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


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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

	$E l e g a n t!$ $T h a n k s M r W s i r!$
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