Proving the Last Gauss Lemma for Real Numbers Using Induction

[Ben2]

Modified from Disquisitiones Arithmeticae, p. 10: Let "*" indicate multiplication and "^" indicate "to the power." For real numbers a, b, c,..., let [a] = a, [a,b] = b*a+1, [a,b,c] = c*[a,b]+[a], [a,b,c,d] = d*[a,b,c]+[a,b], etc. Prove that [a,b,...,l,m]*[b,c,...,l] - [a,b,...,l]*[b,c,...,m] = (-1)^n, where n is the number of elements in the set {a,b,...,l,m}.
Thanks to the moderators for their help, and I won't post this again.

 

[fresh_42]

Modified from Disquisitiones Arithmeticae, p. 10: Let "*" indicate multiplication and "^" indicate "to the power." For real numbers a, b, c,..., let [a] = a, [a,b] = b*a+1, [a,b,c] = c*[a,b]+[a], [a,b,c,d] = d*[a,b,c]+[a,b], etc. Prove that [a,b,...,l,m]*[b,c,...,l] - [a,b,...,l]*[b,c,...,m] = (-1)^n, where n is the number of elements in the set {a,b,...,l,m}.
Thanks to the moderators for their help, and I won't post this again.

In modern notation we have the following situation:
Given two numbers $a,b$ we define:
$$
a_{-1}=0\, , \,a_0=1\, , \,a_1=a\, , \,a_n=b_n\cdot a_{n-1}+a_{n-2} \text{ and } \\
b_1 = a\, , \,b_2=b\, , \,b_n=\dfrac{a_n-a_{n-2}}{a_{n-1}} \\
c_{-2}=1\, , \,c_{-1}=0\, , \,c_0=1\, , \,c_1=b\, , \,c_n=b_{n+1}\cdot c_{n-1}+c_{n-2}
$$
and the statement then reads
$$
a_n\cdot c_{n-2}-a_{n-1}\cdot c_{n-1} = (-1)^n\quad (n \geq 0)
$$
and the rest is induction where the actual value of $b_n$ isn't relevant anymore.