diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index 0a356293e44f36c544007c7664c2983523fb93b6..6a788fe1a17b36974017dad410da13c21267aaea 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -5,12 +5,12 @@ date: "25 juin 2018"
 output: html_document
 ---
 ## Asking the maths library
-My computer tells me that π is approximatively
+My computer tells me that $\pi$ is *approximatively*
 ```{r}
 pi
 ```
-## Buffon’s needle
-Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation
+## Buffon's needle
+Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
 ```{r}
 set.seed(42)
 N = 100000
@@ -19,7 +19,7 @@ theta = pi/2*runif(N)
 2/(mean(x+sin(theta)>1))
 ```
 ## Using a surface fraction argument
-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[X2+Y2≤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 $X\sim U(0,1)$ 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:
 ```{r}
 set.seed(42)
 N = 1000
@@ -28,7 +28,7 @@ df$Accept = (df$X**2 + df$Y**2 <=1)
 library(ggplot2)
 ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
 ```
-It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, X2+Y2 is smaller than 1 :
+It is then straightforward to obtain a (not really good) approximation to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1 :
 ```{r}
 4*mean(df$Accept)
 ```