solutions to 2.6. Truely mind bending!

diff --git a/src/sicp/ex2_6.clj b/src/sicp/ex2_6.clj
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+(ns sicp.ex2_6
+  (:use [sicp utils]
+       [clojure.test]))
+
+;; from the problem
+(def zero
+  (fn [f] (fn [x] 
+           x)))
+
+(defn add-1 [n]
+  (fn [f] (fn [x]
+           (f ((n f) x)))))
+
+(comment
+(((add-1 zero) inc) 0)
+;; => 1
+(((add-1 zero) inc) 1)
+;; => 2
+)
+;; the above examples should be read as follows:
+;; (add-1 zero) returns the function which represents 1.
+;; i.e. You apply inc once to 0 and that yields 1.
+;; in the second example, we apply inc once to 1, and
+;; that yields 2. This is just to convert Churl numeral
+;; definition to roman numeral definition.
+
+;; as said in the hint, using substitution to evaluate (add-1 zero)
+;; tells us that (add-1 zero) == one has one composition of (f x)
+;; and (add-1 (add-1 zero)) == 1 is (f (f x)), so here is their formal
+;; definition.
+(def one (fn [f] (fn [x] (f x))))
+
+(def two (fn [f] (fn [x] (f (f x)))))
+
+(comment
+(((add-1 one) inc) 1)
+;;=> 3
+(((add-1 one) inc) 2)
+;;=> 4
+(((add-1 two) inc) 2)
+;;=> 5
+)
+;; (add-1 one) => two. Apply inc twice on 1 => 3
+;; similarly the other examples.
+
+;; Definition of an add operator
+(defn add [m n]
+  (fn [f] (fn [x]
+           ((m f) ((n f) x)))))
+
+(comment
+(((add one two) inc) 1)
+;;=> 4  
+)
+;; continueing the same logic, (add one two) => three
+;; and inc applied three times to 1 is 4. This proves
+;; that our definition of add is correct.
+
+;; I used the wikipedia article to study Church Numerals:
+;; <http://en.wikipedia.org/wiki/Church_encoding#Computation_with_Church_numerals>
+
+;; comments?
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