Hello friend This is for discrete mathematics Can you solve

Solution

Let, n=kq+r   , 0<=r<k

q and r are integers.

Case 1: r=0

n/k=q

ceiling(n/k)=q

n-1=kq-1

(n-1)/k=q-1/k

floor((n-1)/k)=q-1

HEnce, floor((n-1)/k)+1=q=ceiling(n/k)

Case 2: k>r>=1

n=kq+r

n/k=q+r/k

ceiling(n/k)=q+1

n-1=kq+r-1

(n-1)/k=q+(r-1)/k

q is an integer and

0<=(r-1)/k<1

HEnce, q is integer part of (n-1)/k and (r-1)/k is fractional part

So,

floor((n-1)/k)=q

floor((n-1)/k)+1=q+1=ceiling(n/k)

HEnce proved