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 o f t w o u n e q u a l n u m b e r s {c a n}^{'} t

e x c e e d t h e i r a b s o l u t e

d i f f e r e n c e . P r o v e .

Commented by mr W last updated on 01/Sep/20

n o t t r u e i f b o t h n u m b e r s a r e e q u a l .

Commented by Rasheed.Sindhi last updated on 01/Sep/20

Y e s s i r . {Y o u}^{'} r e r i g h t . I h a v e

a d d e d r e s t r i c t i o n .

Commented by Aina Samuel Temidayo last updated on 01/Sep/20

Does it imply \gcd (n, q) = \gcd (n - q)

where n > q ?

Answered by mr W last updated on 01/Sep/20

s a y p > q

m = g c d (p, q) ⩾ 1

\Rightarrow p = m s

\Rightarrow q = m t

w i t h g c d (s, t) = 1 a n d s > t \Rightarrow s - t ⩾ 1

p - q = m (s - t) ⩾ m

\Rightarrow g c d (p, q) ⩽ p - q

Commented by Rasheed.Sindhi last updated on 01/Sep/20

E l e g a n t!

T h a n k s M r W s i r!