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Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

Exercise 1.1

Question: 1
If a and b are two odd positive integers such that a > b, then prove that one of the two numbers Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10and Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10 is odd and the other is even.
Solution:
Given: If a and b are two odd positive integers such that a > b.
To Prove: That one of the two numbers Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10 and Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10is odd and the other is even.
Proof: Let a and b be any odd positive integer such that a > b. Since any positive integer is of the form q, 2q + 1
Let, a = 2q + 1 and b = 2m + 1, where, q and in are some whole numbers
Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10

Which is a positive integer.
Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10
Given, a > b
2q + 1 > 2m + 1
2q > 2m
q > m
Therefore, 
Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10
Thus, Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10is a positive integer.
We need to prove that one of the two numbers Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10 and Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10is odd and other is even.
Consider, Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10
Also, we know that from the proof above that 
Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10
are positive integers.
We know that the difference of two integers is odd if one of them is odd and another is even. (Also, the difference between two odd and two even integers is even)
Hence it is proved that if a and b are two odd positive integers is even.
Hence, it is proved that if a and b are two odd positive integers such that a > b then one of the two number Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10and Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10is odd and the other is even.

Question: 2
Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
To Prove: that the product of two consecutive integers is divisible by 2.
Proof: Let n – 1 and n be two consecutive positive integers.
Then their product is n (n – 1) = n2 – n
We know that every positive integer is of the form 2q or 2q + 1 for some integer q.
So let n = 2q
So, n2 – n = (2q)2 – (2q)
n2 – n = (2q)2 – (2q)
n2 – n = 4q2 – 2q
n2 – n = 2q (2q – 1)
n2 – n = 2r [where r = q(2q – 1)]
n2 – n is even and divisible by 2
Let n = 2q + 1
So, n2 – n = (2q + 1)2 – (2q + 1)
n2 – n = (2q + 1) (2q + 1) – 1)
n2 – n = (2q + 1) (2q)
n2 – n = 2r[r = q (2q + 1)]
n2 – n is even and divisible by 2
Hence it is proved that the product of two consecutive integers is divisible by 2.

Question: 3
Prove that the product of three consecutive positive integers is divisible by 6.
Solution:
To Prove: the product of three consecutive positive integers is divisible by 6.
Proof: Let n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5.
If n = 6q, n(n + 1)(n + 2) = 6q(6q + 1)(6q + 2), which is divisible by 6
If n = 6q + 1, n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3) n(n + 1)(n + 2) = 6 (6q + 1)(3q + 1)(2q + 1) Which is divisible by 6
If n = 6q + 2, n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4) n(n + 1)(n + 2) = 12(3q + 1)(2q + 1)(2q + 3),
Which is divisible by 6.
Similarly, we can prove others.
Hence it is proved that the product of three consecutive positive integers is divisible by 6.

Question: 4
For any positive integer n, prove that n3 – n divisible by 6.
Solution:
To Prove: For any positive integer n, n3 — n is divisible by 6.
Proof: Let n be any positive integer. n3 — n = (n – 1)(n)(n + I)
Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5
If n = 6q,
Then, (n —1)n(n + 1) = (6q —1)6q (6q + 1)
Which is divisible by 6
If n = 6q + 1,
Then, (n —1)n(n + 1) = (6q)(6q + 1)(6q + 2)
Which is divisible by 6.
If n = 6q + 2,
Then, (n – 1)n(n + 1) = (6q + 1)(6q + 2)(6q + 3)
(n – 1)n(n + 1) = 6(6q + 1)(3q + 1)(2q + 1)
Which is divisible by 6.
Similarly, we can prove others.
Hence it is proved that for any positive integer n, n3— n is divisible by 6.

Question: 5
Prove that if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.
Solution:
To Prove: If a positive integer is of the form 6q + 5 then it is of the form 3q + 2 for some integer q, but not conversely.
Proof: Let n = 6q + 5
Since any positive integer n is of the form of 3k or 3k + 1, 3k + 2
If q = 3k,
Then, n = 6q + 5
n = 18k + 5 (q = 3k)
n = 3(6k + 1) + 2
n = 3m + 2 (where m = (6k + 1))
If q = 3k+ 1,
Then, n = (6q + 5)
n = (6 (3k + 1) + 5)(q = 3k + 1)
n = 18k + 6 + 5
n = 18k + 11
n = 3(6k + 3) + 2
n = 3m + 2 (where m = (6k + 3))
If q = 3k + 2,
Then, n = (6q + 5)
n = (6(3k + 2) + 5)(q = 3k + 2)
n = 18k + 12 + 5
n = 18k + 17
n = 3(6k + 5) + 2
n = 3m + 2 (where m = (6k + 5))
Consider here 8 which is the form 3q + 2 i.e. 3 × 2 + 2 but it can’t be written in the form 6q + 5. Hence the converse is not true.

Question: 6
Prove that any square of any positive integer of the form 5q + 1 is of the same form.
Solution:
To Prove: That the square of a positive integer of the form 5q + 1 is of the same form
Proof: Since positive integer n is of the form 5q + 1
If n = 5q + 1
Then n2 = (5q + 1)2
n2 = (5q)2 + 2 (1) (5q) + 12 = 25q2 + 10q + 1
n2 = 5m + 1 (where m = (5q2 + 2q))
Hence n2 integer is of the form 5m + 1.

