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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 numbersReal 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.

Solution:
Given: If a and b are two odd positive integers such that a > b.
To Prove:That one of the two numbers a + b2 and a - b2is 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

a + b2 = (2q + 1) + (2m + 1)2 = 2((q + m) + 1)2 = a + b2 = (q + m + 1)

Which is a positive integer.

a - b2 = (2q + 1) - (2m + 1)2 = 2(q + m)2 = a - b2 = (q - m)

Given, a > b

2q + 1 > 2m + 1

2q > 2m

q > m

Therefore,

a - b2 = (q - m) > 0

Thus,

a - b2 is a positive integer.

We need to prove that one of the two numbers

a + b2 and a - b2 is odd and the other is even.

Consider,

a + b2 - a - b2 = (a + b) - (a - b)2 = 2b2 = b

Also, we know from the proof above that

a + b2 and a - b2
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 n2integer 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, n2integers 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, n2integers 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 how are they classified?
Ans. Real numbers are the set of numbers that include both rational and irrational numbers. They can be classified into several categories: natural numbers (1, 2, 3,...), whole numbers (0, 1, 2,...), integers (..., -2, -1, 0, 1, 2,...), rational numbers (fractions like 1/2, -3/4), and irrational numbers (like √2, π) which cannot be expressed as simple fractions.
2. How do you represent real numbers on a number line?
Ans. Real numbers can be represented on a number line by marking points corresponding to their values. The line is typically drawn horizontally, with negative numbers on the left, zero in the middle, and positive numbers on the right. Each point on the line corresponds to a real number, allowing for visual comparison of their values.
3. What are the properties of real numbers?
Ans. The properties of real numbers include the following: commutative property (a + b = b + a and ab = ba), associative property ((a + b) + c = a + (b + c) and (ab)c = a(bc)), distributive property (a(b + c) = ab + ac), identity property (a + 0 = a and a × 1 = a), and inverse property (a + (-a) = 0 and a × (1/a) = 1 for a ≠ 0).
4. How do you perform operations with real numbers?
Ans. Operations with real numbers include addition, subtraction, multiplication, and division. To add or subtract real numbers, align their decimal points if they are in decimal form. For multiplication, simply multiply the numbers as you would with integers, and for division, divide the numbers while remembering to handle any decimal points appropriately.
5. Can you provide examples of rational and irrational numbers?
Ans. Yes, rational numbers are numbers that can be expressed as a fraction, such as 1/2, -3, or 0.75. Irrational numbers cannot be expressed as fractions; examples include √3, π, and e. These numbers have non-repeating, non-terminating decimal expansions, distinguishing them from rational numbers.
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