diff --git a/05. Bit manipulation/5.1. Insertion.md b/05. Bit manipulation/5.1. Insertion.md
index d7a3ae5..6800312 100644
--- a/05. Bit manipulation/5.1. Insertion.md	
+++ b/05. Bit manipulation/5.1. Insertion.md	
@@ -67,7 +67,7 @@ int updateBits(int n, int m, int i, int j) {
     int allOnes = ~0;
 
     // 11100000
-    int left = j < 31 ? (allOnes << (j+1) : 0;
+    int left = j < 31 ? (allOnes << (j+1)) : 0;
     // 00000011
     int right = ((1 << i) - 1);
 
diff --git a/05. Bit manipulation/5.2. Binary to string.md b/05. Bit manipulation/5.2. Binary to string.md
new file mode 100644
index 0000000..dab81e9
--- /dev/null
+++ b/05. Bit manipulation/5.2. Binary to string.md	
@@ -0,0 +1,107 @@
+# 5.2. Binary to string
+
+> Given a real number between 0 and 1 (0.72), passed in as a double, print the binary representation. If the number can't be represented with less than 32 characters, print "ERROR"
+
+Hints 143, 167, 173, 269, 297
+
+How am I supposed to represent a real number in a binary form?
+
+## Hint 1
+
+> First do it with integers
+
+```c++
+str binaryToString(int num) {
+    str stringnum = "";
+    int biggestexp = 0;
+    while (num >= pow(2, biggestexp)) {
+        biggestexp += 1;
+    }
+    while (biggestexp >= 0) {
+        if (num >= 0 && num - pow(2, biggestexp) >= 0) {
+            num -= pow(2, biggestexp);
+            stringnum = "1" + stringnum;
+        } else {
+            stringnum = "0" + stringnum;
+        }
+        biggestexp -= 1;
+    }
+    return stringnum;
+}
+```
+
+This should work, takes O(sqrt(num) + sqrt(num)). We need to know the biggest exponent in order to substract it. And then keep substracting it until reaching 0. Maybe some wrong c++ syntax, but should work. Now with decimals??
+
+## Hint 2
+
+> In a number like .893 (in base 10), what does each digit signify? What then does each digit in .10010 signify in base 2?
+
+So just `stringnum = "." + stringnum`?
+
+## Hint 3
+
+> A number such as .893 (in base 10) indicates 8 \* 10^-1 + 9 \* 10^-2 + 3 \* 10^-3. Translate this system into base 2
+
+.10010 = 1 \* 2^-1 + 0 \* 2^-2 + 0 \* 2^-3 + 1 \* 2^-4 + 0 \* 2^-5
+
+Ok so stringnum in the opposite way, '.' added to it. Can we reverse dividing by pow(2,i) and go from low to high?
+
+```c++
+str binaryToString(int num) {
+    str stringnum = "";
+    int i = 0;
+    num -= pow(2, i);
+    while (num > 0) {
+        stringnum = "1" ? num % 2 == 0 : "0" + stringnum;
+        i += 1;
+        num -= pow(2, i);
+    }
+    return "." + stringnum;
+}
+```
+
+This won't work for 11 -> 1011, it will create 111 and then have 6 left and nowhere to put it. So just do the other way and take longer.
+
+## Hint 4
+
+> How would you get the first digit in .893? If you multiplied by 10, you'd shift the values over to get 8.93. What happens if you multiply by 2?
+
+## Hint 5
+
+> Think about what happens for values that can't be represented accurately in binary
+
+I'm officially lost.
+
+## Solution
+
+To print the decimal part of the 0.101 number, we multiply by 2 and check if 2n is >= 1. r = 2<sub>10</sub> * n = 1.01<sub>2</sub>. If r >= 1, we know that n had a 1 right after the decimal point. By doing this continuously, we can check every digit.
+
+```java
+public static String printBinary(double num) {
+    if (num >= 1 || num <= 0) {
+        return "ERROR";
+    }
+
+    StringBuilder binary = new StringBuilder();
+    binary.append(".");
+    while (num > 0) {
+        /* Setting a limit on length: 32 characters */
+        if (binary.length() > 32) {
+            return "ERROR";
+        }
+        double r = num * 2;
+        if (r >= 1) {
+            binary.append(1);
+            num = r - 1;
+        } else {
+            binary.append(0);
+            num = r;
+        }
+    }
+    return binary.toString();
+}
+```
+
+Don't understand this logic at all. Trying with `double num = 0.7`, I get ".1011001001....".
+
+[How to convert decimal number to binary](https://stackoverflow.com/a/39947437/4569908).
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