Real Number / chapter 1 / class - 10 / New St. Mery English School

 What are natural numbers?

Ans:- Counting Numbers are called Natural numbers.

Ex:- 1,    2,    3,    4,    5,    6,    7, .....................................∞

What are whole numbers?

Ans:- All natural numbers including with zero are called whole numbers

Ex:- 0,    1,    2,    3,    4,    5,    6,    7, .....................................∞

What are integers?

Ans:- All positive numbers, all negative numbers including with zero are called integers.

  Ex:- ∞ ...............  -7,    -6,    -5,    -4,    -3,    -2,    -1,    0    1,    2,    3,    4,    5,    6,    7, ..................∞

What is rational number?

Ans:- A number that is in the form of  P / q where q ≠ 0 is called rational number. P and q are always an integers.

Ex: 5,    -9,    0,    √4,    √9,    3√8,    3√27,    2.5,    1.6666 ........ 

👉 All natural numbers are rational numbers.

👉All whole numbers are rational numbers.

👉 All integers are rational numbers.

👉All fractions are rational numbers.

👉 All terminating decimals are rational numbers

👉 All Non-terminating but repeating decimals are rational numbers.

👉 All complete square root numbers and all complete cube root numbers are rational numbers.

What are irrational numbers?

Ans:- A number that is not in the form of  P / q is called rational number.

Ex:- √2,    √5,    3√9,    3√4,    2.50550555...............,    1.02002000200002.............

👉All incomplete square root numbers and all incomplete cube root numbers are irrational numbers. 

👉 All Non-terminating and non-repeating decimals are irrational numbers.

What are real numbers?

Ans:- All rational numbers and irrational numbers constitute real numbers.

Ex- 5,    -9,    0,    √4,    √9,    3√8,    3√27,    2.5,    1.6666 ........

      √2,    √5,    3√9,    3√4,    2.50550555...............,    1.02002000200002.............

👉 ㄫ is an irrational number but 22 / 7 is a rational number.

What is lema?

Ans:- The proven statement that is used to prove other statements is called lema.

What is Euclid's division lema?

Ans:-Let a and b be any two positive integers. than there exist two unique integers 'q' and 'r' such that

a = bq + r where 0 ≤  r < b 

Forms of positive integers when an integer is divided by 2

👊 Let the arbitary positive integer is 'a'

When 'a' is divided by 2 than let the quotient is 'q' and remainder is 'r'

By Euclid's division lema, we can say that

a = 2q + r where 0 ≤  r < 2

than r = 0, 1

When  r = 0 , than

a = 2q + 0

a = 2q

When r = 1 , than

a = 2q + 1

Hence it is proved that When any positive integer is divided by 2 than only two forms of integers is found , 2a  and 2a + 1 

Forms of positive integers when an integer is divided by 3

👊 Let the arbitary positive integer is 'a'

When 'a' is divided by 3 than let the quotient is 'q' and remainder is 'r'

By Euclid's division lema, we can say that

a = 3q + r where 0 ≤  r < 3

than r = 0, 1, 2

When  r = 0 , than

a = 3q + 0

a = 3q

When r = 1 , than

a = 3q + 1

When r = 2 , than

a = 3q + 2

Hence it is proved that When any positive integer is divided by 3 than only three forms of integers is found , 3a  , 3a + 1 and 3a + 2

Show that every positive integer is of the form 2q or 2q + 1, where q is some integer.

👊 Let the arbitary positive integer is a .

On dividing a by 2 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 2q + r where 0 ≤  r < 2

 ∴ r = 0 , 1

Case - 1

When r = 0

a = 2q + 0

a = 2q

Case - 2

When r =1

a = 2q +1

Hence by the case (1) and (2) we can say that every positive integer is of the form 2q or 2q + 1, where q is some integer.

 

Show that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q .

👊 Let the arbitary positive integer is a .

On dividing a by 3 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 3q + r where 0 ≤  r <3

 ∴ r = 0 , 1 , 2

Case - 1

When r = 0

a = 3q + 0

a = 3q

Case - 2

When r =1

a = 3q +1

Case - 3

When r =2

a = 3q +2

Hence by the case (1) , (2) and (3) we can say that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q .

Show that any positive integer is of the form 4q or 4q + 1 or 4q + 2 or 4q + 3 for some integer q .

👊 Let the arbitary positive integer is a .

On dividing a by 4 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 4q + r where 0 ≤  r <4

 ∴ r = 0 , 1 , 2 , 3

Case - 1

When r = 0

a = 4q + 0

a = 4q

Case - 2

When r =1

a = 4q +1

Case - 3

When r =2

a = 4q +2

Case - 4

When r =3

a = 4q +3

Hence by the case (1) , (2) , (3) and (4) we can say that any positive integer is of the form 4q or 4q + 1 or 4q + 2 or 4q + 3  for some integer q .

 

Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, where q is some integer.

👊 Let the arbitary positive integer is a .

On dividing a by 2 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 2q + r where 0 ≤  r < 2

 ∴ r = 0 , 1

Case - 1

When r = 0

a = 2q + 0

a = 2q ( since 2 X integer = even integer)

∴ a = 2q represents even positive integers.

