0
$\begingroup$

In the above statement N , a , b are natural numbers . I was wondering whether the above statement is always true . If it is always true will anyone give me a simple reason or proof for it ? Please guide me .

$\endgroup$
4
  • $\begingroup$ Yes its always true. Consider the fundamental theorem of arithmetic: n has a unique prime decomposition, as do $a$ and $b$. $\endgroup$
    – David Diaz
    Jan 9, 2020 at 16:14
  • $\begingroup$ related and likely inspiration for this question: math.stackexchange.com/questions/3502984/… $\endgroup$
    – gt6989b
    Jan 9, 2020 at 16:19
  • 1
    $\begingroup$ If $ax+by=1$ then $Nax+bny=N$ and each term is divisible by $ab.$ $\endgroup$ Jan 9, 2020 at 16:23
  • $\begingroup$ i.e. $\ ab\mid aN,bN\,\Rightarrow\,ab\mid (aN,bN) = (a,b)N = N\ $ in gcd or ideal language (cf. various forms of Euclid's Lemma) $\endgroup$ Jan 10, 2020 at 0:28

1 Answer 1

0
$\begingroup$

well yes, since: a/N and b/N $\rightarrow$ $N = a* \alpha , N = b*\beta $

we have: $a/N \rightarrow a/b*\beta \rightarrow a/ \beta$ (because a and b are coprime by Gauss theorem)

Hence: $\beta = 0 or \beta = k*a \rightarrow N = a*b*k$

Therefore: a*b/N

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .