diff --git a/src/Advanced/Fast Inverse Root/parts/02 float.typ b/src/Advanced/Fast Inverse Root/parts/02 float.typ
index db056ff..09708b7 100644
--- a/src/Advanced/Fast Inverse Root/parts/02 float.typ	
+++ b/src/Advanced/Fast Inverse Root/parts/02 float.typ	
@@ -28,7 +28,7 @@ Rewrite the following binary decimals in base 10: \
 #v(1fr)
 #pagebreak()
 
-#definition()
+#definition(label: "floatbits")
 Another way we can interpret a bit string is as a _signed floating-point decimal_, or a `float` for short. \
 Floats represent a subset of the real numbers, and are interpreted as follows: \
 #note([The following only applies to floats that consist of 32 bits. We won't encounter any others today.])
@@ -118,13 +118,17 @@ Floats represent a subset of the real numbers, and are interpreted as follows: \
 
 #problem(label: "floata")
 Consider `0b01000001_10101000_00000000_00000000`. \
+#hint([The underscores here do _not_ match those in @floatbits])
+
+#v(2mm)
+
 Find the $s$, $E$, and $F$ we get if we interpret this bit string as a `float`. \
 #note([Leave $F$ as a sum of powers of two.])
 
 #solution([
   $s = 0$ \
-  $E = 258$ \
-  $F = 2^31+2^19 = 2,621,440$
+  $E = 131$ \
+  $F = 2^21+2^19$
 ])
 
 #v(1fr)
@@ -160,9 +164,9 @@ What value do we get if we interpret this bit string as a float? \
 #solution([
   This is 21:
   $
-    2^(131) times (1 + (2^(21) + 2^(19)) / (2^(23)))
-    = 2^(4) times (1 + 0.25 + 0.0625)
-    = 16 times (1.3125)
+    2^4 times (1 + (2^(21) + 2^(19)) / (2^(23)))
+    = 2^(4) times (1 + 2^(-2) + 2^(-4))
+    = 16 + 4 + 1
     = 21
   $
 ])