diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 5c4d0866e0cadfbeaf24627cb9f99cc1735c68ba..e20592e396fbc274178711967ffc6ca1df48308f 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,11 +4,11 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# 1 On the computation of π\n",
+    "# On the computation of $\\pi$\n",
     "\n",
-    "## 1.1 Asking the maths library\n",
+    "## Asking the maths library\n",
     "\n",
-    "My computer tells me that π is approximatively"
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -33,7 +33,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.2 Buffon’s needle\n",
+    "## Buffon’s needle\n",
     "\n",
     "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
    ]
@@ -67,10 +67,9 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.3 Using a surface fraction argument\n",
+    "## Using a surface fraction argument\n",
     "\n",
-    "A method that is easier to understand and does not make use of the sin function is based on the\n",
-    "fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
+    "A method that is easier to understand and does not make use of the sin function is based on the fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
    ]
   },
   {
@@ -112,8 +111,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to *π* by counting how\n",
-    "many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
+    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
     "is smaller than 1:"
    ]
   },