Question: 7
Prove that the square of any positive integer is of form 3m or 3m + 1 but not of form 3m + 2.
Solution:
To Prove: that the square of a positive integer is of form 3m or 3m + 1 but not of form 3m + 2.
Proof: Since positive integer n is of the form of 3q , 3q + 1 and 3q + 2
If n = 3q
n2 = (3q)2
n2 = 9q2
n2 = 3 (3q)2
n2 = 3m(m = 3q)2
If n = 3q + 1
Then, n2 = (3q + 1)2
n2 = (3q)2 + 6q + 1
n2 = 9q2 + 6q + 1
n2 = 3q(3q + 1) + 1
n2 = 3m +1(where m = (3q + 2))
If n = 3q + 2
Then, n2 = (3q + 2)2 = (3q)2 + 12q + 4
n2 = 9q2 + 12q + 4
n2 = 3 (3q + 4q + 1) + 1
n2 = 3m + 1 (where q = (3q + 4q + 1)
Hence, n2 integers are of the form 3m, 3m + 1 but not of the form 3m + 2.

Question: 8
Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
Solution:
To Prove: The square of any positive integer is of the form 4q or 4q + 1 for some integer q. 
Proof: Since positive integer n is of the form of 2q or 2q + 1
If n = 2q
Then, n2 = (2q)2
n2 = 4q2
n2 = 4m (where m = q2)
If n = 2q + 1
Then, n2 = (2q + 1)2
n2 = (2q)2 + 4q + 1
n2 = 4q2 + 4q + 1
n2 = 4q (q + 1) + 1
n2 = 4q + 1 (where m = q (q + 1))
Hence it is proved that the square of any positive integer is of the form 4q or 4q + 1, for some integer q.

Question: 9
Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
Solution:
To Prove: The square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
Proof: Since positive integer n is of 5q or 5q + 1, 5q + 4.
If n = 5q, Then. n2 = (5q)2
n2 = 25q2
n2 = 5 (5q)
n2 = 5m (Where m = 5q)
If n = 5q + 1
Then, n2 = (5q +1)2
n2 = (5q)2 + 10q + 1
n2 = 25q2 + 10q + 1
n2 = 5q (5q + 2) + 1
It n2 = 5q (5q +2) + 1
n2 = 5m +1 (where m = q (5q + 2))
If n = 5q + 2
Then, n2 = (5q + 2)2
n2 = (5q)2 + 20q + 4
n2 = 25q2 + 20q + 4
n2 = 5q (5q + 4) + 4
n2 = 5m + 4 (where m = q (5q + 4))
If n = 5q + 4
Then, n2 = (5q + 4)2
n2 = (5q)2 + 40q + 16
n2 = 25q2 + 40q + 16
n2 = 5 (5q2 + 8q + 3) + 1
n2 = 5m + 1 (where m = 5q2 + 8q + 3 )
Hence it is proved that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Question: 10
Show that any positive odd integer is from 6q +1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
To Prove: the square of any positive integer is of the form 8q + 1 for some integer q.
Proof: Since any positive integer n is of the form 4m + 1 and 4m + 3
If n = m + 1
Then,
n2 = (4m + 1)2
n2 = (4m)2 + 8m + 1
n2 = 16m2 + 8m + 1
n2 = 8m (2m + 1) + 1
n2 + 8q +1 (where q = m (2m + 1))
If n = 4m + 3
Then, n2 = (4m + 3)2
n2 = (4m)2 + 24m + 9
n= 16m2 + 24m + 9
n2 = 8 (2m2 + 3m + 1) + 1
n2 = 8q + 1 (where q = (2m2 + 3m + 1))
Hence, n2 integers are of the form 8q + 1, for some integer q.

Question: 11
Show that any positive odd integer is from 6q +1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
To Show: Any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is any integer.
Proof: Let ‘a’ be any odd positive integer and b = 6.
Then, there exists integers q and r such that a = 6q + r, 0 ≤ r < 6 (by division algorithm)
a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
But 6q or 6q + 2 or 6q + 4 are even positive integers.
So, a = 6q + 1 or 6q + 3 or 6q + 5
Hence it is proved that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is an integer.

The document Real Numbers - 1 RD Sharma Solutions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Real Numbers - 1 RD Sharma Solutions - Mathematics (Maths) Class 10

1. What are real numbers and why are they important in mathematics?
Ans. Real numbers are a set of numbers that include both rational and irrational numbers. They are important in mathematics because they can be used to represent various real-life quantities such as length, time, temperature, and distance. Real numbers also form the foundation for many mathematical operations and concepts.
2. How can I identify if a number is a real number or not?
Ans. Any number that can be expressed as a decimal or a fraction is a real number. If a number cannot be expressed in any of these forms, it is not a real number. For example, 3, -2, 0.5, and √2 are all real numbers, while √-1 (imaginary number) is not a real number.
3. What is the difference between rational and irrational numbers?
Ans. Rational numbers are numbers that can be expressed as the ratio of two integers, while irrational numbers cannot be expressed as a simple fraction. Rational numbers can be terminating or repeating decimals, while irrational numbers are non-terminating and non-repeating decimals. Examples of rational numbers include 1/2, 3/4, and -5, while examples of irrational numbers include √2, π, and e.
4. Can real numbers be negative?
Ans. Yes, real numbers can be both positive and negative. Real numbers include all the positive and negative integers, fractions, and decimals. For example, -2, 1/2, and -0.75 are all real numbers.
5. How are real numbers used in practical applications?
Ans. Real numbers are used in a wide range of practical applications. They are used in finance to represent money and investments, in physics to represent measurements and quantities, in engineering to solve real-world problems, and in many other fields. Real numbers provide a way to quantify and describe various aspects of the real world accurately.
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