"Greater Than" Operator in Racket with 3 Arguments

Question

Looking at > in Racket, the following makes sense:

> (> 5 0)
#t

Why does the following evaluate to false?

> (> 5 0 0)
#f

Answer 1

Because (> 5 0 0) is the same as:

(and (> 5 0) (> 0 0))

...Which is #f. For comparison, evaluate (>= 5 0 0) and you'll see that it returns #t.

Answer 2

Pronounce > as strictly decreasing and >= as non-increasing. Then it becomes easy to see that (> 5 0 0) is false.