inequality is N^(nu-2) <= 1, which holds exactly because
+The middle ineuality is N^(nu-2) <= 1, which holds exactly because
nu <= 2. So
||q x|| <= 1/N <= q^(1-nu).
@@ -142,7 +142,7 @@ N, forcing ||q x|| = 0, that is, q x is an integer and x is
rational. That contradicts the hypothesis. So the q_N are unbounded,
in particular there are infinitely many of them, and each satisfies
||q x|| <= 1/N <= q^(1-nu). By (1.1) each gives a fraction with
-0 < |x - p/q| <= q^(-nu). The strict inequality follows by using
+0 < |x - p/q| <= q^(-nu). The strict ineuality follows by using
nu' = (nu+2)/2 in place of nu and dropping the one possible q = 1.
Done. Two remarks.
@@ -194,7 +194,7 @@ q^2 (2 sqrt 2 + 1). Therefore
for every fraction p/q with |sqrt 2 - p/q| < 1.
Now let nu > 2. For q with q^(nu-2) >= 2 sqrt 2 + 1, that is for
-q >= Q(nu) := (2 sqrt 2 + 1)^(1/(nu-2)), inequality (3.1) gives
+q >= Q(nu) := (2 sqrt 2 + 1)^(1/(nu-2)), ineuality (3.1) gives
|sqrt 2 - p/q| >= q^(-2) / (2 sqrt 2 + 1) >= q^(-nu).
@@ -604,7 +604,7 @@ no information at all if nu0 <= 1.
converges, angles in radians.
Proof. Choose nu0 with mu(pi) < nu0 < 5/2, which is possible because
-the inequality is strict. By the definition of mu in section 1.3 and
+the ineuality is strict. By the definition of mu in section 1.3 and
the remark opening the proof of Lemma 7.2, there is a constant c > 0
with