clojure: If I am creating an operator staircase, how do I define the identity for the operator in general? -

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I was wondering how you made * out of + In the end, then going to the other direction, using the following function (fb) b, and + is out of + By applying the same pattern, get a staircase of hyper-hyper-quality work of more hyperense, so I decided to try an infinite list of increasing hyperparators and try to code. Or. I came with this, which is very close! : 

  (operator-stair [defer operator-ladder [] (repeat (fn [f] (fn [ab] (loop [bottom b result a] (if (> = 0 Below) Results (recurring) (below the DEC) (FA results)))) (FN [AB] (APB))) (DEF OPS (operator-ladder)) ((NHPS) 3 5); ; - & gt; 6 (inc is an arity one so it should ignore one args?) ((Nth ops 1) 3 5) ;; - & gt; 8 (right knee) ((nth ops 2) 3 5) ;; - & gt; 18 (Oops, one stop! Otherwise correctly multiply.) Is basically applicable (fn [a b] (* a b)) ((nth ops 3) 3 5) ;; ---- & gt; 10 9 2 (wow)   
 The only thing I do not know is how to define the initial  results  in a normal way! I've made it  a  because it works, but for this, for example, this should be 0, while the product should be 1.  
 How do I start the initial value in the loop above the  result  so that it works in all cases in normal cases?  
 Thanks in advance!   
 
 
 I think if you want to start with  inc  You can define only oriented operator. So the 0th step is "one plus"; The first step is "N times: one plus", which is basically  # (+ %%) ; The second step is "N Bar: Add N", which is  # (* %%) ; The third step is "N times: multiplied by", which is  # (pow %%)  ... and so on.  
 If you want to define binary operators, I think that you just start from  +  to  inc , and the rest are in the same way Receive what you did as you did.

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