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Question Number 63532 by Rio Michael last updated on 05/Jul/19
prove that there exist unique intergers p and s sucb that a = bp + s with −((∣b∣)/2)< s ≤((∣b∣)/2) hence find p and s given that a=49 and b=26
provethatthereexistuniqueintergerspandssucbthata=bp+swith−∣b∣2<s⩽∣b∣2hencefindpandsgiventhata=49andb=26
Answered by MJS last updated on 05/Jul/19
a=bp+s ⇒ s=a−bp case 1: b>0 −(b/2)<a−bp≤(b/2) (a/b)−(1/2)≤p<(a/b)+(1/2) case 2: b<0 (b/2)<a−bp≤(b/2) (a/b)−(1/2)<p≤(a/b)+(1/2) in both cases: the width of the interval [(a/b)−(1/2); (a/b)+(1/2)] is 1 ⇒ ⇒ there′s exactly one p∈Z∧((a/b)−(1/2)≤p<(a/b)+(1/2)∨(a/b)−(1/2)<p≤(a/b)+(1/2)) ⇒ s is also unique a=49∧b=26 ((49)/(26))−(1/2)≤p<((49)/(26))+(1/2) ((18)/(13))≤p<((31)/(13)) ⇒ p=2 s=a−bp=49−26×2=−3
a=bp+s⇒s=a−bpcase1:b>0−b2<a−bp⩽b2ab−12⩽p<ab+12case2:b<0b2<a−bp⩽b2ab−12<p⩽ab+12inbothcases:thewidthoftheinterval[ab−12;ab+12]is1⇒⇒there′sexactlyonep∈Z∧(ab−12⩽p<ab+12∨ab−12<p⩽ab+12)⇒sisalsouniquea=49∧b=264926−12⩽p<4926+121813⩽p<3113⇒p=2s=a−bp=49−26×2=−3

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