Case - 2

When r =1

a = 2q +1 ( When 1 is added to any even positive integers than it becomes odd positive integer)

∴ a = 2q +1  represents odd positive integers.

Hence by the case (1) and (2) we can say that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, where q is some integer.

 

Show that any positive odd integer is of the form 4q + 1 or 4q + 3 . where q is some integer.

 

👊 Let the arbitary odd positive integer is a .

On dividing a by 4 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 4q + r where 0 ≤  r <4

 ∴ r = 0 , 1 , 2 , 3

Case - 1 

When r = 0

a = 4q + 0

a = 4q ( since 4q = 2 X 2q = even integer )

Here a is odd positive integer but 4q is even positive integer so case 1 contradicts our assumption.

case-1 is not true

∴ a ≠ 4q

Case - 2

When r =1

a = 4q +1 ( 4q + 1 =  odd positive integer)

Here a is odd positive integer and 4q + 1 is also odd positive integer. so case- 2 support our assumption.

∴ case-2 is true

Case - 3

When r =2

a = 4q +2 ( 4q + 2 =  even positive integer)

Here a is odd positive integer but 4q +2 is even positive integer so case-3 contradicts our assumption.

case-3 is not true

∴ a ≠ 4q +2

Case - 4

When r =3

a = 4q +3 ( 4q + 3 =  odd positive integer)

Here a is odd positive integer and 4q + 3 is also odd positive integer. so case- 4 support our assumption.

∴ case-4 is true

Hence by the cases 2 and 4 we can say that  any positive odd integer is of the form 4q + 1 or 4q + 3 . where q is some integer.

 

Show that any positive odd integer is of the form 6q + 1 or 6q + 3 , 6q + 5 . where q is some integer.

 

👊 Let the arbitary odd positive integer is a .

On dividing a by 6 let q is quotient and r is remainder

Than by the euclid's division lema,

a = 6q + r where 0 ≤  r <6

 ∴ r = 0 , 1 , 2 , 3, 4, 5

Case - 1 

When r = 0

a = 6q + 0

a = 6q ( since 6q = 2 X 3q = even integer )

Here a is odd positive integer but 6q is even positive integer so case 1 contradicts our assumption.

case-1 is not true

∴ a ≠ 6q

Case - 2

When r =1

a = 6q +1 ( 6q + 1 =  odd positive integer)

Here a is odd positive integer and 6q + 1 is also odd positive integer. so case- 2 support our assumption.

∴ case-2 is true

Case - 3

When r =2

a = 6q +2 ( 6q + 2 =  even positive integer)

Here a is odd positive integer but 6q +2 is even positive integer so case-3 contradicts our assumption.

case-3 is not true

∴ a ≠ 6q +2

Case - 4

When r =3

a = 6q +3 ( 6q + 3 =  odd positive integer)

Here a is odd positive integer and 6q + 3 is also odd positive integer. so case- 4 support our assumption.

∴ case-4 is true

Case - 5

When r =4

a = 6q +4 ( 6q + 4 =  even positive integer)

Here a is odd positive integer but 6q +4 is even positive integer so case-5 contradicts our assumption.

case-5 is not true

∴ a ≠ 6q +4

Case - 6

When r =5

a = 6q +5 ( 6q + 5 =  odd positive integer)

Here a is odd positive integer and 6q + 5 is also odd positive integer. so case- 6 support our assumption.

∴ case-6 is true

 

Hence by the cases 2 , 4 and 6 we can say that  any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 . where q is some integer.

 

Show that the square of an odd positive integer is of the form 4q + 1 for the some integer q.

 

👊 Let the positive odd integer is x.

Where x = 2m + 1 and m is any positive integer

Now, A / C

x² = ( 2m + 1 )²

     = 4m² + 1 + 4m

    = 4m²  + 4m + 1

    = 4 (m²  + m ) + 1

    = 4q + 1 , where q =  m²  + m

Hence it is proved that the square of an odd positive integer is of the form 4q + 1 for the some integer q.

    

 Show that ( n²  - 1 ) is divisible by 8, if n is an odd positive integer.

 Let the positive odd integer is n.

Where n = (4q + 1) or (4q + 3)

Case-1

When   n = (4q + 1)

squaring both side

n² = (4q + 1)²

  subtracting both side by 1

n²  - 1 = (4q + 1)²  - 1

            =   16q² + 1 + 8q - 1 

            = 16q² + 8q  

 n²  - 1 = 8 ( 2q² + q )                             

∴ when n = 4q + 1 , than n²  - 1 is divisible by 8  

 

Case-2

When   n = (4q + 3)

squaring both side

n² = (4q + 3)²

  subtracting both side by 1

n²  - 1 = (4q + 3 )²  - 1

            =   16q² + 9 + 24q - 1 

            = 16q² + 24q + 8

 n²  - 1 = 8 ( 2q² + 3q + 1 )                             

∴ when n = 4q + 3 , than n²  - 1 is divisible by 8 

 

From the case-1 and case-2 we can say that( n²  - 1 ) is divisible by 8, if n is an odd positive integer.