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I am reading the proof of the following theorem from Apostol's Introduction to Analytic Number Theory.

Theorem 1.2 Given any two integers a and b, there is a common divisor $d$ of a and b of the form $d=ax+by$, where $x$ and $y$ are integers. Moreover, every common divisor of $a$ and $b$ divides this $d$.

I'm stuck on the first step. The author used induction on $n$, where $n=a+b$ and assumed that the theorem has been proved for $0,1,2, \cdots, n-1$.

I think the induction on $n$ should be an independent process where $n$ should be independent on other variables such as $a$ and $b$ here. For example, when we prove that binomial theorem:

$$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$

We would use the induction on $n$, and assume this equality holds for $0,1,2, \cdots, n-1$. However, $a,b$ are all constants, not relevant to $n$.

What's wrong with my thought?

For reference, the following is the whole proof from the book:

enter image description here

Here is a possible relevant: link.

Dr. Mathva
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Hamilton
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1 Answers1

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We are being asked to prove a statement of the form

"$P(a, b)$ is true for all $(a, b)$ in $\{(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), \dots\}$"

To prove this statement, we first chunk this set into smaller groupings based on the height $a + b$: $$\{(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), \dots\} = \{(0, 0)\} \cup \{(1, 0), (0, 1) \} \cup \{(2, 0), (1, 1), (0, 2) \} \cup \cdots.$$

Then, we argue that:

  • If $(a, b)$ is in the 0th chunk $\{(0, 0)\}$, then the proposition is true.
  • If $P(x)$ is true for all $(a, b)$ in the first $n$ chunks, then $P(x)$ is true for all $(a, b)$ in the $(n + 1)$th chunk.

By inducting on $n$, the number of chunks being considered, we can demonstrate that the proposition is true for $(a, b)$ coming from any chunk.

Does this address your questions?

Sam Freedman
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