Thursday, 26 September 2019

elementary number theory - if pmida and pmidb then pmidgcd(a,b)


I would like to prove the following property :




(p,a,b)Z3pa and pbpgcd(a,b)




Knowing that :



Definition




Given two natural numbers a and b, not both zero, their greatest common divisor is the largest divisor of a and b.






  • If Div(a) denotes the set of divisors of a, the greatest common divisor of a and b is gcd(a,b)=max(Div(a)Div(b))

  • d=gcd(a,b){dDiv(a)Div(b)xDiv(a)Div(b):xd

  • (a,b)N2abDiv(a)Div(b)

  • xZDiv(x)=Div(x)

  • If a,bZ, then gcd(a,b)=gcd(|a|,|b|), adding gcd(0,0)=0



Indeed,




Let (p,a,b)Z3 such that pa and pb then :



paDiv(p)Div(a) and pbDiv(p)Div(b) then



Div(p)(Div(a)Div(b))pgcd(a,b)



Am I right?

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