NUMBER SYSTEM - CBSE MATHEMATICS
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10. TWIN PRIMES :- Consecutive prime numbers which are differ by 2 are called twin primes. Eg :- (11, 13), (17, 21), (29, 31)…….etc.
12. ODD NUMBERS:- Natural numbers which are not divisible by 2 are called odd numbers. Eg:- 1, 3, 5, 7, 9 ,........
Note:- "1" is neither a prime number nor a composite number. It is a unit.
Decimals are of three types
\[Decimal\; expansion\: of\: the\: numbers\: like\: \sqrt{3},\; \sqrt{5},\; \sqrt{7}\; is\; called\]\[ non-terminating\; non-repeating\; decimals.\]
Eg:- 1.1010010001......, 2.5050050005......, 3.6060060006.......
a = bq + r , ................ where 0 ≤ r <b.
HCF x LCM = Product of two numbers
Class 10 : Exercise
1.1 Q No. 4: Prove that the square of any positive integer is of the
form 3m or 3m+1 Solution: For two given
positive integers a and b there exist unique integers q and r such that Now taking a = 3q Squaring on both side we get
(a)^{2} = (3q)^{2}
⇒ a^{2} =
9q^{2 } ⇒ a^{2} = 3(3q^{2} )
⇒ a^{2} =
3m_{1} …
.......Where m_{1} = 3q^{2} Now taking a = 3q
+1 Squaring on both side
we get (a)^{2} = (3q + 1)^{2}
⇒ a^{2} =
9q^{2} + 1 + 6q ⇒ a^{2} =
9q^{2} + 6q + 1 ⇒ a^{2} =
3(3q^{2} + 2q) + 1 ⇒ a^{2} = 3m_{2} +1
...........Where m_{2} = 3q^{2} +
2q Now taking a = 3q
+ 2 Squaring on both side we get
(a)^{2} = (3q + 2)^{2}
⇒ a^{2} =
9q^{2} + 4 + 12q ⇒ a^{2} =
9q^{2} + 12q + 4 ⇒ 9q^{2} +
12q + 3 + 1 ⇒ a^{2} = 3(3q^{2} +
4q + 1) + 1 ⇒ a^{2} = 3m_{3} + 1
............Where m_{3} = 3q^{2} + 4q + 1 Here we prove that the square of any
positive integer is of the form 3m or 3m + 1 Q No. 5: Prove that the cube of any positive integer is of the
form 9m or 9m + 1 or 9m + 8 Solution: For two given
positive integers a and b there exist unique integers q and r such that Now taking a = 3q Cubing on both sides we get
(a)^{3} = (3q)^{3}
⇒ a^{3} =
27q^{3 } ⇒ a^{3} =
9(3q^{3} ) ⇒ a^{3} =
9m_{1} …
.......Where m_{1} = 3q^{3} Now taking a = 3q
+1 Cubing on both side we get
(a)^{3} = (3q + 1)^{3}
⇒ a^{3} =
27q^{3} + 1 + 27q^{2} + 9q ⇒ a^{3}
= 27q^{3} + 27q^{2} + 9q +1 ⇒ a^{3} = 9(3q^{3}
+ 3q^{2} + q) +1 ⇒ a^{3} = 9m_{2} +1
...........Where m_{2} = 3q^{3}
+ 3q^{2} + q Now taking a = 3q
+2 (a)^{3} = (3q + 2)^{3}
⇒ a^{3} =
27q^{3} + 8 + 54q^{2} + 36q ⇒ a^{3}
= 27q^{3} + 54q^{2} + 36q +8 ⇒ a^{3} = 9(3q^{3}
+ 6q^{2} + 4q) +8 ⇒ a^{3} = 3m_{3} +8 ...........Where m_{3} = 3q^{3}
+ 6q^{2} + 4q Here we prove that the cube of any positive
integer is of the form 9m or 9m + 1 or 9m + 8 |
Class 10 Chapter 1
Exercise 1.2 Q. 5 Check whether 6^{n} can ends with digit 0 for any natural number n. Solution: If
6^{n} ends with 0 then its prime factorization must contain the power
of 2 and 5. Prime factors of 6 are 2 x 3 and by the fundamental theorem of arithmetic
this factorization is unique. This shows that prime factors of 6^{n}
does not contain 5. Hence 6^{n}
never ends with zero for any value of n |
Class 10 : Exercise 1.3 Question 1) Prove
that √5 is an irrational number Solution: Let
√5 is a rational number Then
there exist co-prime a and b such that √5
= a/b Cross
multiply here we get √5b
= a Now
squaring on both side we get 5b^{2} =
a^{2} ⇒ b^{2} =
a^{2}/5 ……………..(1) ⇒ 5
divides a^{2} ⇒ 5
divides a ………(2) Let
a = 5c , Squaring on both sides, we get a^{2} =
25c^{2} ………………(3) Putting
eqn.(3) in equation (1) we get ⇒ b^{2
}= 25c^{2}/5 , ⇒ b^{2} =
5c^{2} ⇒ b^{2}/5
= c^{2} 5
divides b^{2} ⇒ 5
divides b…..(4) From
(2) and (4) 5
divides a and b both ⇒ a
and b are not co-prime But
it is given in the starting that a and b are co-prime numbers. So
there is a contradiction and our assumption is wrong Hence
√5 is an irrational number Question: Prove that 6 - 2√5 is an irrational number Solution: Let 6 - 2√5 is a rational number. Then there exist unique integers a and b such that \[6-2\sqrt{5}=\frac{a}{b}\]\[-2\sqrt{5}=\frac{a}{b}-6\]\[-2\sqrt{5}=\frac{a-6b}{b}\]\[\sqrt{5}=\frac{a-6b}{-2b}\] a, b, 2 and 6 all are integers \[\Rightarrow \frac{a-6b}{-2b}\: \: is\: \: a\: \: rational\: \: number\] But √5 is an irrational number. So there is a contradiction and our assumption is wrong Hence 6 - 2√5 is an irrational number. |
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