properties of integers

1. TOPIC: PROPERTIES OF INTEGERS
NAME: FARHEEN PATEL
STD : FYBCA
DIV : I
ROLL NO : 29

2. WHAT IS AN INTEGERS?
An integer is a whole number (not a fractional
number) that can be positive, negative, or
zero.
Examples of integers are: -5, 1, 5, 8, 97, and
3,043.
Examples of numbers that are not integers are:
-1.43, 1 3/4, 3.14, .09, and 5,643.1.
The set of integers, denoted Z, is formally
defined as follows:
Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

3. PROPERTIES OF ADDITION AND
MULTIPLICATION
1.BINARY OPERATION : The operation is called binary
operation if a, b, ∈ Z => a o b ∈ Z.
For e.g. : a, b, ∈ Z => a + b ∈ Z
a, b, ∈ Z => ab ∈ Z
2.COMMUTATIVE PROPERTY : The operation o is
called commutative property if a o b => b o a.
For e.g. : If a, b, ∈ Z then a+b => b+a
If a, b, ∈ Z then ab => ba

4. 3. ASSOCIATIVE PROPERTY : The operation o is called
associative property if (aob)oc = ao(boc).
For e.g. : a, b ∈ Z => (a+b)+c =a+(b+c)
a, b ∈ Z => (a.b).c =a.(b.c)
4. DISTRIBUTIVE PROPERTY : The operation o is called
distributive property if ao(boc)=(aob)o(aoc).
For e.g. : a(b+c)=ab+ac
5. ADDITIVE IDENTITY: Z contains an element o such that ∀a
∈ Z a+o=o+a=a.
6. MULTIPLICATIVE IDENTITY : Z contains an element 1
such that ∀a ∈ Z a.1=1.a=a
7. ADDITIVE INVERSE : For each element a ∈ Z there is an
element b ∈ Z such that a+b=o=b+a.

5. ABSOLUTE VALUE :
 For any integer x , we define absolute value |x| as follows :
|x|= x if x ≥ 0
= x if x ≥ 0
NOTE : 1) For a ∈ Z , |a|≥ 0.
2) For a ∈ Z , |a|=max{a,-a}=|-a|
3) For a ∈ Z , |a|≤ 𝑎 ≤ |a|
Also, -|a|≤ −𝑎 ≤ 𝑎
THEOREMS : 1) |a+b| ≤ |a|+|b| for a, b ∈ Z
Proof:
If a+b≥ 0 , then |a+b|= a+b
≤ |a|+|b|
If a+b < 0 , then |a+b|= -(a+b)
= -a – b
≤ |a|+|b|
∴ 𝑎 + 𝑏 ≤ |𝑎| + |b|.

6. 2)|ab|=|a||b|
If a>0 , b>0 then |ab|= ab
Also, |a||b|= ab
∴ 𝑎𝑏 = 𝑎 𝑏
If a>0 , b<0 then |ab|= -ab
Also, |a||b|= a(-b)= -ab
∴ 𝑎𝑏 = 𝑎 𝑏
If a<0 , b>0 then |ab|= -ab
Also, |a||b|= (-a)b = -ab
∴ 𝑎𝑏 = 𝑎 𝑏

7. DIVISIBILITY IN Z : For integers a, b we say a divides
b if a≠0 and there exists integer c such that b=ac . Notation a|b.
 THEOREMS :
1) If a|b and b|c then a|c where a , b, c ∈ Z
Proof :
Let a|b & b|c
∴ there exists 𝑘1& 𝑘2 such that b= a𝑘1 & 𝑐 = 𝑏𝑘2
∴ 𝑐 = b𝑘2 = 𝑎𝑘1 𝑘2 = 𝑎 𝑘1 𝑘2
∴ 𝑎|𝑐
2) If c|a , c|b then c|ma+nb
Proof:
Since c|a & c|b
∴there exists 𝑘1, 𝑘2 ∈ 𝑍 such that a=c𝑘1 & 𝑏 = 𝑐𝑘2
ma+nb= m(c𝑘1) + 𝑛 𝑐𝑘2 = 𝑐(𝑚𝑘1 + 𝑛𝑘2)
∴ c|ma+nb.

8. DIVISION ALGORITHM:
For given integers a and b with b>0 ,
there exists unique integers q and r such
that
a = bq + r & 0 ≤ 𝑟 < h

9. GREATEST COMMON DIVISOR :
Let a and b be two non- zero integers d is
said to be greatest common divisor of a & b
if 1) d|a & d|b and
2) if c|a & c|b then c|d

10. THEOREM : Any two non-zero integers a and b have a unique
positive g.c.d. This g.c.d can be expressed in the form ma+nb where
m, n∈ Z .
 THEOREM :
If a=bq+r & r ≠ 0 then (a , b) = (b , r)
Proof : Let d = (a , b) & 𝑑1 = 𝑏 , 𝑟
Then d|a & d| b
∴ 𝑑|𝑎 − 𝑏𝑞
∴ 𝑑 | 𝑟
Thus , d divides both b & r
∴ d divides g.c.d of b & r i.e. d | 𝑑1
Now , 𝑑1 = (𝑏 , 𝑟)
∴ 𝑑1 𝑏 & 𝑑1 𝑟
∴ 𝑑1 | 𝑏𝑞 + 𝑟
∴ 𝑑1 | 𝑎
∴ 𝑑1 divides both a & b
∴ 𝑑1 divides g.c.d of a & b i.e. d | 𝑑1 ∵ 𝑑 & 𝑑1 𝑎𝑟𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑑 = 𝑑1

11. THEOREM : Two non-zero integers a and b are co-
prime if and only if there exists integers m , n ∈ Z such
that ma+nb=1.
 Proof :
If a & b are co-prime then (a , b) = 1
∴ 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑚 , 𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑚𝑎 + 𝑛𝑏 = 1
Conversely ,
Suppose that ma+nb = 1 for m , n ∈ Z & let if possible (a , b ) =
d
Thus , d which is common divisor of a & b also divides ma + nb
.
∴ 𝑑 | 𝑚𝑎 + 𝑛𝑏
∴ 𝑑 | 1
As d is positive , we have d = 1 .

12. THEOREM - EUCLID’S LEMMA :
If (a , b ) = 1 & b|ac then b | c .
Proof :
∵ 𝑎 , 𝑏 = 1 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑚 , 𝑛 ∈ 𝑍 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
1 = ma + nb
∵ 𝑐 = 𝑚𝑎𝑐 + 𝑛𝑏𝑐
∵ 𝑏 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎𝑐 , 𝑏 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑚𝑎𝑐 + 𝑛𝑏𝑐
∵ 𝑏 | 𝑐

13. E.g. : Find the GCD of 2329 & 3151 , Express it in the form
,2329m + 3151n ?
solution :
3151 = 2329 x 1 + 822
2329 = 822 x 2 + 685
822 = 685 x 1 + 137
685 = 137 x 5 + 0
∴ 2329 , 3151 = 137
Now , 137 = 822 – 685 x 1
= 822 – (2329 – 822 x 2 ) x 1
= 822 x 1 – 2329 x 1 + 822 x 2
= 822 x 3 – 2329 x 1
= (3151 – 2329 x 1) x 3 – 2329 x 1
= 3151 x 3 – 2329 x 3 – 2329 x 1
= 3151 x 3 – 2329 x 4
= 3151 x 3 + 2329 (- 4)
∴ 3151𝑚 + 2329𝑛 = 3151 x 3 + 2329 (-4 )