Add README in english

src/algorithms/math/power-of-two-greater-than-number/README.md

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+# Power of two greater than a given Number
+
+We want to calculate the power of two which is greater than a given
+number
+
+**Naive solution**
+
+In this case, we need to use a loop.
+
+Let's set `number, power` which are two variables that represente respectively
+the number given to the function and the power of two obtained after each iteration.
+The variable `power` is initialize to 1 before the loop.
+As condition of our loop, we must check out if power is less than number.
+If it verified, we continue to run the loop and `power <- power*2` else we break it 
+and return the value of the variable power.
+
+For exemple: consider number = 5
+
+```
+1. number = 5 and power = 1	(power < number) we continue
+2. number = 5 and power = 2 (power < number) we continue
+3. number = 5 and power = 4 (power < number) we continue
+4. number = 5 and power = 8 (power > number) we do not continue and we return 8
+```
+This method resolve the probleme but the complexity is great.
+
+**Optimal solution**
+
+The best solution for this problem is to use the logarithme function in basis 2.
+
+Note: log2(number) = ln(number) / ln(2)
+
+This expression return a real number and, if we consider the entire part of this
+and adding 1 of it we obtain a number. Assuming that this number is n, then if
+we calculate 2 power n we obtain the power of two which is directly greater than
+number.
+
+For exemple: consider number = 5
+
+```
+	n = E(log2(number = 5)) + 1
+	n = E(ln(5) / ln(2)) + 1
+	n = E(2.3219...) + 1
+	n = 2 + 1
+	n = 3
+
+	then 2 power n=3 equals 8
+```
+
+#reference
+"Logarithme binaire - Wikipédia"  https://fr.m.wikipedia.org/wiki/Logarithme_binaire
+
+