Here's the proof.

**Proof of the Quotient Remainder Theorem**

We want to prove:

Given any integer A, and a positive integer B, there exist unique integers Q and R such that: A= B * Q + R where 0 ≤ R < B

We have to prove two things:

Given any integer A and positive integer B:

1) Q and R exist

2) Q and R are unique

**Proof that Q and R exist**

(One approach is to use the Well Ordering Principle of Integers, but I will use an approach that is arguably simpler)

Suppose we have an integer A and a positive integer B.

Given any real number A and any positive real number B if I divide A by B, I will have an integer part (before the decimal),w, and a fractional part,f, after the decimal.

If A is >= 0 then w and f are non-negative:

A/B = w + f where 0 ≤ f < 1

A = B * w + B * f

w is an integer (it is the whole part) we can simply label it as Q, i.e. Q = w

by multiplying 0 ≤ f < 1 by B we find that

B * 0 ≤ B * f < B * 1

0 ≤ B * f < B

we can simply label the term B * f as R i.e. R= B * f

We have shown that 0 ≤ R < B

To show that R is an integer we can say that:

A = B * w + B * f

A = B * Q + R (using our new labels)

we can rearrange this to:

R = A - B * Q

but A, B, Q are integers and the result of any integer minus the product of integers is still an integer (this a property of integers)

so R must be an integer

so we have shown that given any integer A >= 0 and any positive integer B, there exist integers Q and R such that: A= B * Q + R where 0 ≤ R < B

if A is negative then

A/B = w + f where -1 < f ≤ 0

A = B * w + B * f

if f = 0 then label w as Q and B * f as R

A = B * Q + R

since B * f = R = 0 we can say that R satisfies 0 ≤ R < B

and that R is an integer

if -1 < f < 0

A = B * w + B * f

A = B * ( w - 1) + B * (f + 1)

label w-1 as Q, which is an integer

add 1 to -1 < f < 0

0 < f + 1 < 1

multiply by B

0 < B * ( f + 1 ) < B

we can simply label the term B * (f + 1) as R

We have shown that 0 < R < B which satisfies 0 ≤ R < B

To show that R is an integer, in this case, we can say that:

A = B * ( w - 1) + B * (f + 1)

A = B * Q + R (using our new labels)

we can rearrange this to:

R = A - B * Q

but A, B, Q are integers and the result of any integer minus the product of integers is still an integer (this a property of integers)

so R must be an integer

so we have shown that given any integer A < 0 and any positive integer B, there exist integers Q and R such that: A= B * Q + R where 0 ≤ R < B

so we have shown that given any integer A and any positive integer B, there exist integers Q and R such that: A= B * Q + R where 0 ≤ R < B

**Proof that Q and R are unique**

Suppose we have an integer A and a positive integer B.

We have shown before that Q and R exist above.

So we can find at least one pair of integers, Q1 and R1, that satisfy

A= B * Q1 + R1 where 0 ≤ R1 < B

And we can find at least one pair of integers, Q2 and R2, that satisfy

A= B * Q2 + R2 where 0 ≤ R2 < B

For labeling purposes, R2 is >= than R1 (if not we could just switch the integer pairs around).

We will show that Q1 must equal Q2 and R1 must equal R2 i.e. Q and R are unique

We can set the equations equal to each other

B * Q1 + R1 = B * Q2 + R2

B * (Q1 - Q2) = (R2 - R1)

(Q1 - Q2) = (R2 - R1)/ B

since R2 >= R1 we know that R2 - R1 is >= 0

since R2 <B and R1 >= 0 we know that R2-R1 < B

So we can say that 0 ≤ R2 - R1 < B

divide by B

0 ≤ (R2 - R1)/B < 1

but from above we showed that

(Q1 - Q2) = (R2 - R1)/ B

and Q1 - Q2 must be an integer since an integer minus an integer is an integer

so (R2 - R1)/B must be an integer but its value is >= 0 and < 1.

The only integer in that range is 0.

So (R2- R1)/B= 0 ,thus R2-R1 =0 ,thus R2 = R1

also

Q1-Q2 = 0 thus Q1 = Q2

Thus we have shown that Q1 must equal Q2 and R1 must equal R2 i.e. Q and R are unique

Hope this makes sense