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


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Question Number 63532 by Rio Michael last updated on 05/Jul/19

$p r o v e t h a t t h e r e e x i s t u n i q u e i n t e r g e r s p a n d s s u c b t h a t$ $a = b p + s w i t h - \frac{∣ b ∣}{2} < s ⩽ \frac{∣ b ∣}{2}$ $h e n c e f i n d p a n d s g i v e n t h a t a = 49 a n d b = 26$
	Answered by MJS last updated on 05/Jul/19

	$a = b p + s \Rightarrow s = a - b p$ $case 1 : b > 0$ $- \frac{b}{2} < a - b p ⩽ \frac{b}{2}$ $\frac{a}{b} - \frac{1}{2} ⩽ p < \frac{a}{b} + \frac{1}{2}$ $case 2 : b < 0$ $\frac{b}{2} < a - b p ⩽ \frac{b}{2}$ $\frac{a}{b} - \frac{1}{2} < p ⩽ \frac{a}{b} + \frac{1}{2}$ $in both cases :$ $the width of the interval [\frac{a}{b} - \frac{1}{2}; \frac{a}{b} + \frac{1}{2}] is 1 \Rightarrow$ $\Rightarrow {there}^{'} s exactly one p \in Z \land (\frac{a}{b} - \frac{1}{2} ⩽ p < \frac{a}{b} + \frac{1}{2} \lor \frac{a}{b} - \frac{1}{2} < p ⩽ \frac{a}{b} + \frac{1}{2})$ $\Rightarrow s is also unique$ $a = 49 \land b = 26$ $\frac{49}{26} - \frac{1}{2} ⩽ p < \frac{49}{26} + \frac{1}{2}$ $\frac{18}{13} ⩽ p < \frac{31}{13} \Rightarrow p = 2$ $s = a - b p = 49 - 26 \times 2 = - 3$
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