Finding the Greatest Common Factor: Proving the Property (xa,xb)=x(a,b)

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SUMMARY

The discussion centers on proving the property (xa, xb) = x(a, b) where x is a constant. The user struggles to demonstrate that the greatest common factor of xa and xb can be expressed in terms of the greatest common factor of a and b, multiplied by x. Key steps include recognizing that if e = (a, b), then e can be expressed as a linear combination of a and b, leading to the conclusion that (xa, xb) is divisible by x(a, b). The user successfully identifies critical steps in the proof, particularly the implications of divisibility from equations (1) and (2).

PREREQUISITES
  • Understanding of greatest common factors (GCF)
  • Familiarity with linear combinations in number theory
  • Basic knowledge of divisibility rules
  • Experience with algebraic manipulation of equations
NEXT STEPS
  • Study the properties of greatest common factors in number theory
  • Learn about linear combinations and their applications in proofs
  • Explore advanced divisibility concepts and their implications
  • Review algebraic techniques for manipulating equations in proofs
USEFUL FOR

Students studying number theory, mathematicians interested in algebraic proofs, and educators teaching properties of greatest common factors.

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Homework Statement



is this suppose to be very obvious?

(xa,xb)=x(a,b) , x are constant

i can't see it T_T

nevertheless, i can't even prove it

to prove (xa,xb) l x(a,b)

let e=(a,b), then there exist integer m,n such that

e=ma+nb

xe=max+nbx

since (xa,xb) l ax and (xa,xb) l bx, then (xa,xb) l xe

to prove x(a,b) l (xa,xb)

i'm stuck here, many method i used but all halfway, can someone give me clue, and tell me is that suppose to be obvious? I'm rushing to class now, thanks in advance
 
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[tex](a,b)=e\Rightarrow{xe=xam+xbn}[/tex]

(1) [tex]e | a \Rightarrow{es=a}\Rightarrow{xes=xa}\Rightarrow{xe | xa}[/tex]

(2) [tex]e | b \Rightarrow{et=b}\Rightarrow{xet=xb}\Rightarrow{xe | xb}[/tex]

(3) [tex]d |xa, d | xb \Rightarrow{di=xa,dj=xb}\Rightarrow{xe=xam+xbn=d(im+jn)}\Rightarrow{d | xe}[/tex]

[tex]\(1\),\(2\),\(3\)\Rightarrow{(ax,bx)=ex}[/tex]
 
huaaaaa, thankss, i didn't noticed the (1) and (2), it really helpful, thanks
 

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