diff --git a/sicp-2.5-notes b/sicp-2.5-notes
deleted file mode 100644
index 941bc3b..0000000
--- a/sicp-2.5-notes
+++ /dev/null
@@ -1,41 +0,0 @@
-(car (cons x y))
-
-(car (lambda (m) (m x y)))
-
-((lambda (m) (m x y)) (lambda (p q) p)))
-
-(lambda (x y) x)))
-
-x
-
-(define (cdr z) (z (lambda (x y) y)))
-
-
-
-2.5 - Given that we know the bases - 2, and 3, we only need to find the exponents.
-
-For example, 6 can only be represented one way - 2^1 3^1, meaning our pair of numbers a and b are 1
-
-So, cons is simple:
-
-(define (cons x y) (* (expt 2 x) (expt 3 y)))
-
-Now we're left solving for a and b
-
-log(2^a * 3^b) = alog(2) + blog(3)
-
-call our "cons" value x
-
-Because integer factorizations are unique we only need to find the biggest power of two and three that 
-go into the number.
-
-(define (power-of-x value base) 
-  (define (iter current-power)
-    (if (< ((expt base current-power) value)
-	(iter (+ current-power 1))
-	(current-power))))
-
-Then with this we can write car and cdr as
-
-(define (car int) (power-of-x int 2))
-(define (cdr int) (power-of-x int 3))