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If $(a,b)=1$ then $\exists s,t \in \mathbb {Z}$ s.t. $as +bt = 1$.

In the book by Pollard, Diamond titled "The Theory of Algebraic Numbers", there is a proof in the initial pages for proving the above theorem.

Proof (Th. 1.2) takes for the set of numbers given by $ax + by$, the smallest value positive combination with $x = s, y = t$, and the value of that $d = (as +bt)$. In the next step, the proof takes $b = dq' +r', 0 \le r' \lt d$. For constructing this last equality, the proof takes help of the division algorithm in Th. 1.1, that states : $a = bq +r, 0 \le r \lt b$.

The issue is how can it be directly taken to hold that : $b = dq' + r'$ from the division algo. without introducing proof for : $d \le b$. I mean that at that point, it is not shown that $d$ can attain values less than $b$.

jitender
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  • This is a result of Bézout's identity – rtybase Dec 27 '17 at 10:59
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    In the division algorithm these are different $a$'s and $b$' s ! I think you're confusing the notations because they're the same but since it's different proofs/theorems, the notations represent dofferent things – Maxime Ramzi Dec 27 '17 at 11:06
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    If $b<d$, then $b=d\cdot0+b$. There's no need to assume $d\le b$ (provided, of course, that $b\ge0$). – egreg Dec 27 '17 at 23:05
  • According to the division algorithm, if $b$ is any integer and $d$ is any positive integer, then we can find integers $q$ and $r$ such that $b=dq+r$ and $0\le r\lt d.$ – bof Dec 27 '17 at 23:10
  • @egreg I 'earlier' (not after your responses) supposed that author should have given examples with a lattice or by numerical examples to instill this confusing idea, as I know $b$ can be bigger than $d$. What I mean is for a beginner, it is confusing to imagine a linear combination is less than the smaller of the two values (of which it is a linear combination), In fact, I should have framed question better, it is about not instilling details about the value set taken by $d$. – jitender Dec 27 '17 at 23:12
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    @jitender Actually, I find this a very bad proof of Bézout”s identity: it is nonconstructive, for one thing, whereas a constructive proof is much more useful, with the introduction of the extended Euclidean algorithm. I'll add a comment to my answer, though. – egreg Dec 27 '17 at 23:21
  • I hope that an approach that is constructive will automatically take at least one example or symbols (for a general sense), and I feel that this is what you mean too. This is lacking in non-constructive approach, and cannot be filled unless a change of approach to constructive is done. – jitender Dec 27 '17 at 23:38

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The division with remainder $A=Bq+r$, with $0\le r<B$ doesn't need that $A\ge B$. Indeed, if $0\le A<B$, we directly have $$ A=B\cdot 0+A $$ and $A$ has the required property for the remainder by assumption.

So, you don't need to prove that $d\le b$. The division $b=dq'+r'$, with $0\le r'<d$ is always possible provided $d>0$, which it is by construction.

Actually, in the case when both $a$ and $b$ are positive (otherwise the proof is trivial), it is obvious that $d\le b$, because $b$ is a positive linear combination of $a$ and $b$, since $b=a\cdot0+b\cdot 1>0$.

I don't agree with books that present this proof, which is highly nonconstructive. Using the extended Euclidean algorithm, not only the existence of the greatest common divisor is established, but also an efficient method for computing it is provided.

egreg
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  • I hope that an approach that is constructive will automatically take at least one example or symbols (for a general sense), and I feel that this is what you mean too. This is lacking in non-constructive approach, and cannot be filled unless a change of approach to constructive is done. – jitender Dec 27 '17 at 23:50
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    @jitender When I present the Euclidean algorithm, I often use the two numbers $1837100231809$ and $43126750060172161$ (obtained by mixing the prime factors of $2^{32}+1$ and $2^{64}+1$, the first two nonprime Fermat numbers). Computing the gcd with the Euclidean algorithm takes just a few minutes with pencil and paper (or, maybe, a pocket calculator). Since the least prime factor is $641$, factorization would take much longer. – egreg Dec 27 '17 at 23:59
  • So, you mean your example of using first two composite Fermat numbers to show finding Bezout's coefficients using EEA, to illustrate the significance of the constructive approach. – jitender Dec 28 '17 at 00:30