Prove that product of two consecutive integers is positive when divide

1.	Prove that product of two consecutive integers is positive when divided by 2
Answer» $\huge{\bf{\underline{\red{Solution :}}}}$ ⇒Suppose the two consecutive positive integers be z and (z +1) Now ACCORDING to Question : ⇒Products of two consecutive positive integers = z(z +1) = $\bold{z^{2} + z}$ $\bf{\boxed{\bold{ Case - i) \ z \ is \ even \ number }}}$ Suppose z = 2k $\implies \bold{ z^{2} + z }$ $\implies \bold{ (2k)^{2} + 2k}$ $\implies \bold{4k^{2} + 2k}$ $\implies \bold{2k(2k + 1)}$ Therefore , the PRODUCT is DIVISIBLE by 2 $\bf{\bold{\red{Case - ii) \ z \ is \ odd \ number}}}$ Suppose z = 2k + 1 $\implies \bold{z^{2} + z }$ $\implies \bold{(2k + 1)^{2} + (2k + 1)}$ $\implies \bold{4k^{2} + 4k + 1 + 2k + 1}$ $\implies \bold{4k^{2} + 6k + 2}$ $\implies \bold{2(2k^{2} + 3k + 1)}$ Therefore , the product is divisible by 2 ( From both Condition the proved that the product of two consecutive integers is positive when DIVIDED by 2) $\rule{200}2$