We attempt a naïve multiplicative inverse which works most of the time! From our initial development of Clifford multiplication we copy the relationship of a Cn number being the product of two other Cn numbers. Each of these three numbers is expressed in terms of two Cn−1 numbers.
(a + bγn)(c+ dγn) = (ac − bd*) + (ad + bc*n
We take a and b as given and attempt to solve for c and d taking the product to be 1. There must be componentwise equality which yields:
ac − bd* = 1
ad + bc* = 0
Proceeding as a first year algebra student eliminating unknowns but appealing to * as automorphism, using (a*)* = a, and being careful not to commute, we get:
bc* + ad = 0
a*c* − b*d = 1
a−1bc* + d = 0
a*c* + b*a−1bc* = 1
(a* + b*a−1b)c* = 1
(a + ba−1*b*)c = 1
c = (a + ba−1*b*)−1 = (a + b(a−1b)*)−1
d = −(a−1b)c*

Now these equations will invert most Clifford numbers but will fail for γn (a = 0, b = 1), (and certain other elements of V) which has a fine inverse, namely −γn. 0 has no inverse, no matter what rank we are at. There being no automorphism m such that x m(x) is real (as in division algebras), we resort to contingently eliminating c first which requires inverting b instead of a. If neither a nor b is invertible we report “no inverse” to our caller. (Presumably there are Clifford numbers beside 0 that have no inverse and these would cause the 2nd inverse operation to fail. I have found none. (PS: See some zero divisors.))
b−1bc* + b−1ad = 0
c* + b−1ad = 0
−a*b−1ad − b*d = 1
(a*b−1a + b*)d = −1
d = −(a*b−1a + b*)−1
c = − (b−1ad)*

Note that there is another inverse from the lower algebra that we must have. If that fails we have not solved the equations. Note, however, that when we do get a solution it is unique for by induction on the rank of the algebra, our construction is unique; We have assumed a solution and deduced what it must be.

We read off these last two equations to define the inductive clifford inverse:

(define (G f)
(apply (lambda (/ tr bar alpha sg zer zer? one + - * rls basis) (list
  (lambda (x) (let* ((a (car x))(b (cdr x))(ai (/ a)))
    (if ai (let* ((aib (* ai b))(c (/ (+ a (* b (alpha aib)))))
        (d (- (* aib (alpha c))))) (cons c d))
    (let ((bi (/ b)))
       (if bi (let* ((bia (* bi a))(d (- (/ (+ (* (alpha a) bia) (alpha b)))))
         (c (alpha (- (* bia d))))) (cons c d))
  (lambda (x) (cons (tr (car x)) (bar (cdr x)))) ; tr
  (lambda (x) (cons (bar (car x)) (- (tr (cdr x))))) ; bar
  (lambda (x) (cons (alpha (car x)) (- (alpha (cdr x))))) ; alpha
  (lambda () (cons (sg) (sg))) ; sg
  (cons zer zer) ; zer
  (lambda (a) (and (zer? (car a)) (zer? (cdr a)))) ; zer?
  (cons one zer) ; one
  (lambda (a b) (cons (+ (car a)(car b))(+ (cdr a)(cdr b)))) ; +
  (lambda (a) (cons (-(car a))(-(cdr a)))) ; - (negation)
  (lambda (a b) (cons (+ (* (car a)(car b))(-(* (cdr a)(alpha (cdr b))))) ; *
                     (+ (* (car a)(cdr b))(* (cdr a)(alpha (car b))))))
  (lambda (x)(cons (rls x) zer)) ; rls
  (cons (cons zer one) (map (lambda (x) (cons x zer)) basis)) ; basis
  )) f))
; Use definitions for Do and grc4 for a pseudo random number generator.
(define rr (let ((ig (grc4 "vjoe"))) (lambda ()(/ (ig 1)(+ 1 (ig 1))))))
; As in the division algebras we boot the process with the reals:
(define reals (let ((i (lambda (x) x))) (list 
 (lambda (x) (if (zero? x) #f (/ x)))
 i i i rr 0 zero? 1 + - * i '())))
; Now we build the 4th order Clifford algebra C4 thus:
(define P (G (G (G (G reals)))))
; and unpack the tools from the result p:
(define C/ (car P)) ; multiplicative inverse
(define tr (cadr P)) ; transpose
(define bar (caddr P)) ; conjugate (bar)
(define Ha (cadddr P)) ; main involution
(define p (cddddr P)) ; rest
(define Cr (car p)) ; sample generator
(define C0 (cadr p)) ; 0
(define C0? (caddr p)) ; 0 predicate
(define q (cdddr p)) ; rest of tools
(define C1 (car q)) ; multiplicative identity
(define C+ (cadr q)) ; addition
(define C- (lambda (x y) (C+ x ((caddr q) y)))) ; subtraction
(define C* (cadddr q)) ; multiplication
(define r (cddddr q)) ; rest of tools
(define Hrls (car r)) ; returns Clifford number corresponding to real.
(define basis (cadr r)) ; list of basis vectors of V as Clifford numbers.
; Define.
(define (even a)(C* (Hrls 1/2)(C+ a (Ha a)))) ; even part of Clifford number
(define (odd a)(C* (Hrls 1/2)(C- a (Ha a)))) ; odd part of Clifford number
(define (rv) (let rv ((x basis)) (if (null? x) C0
   (C+ (C* (Hrls (rr)) (car x)) (rv (cdr x)))))) ; a random vector generator

(define g3 (car basis)) ; individual Clifford number basis elements for V in C.
(define g2 (cadr basis))
(define g1 (caddr basis))
(define g0 (cadddr basis))

(let ((a (Cr)))(list (C* a (C/ a)) a))
(let ((a (rv))) (equal? (Ha a)(C- C0 a))) ; => #t

(let ((a (Cr))(b (Cr))) (list
(C0? (C- (C* (tr a)(tr b))(tr (C* b a))))
(C0? (C- (C+ (tr a)(tr b))(tr (C+ a b))))
(C0? (C- (C* (Ha a)(Ha b))(Ha (C* a b))))
(C0? (C- (C+ (Ha a)(Ha b))(Ha (C+ a b))))
(C0? (C- (C* (bar a)(bar b))(bar (C* b a))))
(C0? (C- (C+ (bar a)(bar b))(bar (C+ a b)))))) ; => (#t #t #t #t #t #t)
The expression above corroborates that Ha is an automorphism and that tr and bar are antiautomorphisms (note the reversed (C* b a)).