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GCD-of-two-unequal-numbers-can-t-exceed-their-absolute-difference-Prove-




Question Number 110984 by Rasheed.Sindhi last updated on 01/Sep/20
GCD of two unequal  numbers can′t   exceed their absolute  difference.  Prove.
$$\mathrm{GCD}\:{of}\:{two}\:{unequal}\:\:{numbers}\:{can}'{t}\: \\ $$$${exceed}\:{their}\:{absolute} \\ $$$${difference}.\:\:{Prove}. \\ $$
Commented by mr W last updated on 01/Sep/20
not true if both numbers are equal.
$${not}\:{true}\:{if}\:{both}\:{numbers}\:{are}\:{equal}. \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/20
Yes sir. You′re right. I have   added restriction.
$${Yes}\:{sir}.\:{You}'{re}\:{right}.\:{I}\:{have} \\ $$$$\:{added}\:{restriction}. \\ $$
Commented by Aina Samuel Temidayo last updated on 01/Sep/20
Does it imply gcd(n,q) = gcd(n−q)  where n>q ?
$$\mathrm{Does}\:\mathrm{it}\:\mathrm{imply}\:\mathrm{gcd}\left(\mathrm{n},\mathrm{q}\right)\:=\:\mathrm{gcd}\left(\mathrm{n}−\mathrm{q}\right) \\ $$$$\mathrm{where}\:\mathrm{n}>\mathrm{q}\:? \\ $$
Answered by mr W last updated on 01/Sep/20
say p>q  m=gcd(p,q)≥1  ⇒p=ms  ⇒q=mt  with gcd(s,t)=1 and s>t ⇒s−t≥1  p−q=m(s−t)≥m  ⇒gcd(p,q)≤p−q
$${say}\:{p}>{q} \\ $$$${m}={gcd}\left({p},{q}\right)\geqslant\mathrm{1} \\ $$$$\Rightarrow{p}={ms} \\ $$$$\Rightarrow{q}={mt} \\ $$$${with}\:{gcd}\left({s},{t}\right)=\mathrm{1}\:{and}\:{s}>{t}\:\Rightarrow{s}−{t}\geqslant\mathrm{1} \\ $$$${p}−{q}={m}\left({s}−{t}\right)\geqslant{m} \\ $$$$\Rightarrow{gcd}\left({p},{q}\right)\leqslant{p}−{q} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/20
Elegant!  Thanks Mr W sir!
$$\mathcal{E}{legant}! \\ $$$$\mathcal{T}{hanks}\:{Mr}\:\mathcal{W}\:{sir}! \\ $$

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