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I am revising some elementary number theory using Nathanson's book. My answer to question $14$ on page $14$ (see image) is $u = 5u' -3v'$, $v=-3u'+2v'$. Nathanson suggests that $u=u'$, $v=v'$.

Did I hit an error ( there are quite a lot, see Teaching Blog, or am I wrong? If I am wrong then please explain Nathanson's answer.

exercises page 14 Nathanson EM in NT

Praveen
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nilo de roock
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1 Answers1

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$$u = pu' +qv'$$ $$v=ru'+sv'$$

Note $\,\,\gcd(u',v')|u\,\,$ and $\,\,\gcd(u',v')|v\,\,$. So, $\,\,\gcd(u',v')|\gcd(u,v)$

Similarly you can show this the other side.

Swapnil Tripathi
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  • Correct, I misread the GCD notation, am working in different books... – nilo de roock Dec 23 '14 at 13:59
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    @ndroock1 A crucial point is that $,p,q,r,s,$ are not just any rationals but are integers (why?). For the argument to be convincingly correct this must be explicitly mentioned (and justified). See this answer for details, and a generalization, viz. Theorem $\ $ If $\rm,(x,y)\overset{A}\mapsto (X,Y),$ is linear then $: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y),,\ \Delta = \det A\ \ $ – Bill Dubuque Dec 23 '14 at 18:20
  • What am I missing that allows one to write u and v in this way? – Derek Luna Apr 22 '19 at 17:26