CSCI 2824: Lecture 7

In this lecture we will cover:

Divisibility of numbers and modular arithmetic.
Some proofs involving divisibility.

Topics covered: Chapter 2.2 of the book.

Proving Properties of Numbers

Let us start with divisibility.

An integer is divisible by a non-zero integer , if can be written as n = k times q for some integers and . In logic, we write:

mbox{isDivisible}(n,k): k not= 0 mbox{AND} (exists q in mathbb{Z} n = k times q ) ,.

The following concepts are all equivalent:

is divisible by
divides ,
is a factor of ,
, and
is a multiple of .

Divisibility by has a special name: If is divisible by , we call it even, and other wise we call it odd.

Theorem -1: If m,n are divisible by then m times n is divisible by .

Proof

For your convenience, we split the proof step by step to make the flow of reasoning steps clear.

Let be any two given numbers such that are divisible by .
Therefore, we write , for some integer .
And, since is divisible by , for some integers .
We note that .
Therefore is divisible by .

Important Note

Whenever in a proof, a number can be written as n = 7 cdot m , we conclude that is divisible by . Similarly, if you are told that a number is divisible by , it makes sense to write it as n = 7 l for some .

Theorem-2 If is divisible by and by then m^2 times n is divisible by .

Proof

Let be any given numbers such that divides and divides .
Therefore, we write for some integer , and
for some integer .
The number may be written as: .
Therefore is divisible by .

Proofs on Consecutive Numbers

Theorem-3 The product of two consecutive numbers is always even.

Proof

Let be any given number and be the next consecutive numbers.
We will show that is even.
We note that there are two cases based on whether is odd or even.

- Case-1: is odd. In this case, we have being even. Therefore = which is EVEN.
- Case-2: is even. In this case is even too.

In either case, is even. QED.

Proofs by Case-Splitting

A number n in mathbb{N} is a perfect square if n = m^2 for some m in mathbb{N} . Examples of perfect squares include 1,4,9,16,225, 1600 and so on.

Theorem-4 If a perfect square is even, then it is divisible by .

Proof

Let be any given perfect square that is even.
Since is a perfect square, can be written as for some .
We consider two cases based on whether is odd or even.

Case-1: is odd.

1. In this case, we know that is a product of two odd numbers is odd.
2. However, this conflicts with our assumption that is even.
3. Therefore, this case can never happen.

Case-2: is even.

1. In this case, for some integer .
2. is divisible by .

Therefore, we conclude that must be divisible by based on case 2, which is the only case that can happen.

QED.

There are other ways of proving this theorem that we will revisit when we study proofs by contradiction.

Modular Arithmetic

We have been using the “mod” operator so far in some proofs. Let us study the properties of the “mod” operator in more detail.

Modulo Operator

Let be some integer. For a integer k > 0 say that n mod k = r iff can be written as n = k q + r where is the quotient and is the remainder when is divided by .

Let us look at examples.

Example

We write 15 mod 7 = 1 . Since 15 = 2 * 7 + 1 .

As a convention, the result of a modulo operation a mod b is always between and b -1 .

Also, we always apply over positive, but can be positive or negative What is -22 mod 5 ? We write -22 = -5 * 5 + 3 . Therefore -22 mod 5= 3 .

Note:

is undefined.
We will not bother defining for negative .
We will only apply mod to integers for now. It can be defined on reals but that is not very important to us now.

Odd number: A number is odd iff n mod 2 = 1 .

We can distribute mbox{mod} over +,* .

(m + n) mod k = ((m mod k) + (n mod k))mod k

Secondly,

(m cdot n) mod k = ( (m mod k) cdot (n mod k) ) mod k

Let us prove these facts:

Theorem: Suppose number can be written as n = a k + b for k > 0 . It follows that n mod k = b mod k .

Proof

Let b mod k = r . Therefore b = k q + r for some . Note that 0 leq r < k .

n = (ak + b) = (ak + kq) + r = (k (a+q) + r)

Therefore can be written as (a+q)k +r for 0 leq r < k . We conclude therefore that a+q is the quotient when is divided by and is the remainder. Therefore,

n mod k = r

QED.

Theorem: For any integers m,n and integer k > 0 , the following statement is true:

(m + n) mod k = ((m mod k) + (n mod k))mod k

Proof

Let us assume m mod k = r_1 and n mod k = r_2 . Therefore, we may write as m = k q_1 + r_1 for some q_1 . We also write as n = k q_2 + r_2 for some q_2 .

Therefore, m + n = k q_1 + k q_2 + r_1 + r_2 .

m + n mod k = k (q_1 + q_2) + (r_1 + r_2) mod k = (r_1 + r_2) mod k = (m mod k + n mod k) mod k

QED.

Similarly, we can prove that modulo operator distributes over multiplication.

Theorem For any integers m,n , and natural number k > 0 , the following statement is true:

(m times n) mod k = (( m mod k) times (n mod k)) mod k

Proof

Let m mod k = r_1 and n mod k = r_2 , where 0 leq r_1 < k and 0 leq r_2 < k . We can write

m = k q_1 + r_1 and n = k q_2 = r_2 .

Therefore, using the distributivity of modulo operator over addition (proved in the previous theorem),

m times n = (k q_1 + r_1) times (k q_2 +r_2) = k^2 q_1 q_2 + k q_1 r_2 + k q_2 r_1 + r_1 r_2

Therefore,

$begin{array}{rcl} (m times n ) mod k & =& (k^2 q_1 q_2 mod k) + (k q_1 r_2 mod k) + (k q_2 r_2 mod k) + (r_1 r_2 mod k) & =& r_1 r_2 mod k end{array}$

This shows that

(m times n) mod k = (( m mod k) times (n mod k)) mod k

(QED)

Computing modulo

Let us try some interesting problems involving modulo arithmetic.

Example 1

For instance, let us try and compute 10^n mod 11 for various values of .

$begin{array}{rcl} 10^4 mod 11 &=& ( (10^3 mod 11) times (10 mod 11) ) mod 11 &=& (10 times 10) mod 11 = 1 end{array}$ .

As you can see the pattern that emerges is

$10^{n} mod 11 = 1$ if is even.
if is odd.

Example 2

We know 1000! is a really large number. The question is very simple. Suppose we wrote 1000! in hexadecimal, what would the last digit be?

In general, the last digit of a number in hexadecimal is simply computed by finding out n mod 16 . So what is 1000! mod 16 ?

Answer

1000! mod 16 = 0 , because 1000! = 1 times 2 times cdots times 16 times cdots times 1000 = 16 n for some n in mathbb{N} .

Example 3

Find the smallest natural number that leaves a remainder of when divided by and a remainder of when divided by .

Let our mystery number be . We require n mod 3 = 2 . Therefore, n = 3k +2 for some .

Now consider nmod 7 = (3k + 2) mod 7 = (3 times (k mod 7) + 2 ) mod 7 = 5 .

Therefore, we note that 3 times (k mod 7) = 3 mod 7 or in other words k mod 7 = 1 and k = 7l +1 for some .

Let us choose l= 0 to yield k = 1 and therefore n = 5 .

The smallest number is indeed n = 5 .

We will examine more problems of this sort when we look at the Chinese remainder theorem.