Question

Does m^2 14 years ago

Answer 1

saubhik i can prove
m-1/2<sqrt(n+1/4) and
m+1/2>sqrt(n-3/4)
will it help u?

Answer 2

Can u bring the greatest integer function (GIF) in ur results?

Answer 3

m^2<m+N<(m+1)^2
m^2-m<n<(m+1)^2-m
m^2-m<n<m^2+m+1
m^2-m<n<m^2+m
(m-1/2)^2<n+1/4<(m+1/2)^2

m+1/2<sqrt(n+1/4)<m+1/2
now as m,n both integer so the notion of [] will come and the 1/2 associated with m-1/2 or m+1/2 will make [sqrtn+1/2] but how it will happen i have to check it...

Answer 4

again sqrt(n+1/4)+1/2 will be integer as m and m+1 integer...

Answer 5

how do u get
m^2-m<n<m^2+m+1
to imply m^2-m<n<m^2+m ?

Answer 6

m^2-m<n<m^2+m+1
(m-1/2)^2-1/4<n<(m+1/2)^2+3/4

(m-1/2)<sqrt(n+1/4) and (m+1/2)>sqrt(n-3/4)
sqrt(n-3/4)-1/2<m<sqrt(n+1/4)+1/2
sqrt(4n-3)-1<2m<sqrt(4n+1)+1
sqrt(4n)-sqrt(3)-1<2m<sqrt(4n)+2
sqrtn -(sqrt (3)+1)/2<m<sqrtn+1

Answer 7

let [sqrtn]=a
sqrtn=a+f 0<=f<1

Answer 8

a+(2f-sqt3 -1)/2<m<a+f+1

so m is an integer bet a and a+1

so m=a or a+1

[sqrt(n+1/2)] = a or a+1
so m=[sqrt(n)+1/2]