Show that if a is an integer and d is an integer greater tha

Show that if a is an integer and d is an integer greater than 1, then the quotient q and remainder r when a is divided by d are given by:

q = a/d
r = ad·a/d

q = [a/d]

r = a - d*[a/d]

=> r = a - qd

=> a = qd + r .......... (1)

Thus we have to prove that a = qd + r where a is an integer and d is an integer greater than 1

Existence

Consider the following progression: .....a - 3d , a - 2d , a - d , a , a + d , a + 2d , a + 3d ....

Then by the Well-Ordering Principle, there must exist a smallest non-negative element, denoted by r.

So r = aqd for some qZ.

r must be in the interval [0,d) because otherwise rd would be smaller than r and a non-negative element in the progression

Uniqueness

Suppose we have another pair q₀ and r₀ such that a = bq₀ + r₀, with 0 r₀ < d.

Then dq+r = dq₀ + r₀.

Factoring we see that rr₀ = d(q₀q), and so d(rr₀).

Since 0r<d and 0r₀<d, we have that d<rr0<d.

Hence, rr₀ = 0 r = r₀.

So now rr₀ = 0 = d(q₀q) which implies that q=q₀.

Therefore the solution is unique.

Hence (1) is proved