I have some doubt about the following theorem from Apostol's Mathematical Analysis [2nd Edition] that I found when I was trying to redo it from my understanding of it. I have taken the following statement of the proof from another question: Apostol Analysis Book Theorem 1.6.
Proof
First assume that 𝑎≥0,𝑏≥0 and use induction on 𝑛=𝑎+𝑏.
If 𝑛=0 then 𝑎=𝑏=0, and we can take 𝑑=0 with 𝑥=𝑦=0.
Assume then that the theorem has been proved for 0,1,2,3,4,…,𝑛−1.
By symmetry we can assume 𝑎≥𝑏.
If 𝑏=0 take 𝑑=𝑎,𝑥=1,𝑦=0.
If 𝑏≥1 we can apply the induction hypothesis to 𝑎−𝑏 and 𝑏,since their sum is 𝑎=𝑛−𝑏≤𝑛−1.
Hence there is a common divisor 𝑑 of 𝑎−𝑏 and 𝑏 of the form𝑑=(𝑎−𝑏)𝑥+𝑏𝑦.
This 𝑑 also divides (𝑎−𝑏)+𝑏=𝑎, so 𝑑 is a common divisor of 𝑎and 𝑏 and we have 𝑑=𝑎𝑥+(𝑦−𝑥)𝑏, a linear combination of 𝑎and 𝑏.
To complete the proof we need to show that every common divisordivides 𝑑. Since a common divisor divides 𝑎 and 𝑏, it alsodivides the linear combination 𝑎𝑥+(𝑦−𝑥)𝑏=𝑑.
This completes the proof if 𝑎≥0 and 𝑏≥0.
In the first step of the induction the divisor d is taken to be 0; though can 0 even be a divisor? I understand a divisor as a number that divides another and since division by 0 is generally undefined I am questioning its validity as a divisor. Is the induction from n=0 then unsound? And would an induction from n=1 resolve this, disregarding the case of n=0?