diff --git a/notebooks/problem0072.ipynb b/notebooks/problem0072.ipynb
index a57c95a..6836f44 100644
--- a/notebooks/problem0072.ipynb
+++ b/notebooks/problem0072.ipynb
@@ -50,7 +50,7 @@
    "id": "8e268820",
    "metadata": {},
    "source": [
-    "If you try to implement the totient function yourself, you might find it difficult to make this approach fast enough. An alternative is to sieve the values of totient - see [problem 70](https://projecteuler.net/problem=70). However, there's a few more *very* interesting methods to compute totient sums.\n",
+    "If you try to implement the totient function yourself, you might find it difficult to make this approach fast enough. An alternative is to \"sieve\" the values of totient - see [problem 70](https://projecteuler.net/problem=70). However, there's a few more *very* interesting methods to compute totient sums.\n",
     "\n",
     "## Recursive totient sum\n",
     "\n",
@@ -159,7 +159,7 @@
     "\n",
     "$$M(n) = 1 + \\lfloor b\\rfloor M(\\lfloor a\\rfloor) - \\sum_{x=1}^a  \\mu(x)\\left\\lfloor \\frac{n}{x}\\right\\rfloor - \\sum_{y=2}^b M\\left(\\left\\lfloor \\frac{n}{y}\\right\\rfloor\\right)$$\n",
     "\n",
-    "Now we have a recursive implementation of $M(n)$. All that's left is to calculate the values of $\\mu(n)$ that we need. We can do this with a sieve, since $\\mu$ is a [multiplicative function](https://en.wikipedia.org/wiki/Multiplicative_function)."
+    "Now we have a recursive implementation of $M(n)$. All that's left is to calculate the values of $\\mu(n)$ that we need. By taking advantage of $\\mu$ being [multiplicative](https://en.wikipedia.org/wiki/Multiplicative_function), we can compute values using a similar strategy to the sieve of Eratosthenes - see [problem 10](https://projecteuler.net/problem=10)."
    ]
   },
   {
@@ -169,9 +169,7 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "from math import isqrt\n",
-    "\n",
-    "def mobius_sieve(limit):\n",
+    "def mobius_range(limit):\n",
     "    ms = [n for n in range(0, limit)]\n",
     "\n",
     "    for n in range(0, limit):\n",
@@ -193,7 +191,7 @@
    "id": "f12a8b30",
    "metadata": {},
    "source": [
-    "We'll use $a = \\sqrt{1000000} = 1000$ as our upper bound on the sieve."
+    "We'll use $a = \\sqrt{1000000} = 1000$ as our upper bound."
    ]
   },
   {
@@ -203,7 +201,8 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "mu = list(mobius_sieve(isqrt(1000000) + 1))"
+    "from math import isqrt\n",
+    "mu = list(mobius_range(isqrt(1000000) + 1))"
    ]
   },
   {