Given a set S={a_1 , ..., a_n } a_k ∈Z a_1 =a, a_n =a_(n−1) +1, a_n =b ∴S contains all integers between a and b. Are there more even or odd values?


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Question Number 3055 by Filup last updated on 04/Dec/15
$Given a set S={a_1 , ..., a_n } a_k ∈Z a_1 =a, a_n =a_(n−1) +1, a_n =b ∴S contains all integers between a and b. Are there more even or odd values?$
$Given a set S = {a_{1}, . . ., a_{n}}$ $a_{k} \in Z$ $a_{1} = a, a_{n} = a_{n - 1} + 1, a_{n} = b$ $∴ S contains all integers between$ $a and b .$ $Are there more even or odd values ?$
	Answered by 123456 last updated on 04/Dec/15

	$a_{1} = a$ $a_{n} = a_{n - 1} + 1$ $a_{n} = b (b ⩾ a)$ $a_{1} = a$ $a_{2} = a + 1$ $⋮$ $a_{n} = b = a + (b - a)$ $n = b - a + 1$ $if n = b - a + 1 \equiv 0 (\mod 2) \Rightarrow b - a \equiv 1 (\mod 2)$ $a_{1} \equiv a_{3} \equiv a_{5} \equiv . . . \equiv a_{n - 1}$ $a_{2} \equiv a_{4} \equiv a_{6} \equiv . . . \equiv a_{n}$ $so there same quantity of even and odd$ $if n = b - a + 1 \equiv 1 (\mod 2) \Rightarrow b - a \equiv 0 (\mod 2)$ $a_{1} \equiv a_{3} \equiv . . . \equiv a_{n - 2} \equiv a_{n}$ $a_{2} \equiv a_{4} \equiv . . . \equiv a_{n - 1}$ $so its not have same quantinty, also$ $a_{1} determines wich will have more$ $wich mean$ $there more numbers of parity of a_{1}$
	Answered by Rasheed Soomro last updated on 04/Dec/15

	$C a s e - 1 : a, b \in E$ $M o r e e v e n n u m b e r s$ $C a s e - 2 : a, b \in O$ $M o r e o d d n u m b e r s$ $C a s e - 3 : O n e o f a a n d b i s o d d o t h e r i s e v e n$ $B o t h a r e e q u a l$
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