(defn abs "(abs n) is the absolute value of n" [n]
  (cond
   (not (number? n)) (throw (IllegalArgumentException.
           "abs requires a number"))
   (neg? n) (minus n)
   :else n))

(defn gcd "(gcd a b) returns the greatest common divisor of a and b" [a b]
  (if (or (not (integer? a)) (not (integer? b)))
    (throw (IllegalArgumentException. "gcd requires two integers"))  
    (loop [a (abs a) b (abs b)]
      (if (zero? b) a,
    (recur b (mod a b))))))

(defn lcm
  "(lcm a b) returns the least common multiple of a and b"
  [a b]
  (when (or (not (integer? a)) (not (integer? b)))
    (throw (IllegalArgumentException. "lcm requires two integers")))
  (cond (zero? a) 0
        (zero? b) 0
        :else (abs (mult b (quot a (gcd a b))))))

(defn expt
  [base pow]
  (if (and (not (float? base)) (integer? pow))　
    (cond
         ... ;; 不是浮點數為底且指數次方為整數者，接著處理 ( -> 進入第二層 )
　　　)
    (Math/pow base pow))) ;; 太複雜的就直接 pow 函數來吧!

...
(cond
 (pos? pow) (expt-int base pow) ;; 正的次方 -> 用專門算整數次方的函數計算
     (zero? pow) (cond   
                      ...       ;;  0 的次方 -> 0 的次方基本上是 1 ( -> 進入第三層 )
                 )
     :else (/ 1 (expt-int base (minus pow)))) ;; 負的次方 -> 轉成正的次方處理，並做個倒數
...

;; 零的次方
...
(cond
  (= (type base) BigDecimal) 1M
  (= (type base) java.math.BigInteger) (java.math.BigInteger. "1")
  (when-available clojure.lang.BigInt (= (type base) clojure.lang.BigInt))
  (when-available clojure.lang.BigInt (bigint 1))
  :else 1)
...

(defprotocol MathFunctions
  (floor [n] "(floor n) returns the greatest integer less than or equal to n.
If n is an exact number, floor returns an integer, otherwise a double.")
  (ceil [n] "(ceil n) returns the least integer greater than or equal to n.
If n is an exact number, ceil returns an integer, otherwise a double.")
  (round [n] "(round n) rounds to the nearest integer.
round always returns an integer.  Rounds up for values exactly in between two integers.")
  (integer-length [n] "Length of integer in binary")
  (sqrt [n] "Square root, but returns exact number if possible."))

;; 整數的 extend-type : 整數做 floor, ceil 與 round 根本不變
(extend-type
 Integer MathFunctions
 (floor [n] n)  (ceil [n] n)  (round [n] n) 
 (integer-length [n] (- 32 (Integer/numberOfLeadingZeros n)))
 (sqrt [n] (sqrt-integer n))) ;; 處理整數的方根問題，另代函數處理

;; 雙精度浮點數的 extend-type : 直接套公式
(extend-type
 Double MathFunctions
 (floor [n] (Math/floor n))  (ceil [n] (Math/ceil n)) 
 (round [n] (Math/round n))  (sqrt [n] (Math/sqrt n)))