diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 03ff4bc425b6c83e4014f017f2bf96b70a3ed6c5..03a662615405639d8065959f91a14e26e8c2c9a9 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,21 +4,15 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# 1 On the computation of $\\pi$"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.1 Asking the maths library"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "My computer tells me that $\\pi$ is <i>approximatively</i>"
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -43,14 +37,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.2 Buffon's needle"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the <b>approximation</b>"
+    "## Buffon's needle\n",
+    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
@@ -82,14 +70,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.3 Using a surface fraction argument"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "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 $P[X^2 + Y^2 ≤ 1] = π/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
+    "## Using a surface fraction argument\n",
+    "A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {