Question: I'm working through Clojure Koans and I'm up to the recursion koans.

I don't understand how to solve `is-even?` using recursion. The exercise partially defines this function as:

```
(defn is-even? [n]
    (if (= n 0)
        true
        (__ (is-even? (dec n)))))

```

If I don't want to use recursion then I would define it as `(defn is-even? [n] (= (mod n 2) 0))` but that goes against the point of the exercise.

Answer: 
A number `n` is even if either:

1. `n` is 0
2. `n-1` is NOT even

So really, `not` should be enough to fill in that blank. Eventually you wind up with N `not`s around `(= 0 0)`, and most of them cancel out.
Consider decrementing by 2 for each recursion. The rest should be obvious: The number is even only if the function ends up with zero.

EDIT: Apparently I missed the memo about filling in the blank. Here is the tail call optimizable solution I had in mind for positive integers:

```
(defn is-even? [n]
    (if (<= n 1)
        (= n 0)
        (is-even? (- n 2))))

```