hopefully fix math blocks?

1 files changed, 14 insertions, 8 deletions

diff --git a/fib.html b/fib.html
index 29d5e71..a5121d7 100644
--- a/fib.html
+++ b/fib.html
@@ -126,7 +126,7 @@
 			<hr>
 
 			<p>Okay, so now that I had a way to split the sequences into pieces, I just needed a way to count the number of sequences for each value of <math><mi>t</mi></math>, and sum those together.</p>
-			<p>The sum was the easy part. I knew the final number would be the the sum of all of the <math><msub><mi>m</mi><mi>t</mi></msub></math>s, or:</p>
+			<p>The sum was the easy part. I knew the final number would be the the sum of all of the <math><msub><mi>m</mi><mi>t</mi></msub></math>s, or:
 			<math display="block">
 				<mrow>
 					<munderover>
@@ -144,7 +144,8 @@
 					<msub><mi>m</mi><mi>t</mi></msub>
 				</mrow>
 			</math>
-			<p>Plugging in our definition for <math><msub><mi>t</mi><mi>max</mi></msub></math>,</p>
+			</p>
+			<p>Plugging in our definition for <math><msub><mi>t</mi><mi>max</mi></msub></math>,
 			<math display="block">
 				<mrow>
 					<munderover>
@@ -166,7 +167,7 @@
 					<msub><mi>m</mi><mi>t</mi></msub>
 				</mrow>
 			</math>
-
+			</p>
 			<p>After thinking about the sequences for a while, I realized that the number of sequences boils down to the question "how many ways can you uniquely arrange a list of <math><mi>t</mi></math>
 				special items out of <math><mi>k</mi></math>
 				total items?"</p>
@@ -178,7 +179,7 @@
 				for some <math><mi>t</mi></math>
 				was equivelant to counting the number of ways to choose <math><mi>t</mi></math>
 				elements from a larger set of <math><mi>k</mi></math>
-				elements. By this definition,</p>
+				elements. By this definition,
 			<math display="block">
 				<mrow>
 					<msub><mi>m</mi><mi>t</mi></msub>
@@ -193,7 +194,8 @@
 					<mo>)</mo>
 				</mrow>
 			</math>
-			<p>Plugging in the prior definition for <math><mi>k</mi></math>, and using the summation from above, I arrived at a final equation:</p>
+			</p>
+			<p>Plugging in the prior definition for <math><mi>k</mi></math>, and using the summation from above, I arrived at a final equation:
 			<math display="block">
 				<mi>f</mi><mo>(</mo><mi>n</mi><mo>)</mo>
 				<mo>=</mo>
@@ -229,6 +231,7 @@
 					</mrow>
 				</mrow>
 			</math>
+			</p>
 			<p>Yay!</p>
 			<hr>
 
@@ -239,7 +242,7 @@
 			<p>The formula on ProofWiki is for the <math><mi>n</mi></math>th fibonacci number, so here's how you could arrive at the real Fibonacci Number as Sum of Binomial Coefficients function:</p>
 			<p>Begin with the derived formula, replacing <math><mi>n</mi></math>
 				for <math><mi>a</mi></math>
-				to avoid variable conflicts later.</p>
+				to avoid variable conflicts later.
 			<math display="block">
 				<mi>f</mi><mo>(</mo><mi>a</mi><mo>)</mo>
 				<mo>=</mo>
@@ -263,7 +266,8 @@
 					</mrow>
 				</mrow>
 			</math>
-			<p>Since this is really for the <math><mi>n</mi><mo>+</mo><mn>1</mn></math>th Fibonacci number, substitute <math><mi>a</mi><mo>=</mo><mi>n</mi><mo>-</mo><mn>1</mn></math>.</p>
+			</p>
+			<p>Since this is really for the <math><mi>n</mi><mo>+</mo><mn>1</mn></math>th Fibonacci number, substitute <math><mi>a</mi><mo>=</mo><mi>n</mi><mo>-</mo><mn>1</mn></math>.
 			<math display="block">
 				<mi>f</mi><mo>(</mo><mi>n</mi><mo>)</mo>
 				<mo>=</mo>
@@ -287,8 +291,9 @@
 					</mrow>
 				</mrow>
 			</math>
+			</p>
 			<p>The original function uses <math><mi>k</mi></math>
-				instead of <math><mi>t</mi></math>.</p>
+				instead of <math><mi>t</mi></math>.
 			<math display="block">
 				<mi>f</mi><mo>(</mo><mi>n</mi><mo>)</mo>
 				<mo>=</mo>
@@ -312,6 +317,7 @@
 					</mrow>
 				</mrow>
 			</math>
+			</p>
 			<p>This is the exact same function that appears on ProofWiki!</p>
 		</article>
 	</body>