Answer to Question #155593 in Calculus for nanu

Question

If x is an arbitrary real number, show that there exists a unique n ∈ Z such that n ≤ x < n + 1. (This is called the greatest integer in x and denoted by [x].) 

Answer 1

$Solution: Let ~A=\{n\in Z| n\leqslant x \}.~\\ We~know~that, If~x~is ~an~arbitrary~real~number,~there~exist~integers~m~and~n~\\such~that~m<x<n.$

$\Rightarrow ~A ~is~non-empty.\\It ~is~also~bounded~ above,~So~by~completness~axiom,~s=sup(A)~exists.$

$Now,~by~property~of~sup,~there~exists~n\in A~such~that~s-1<n\leqslant s. \\Clearly~,~n\leqslant x.~Now,s-1<n~implies ~taht~n+1~does~not ~belong~to~A.$

$Hence~x<n+1.~This ~show~that~there~ is~an~ n~ which~ satisfies~ n\leqslant x<n+1.$

$~~~~Now, ~to~show~uniqueness,~assume~on~the~contrary~that~there~exists~\\m\neq n~such~that~m\leqslant x<m+1.~Then~m \in A~and ~we~must~have~m\leqslant n$

$(Otherwise~,~if~m>n~,~then~m\geqslant n+1>x,~so~m~can~not~satisfy~~the \\~said~inequality).~If ~m<n~ then ~(since~ both~ are~ integers)~(n-m)\geqslant1.$

$Hence~,m+1\leqslant n \leqslant x,~wich~contradicts~the~assumption~that~m~satisfies~the ~\\given~inequality.$

$Hence, there~ exists ~a~ unique~ n ∈ Z ~such ~that~ n ≤ x < n + 1,~where~x~ is ~an~ \\arbitrary ~real~ number.$