Tenth (10th) math
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Chapter-1 /Real Number /Class - 10 /New St. Mery English School
Chapter-2 /polynomial /Class - 10 /New St. Mery English School
Chapter-3 /Linear equation in two variables /Class - 10 /New St. Mery English School
Chapter-1 /Real Number /Class - 10 /New St. Mery English School
Chapter-1 /Real Number /Class - 10 /New St. Mery English School
Chapter-1 /Real Number /Class - 10 /New St. Mery English School
Real Number / chapter 1 / class - 10 / New St. Mery English School
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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
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 = 3qCase - 2
When r =1
a = 3q +1
Case - 3
When r =2
a = 3q +2
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 = 4qCase - 2
When r =1
a = 4q +1
Case - 3
When r =2
a = 4q +2
Case - 4
When r =3
a = 4q +3
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.
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
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 +2Case - 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
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
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 +2Case - 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 +4Case - 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
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
Show that ( n2 - 1 ) is divisible by 8, if n is an odd positive integer.
Let the positive odd integer is n.n2 = (4q + 1)2
subtracting both side by 1n2 - 1 = (4q + 1)2 - 1
= 16q2 + 1 + 8q - 1
= 16q2 + 8q
n2 - 1 = 8 ( 2q2 + q )
n2 = (4q + 3)2
subtracting both side by 1n2 - 1 = (4q + 3 )2 - 1
= 16q2 + 9 + 24q - 1
= 16q2 + 24q + 8
n2 - 1 = 8 ( 2q2 + 3q + 1 )