More improvements to knowledge base

1 files changed, 3 insertions, 3 deletions

diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md
index c954272..fdc90be 100644
--- a/source/know/concept/fredholm-alternative/index.md
+++ b/source/know/concept/fredholm-alternative/index.md
@@ -14,7 +14,7 @@ It is an *alternative* because it gives two mutually exclusive options,
 given here in [Dirac notation](/know/concept/dirac-notation/):
 
 1.  $$\hat{L} \Ket{u} = \Ket{f}$$ has a unique solution $$\Ket{u}$$ for every $$\Ket{f}$$.
-2.  $$\hat{L}^\dagger \Ket{w} = 0$$ has non-zero solutions.
+2.  $$\hat{L}^\dagger \Ket{w} = 0$$ has nonzero solutions.
     Then regarding $$\hat{L} \Ket{u} = \Ket{f}$$:
     1.  If $$\Inprod{w}{f} = 0$$ for all $$\Ket{w}$$, then it has infinitely many solutions $$\Ket{u}$$.
     2.  If $$\Inprod{w}{f} \neq 0$$ for any $$\Ket{w}$$, then it has no solutions $$\Ket{u}$$.
@@ -31,7 +31,7 @@ this theorem can alternatively be stated as follows using the determinant:
 1.  If $$\mathrm{det}(\hat{L}) \neq 0$$, then $$\hat{L} \vec{u} = \vec{f}$$
     has a unique solution $$\vec{u}$$ for every $$\vec{f}$$.
 2.  If $$\mathrm{det}(\hat{L}) = 0$$,
-    then $$\hat{L}^\dagger \vec{w} = \vec{0}$$ has non-zero solutions.
+    then $$\hat{L}^\dagger \vec{w} = \vec{0}$$ has nonzero solutions.
     Then regarding $$\hat{L} \vec{u} = \vec{f}$$:
     1.  If $$\vec{w} \cdot \vec{f} = 0$$ for all $$\vec{w}$$, then it has
         infinitely many solutions $$\vec{u}$$.
@@ -48,7 +48,7 @@ Then for the equation $$\hat{M} \Ket{u} = \Ket{f}$$, we can say that:
 1.  If $$\lambda$$ is *not* an eigenvalue,
     then there is a unique solution $$\Ket{u}$$ for each $$\Ket{f}$$.
 2.  If $$\lambda$$ is an eigenvalue, then $$\hat{M}^\dagger \Ket{w} = 0$$
-    has non-zero solutions. Then:
+    has nonzero solutions. Then:
     1.  If $$\Inprod{w}{f} = 0$$ for all $$\Ket{w}$$, then there are
         infinitely many solutions $$\Ket{u}$$.
     2.  If $$\Inprod{w}{f} \neq 0$$ for any $$\Ket{w}$$, then there are no