Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1

   

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 2
3.14 ∈ Q
0, 4 are integers and 0 ∈ Z, 4 ∈ N, Z, Q
\(\frac{22}{7} \in \mathrm{Q}\)

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1

Question 2.
Prove that \(\sqrt{3}\) is an irrational number.
(Hint: Follow the method that we have used to prove \(\sqrt{2}\) ∉ Q.
Solution:
Suppose that \(\sqrt{3}\) is rational P
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 3
⇒ 3 is a factor of q also
so 3 is a factor ofp and q which is a contradiction.
⇒ \(\sqrt{3}\) is not a rational number
⇒ \(\sqrt{3}\) is an irrational number

Question 3.
Are there two distinct irrational numbers such that their difference is a rational number? Justify.
Solution:
Taking two irrational numbers as 3 + \(\sqrt{2}\) and 1 + \(\sqrt{2}\)
Their difference is a rational number. But if we take two irrational numbers as 2 – \(\sqrt{3}\) and 4 + \(\sqrt{7}\).
Their difference is again an irrational number. So unless we know the two irrational numbers we cannot say that their difference is a rational number or irrational number.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1

Question 4.
Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number.
Solution:
(i) Let the two irrational numbers as 2 + \(\sqrt{3}\) and 3 – \(\sqrt{3}\)
Their sum is 2 + \(\sqrt{3}\) + 3 – 3\(\sqrt{3}\) which is a rational number.
But the sum of 3 + \(\sqrt{5}\) and 4 – \(\sqrt{7}\) is not a rational number. So the sum of two irrational numbers is either rational or irrational.

(ii) Again taking two irrational numbers as π and \(\frac{3}{\pi}\) their product is \(\sqrt{3}\) and \(\sqrt{2}\) = \(\sqrt{3}\) × \(\sqrt{2}\) which is irrational, So the product of two irrational numbers is either rational or irrational.

Question 5.
Find a positive number smaller than \(\frac{1}{2^{1000}}\). Justify.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 4
There will not be a positive number smaller than 0.
So there will not be a +ve number smaller than \(\frac{1}{2^{1000}}\)

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 Additional Questions

Question 1.
Prove that \(\sqrt{5}\) is an irrational number.
Solution:
Suppose that \(\sqrt{5}\) is rational
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 5
So let p = 5c
substitute p = 5c in (1) we get
(5c)2 = 5q2 ⇒ 25c2 = 5q2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 6
⇒ 5 is a factor of q also
So 5 is a factor of p and q which is a contradiction.
⇒ \(\sqrt{5}\) is not a rational number
⇒ \(\sqrt{5}\) is an irrational number

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1

Question 2.
Prove that 0.33333 = \(\frac{1}{3}\)
Solution:
Let x = 0.33333….
10x = 3.3333 ….
10x – x = 9x = 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1 7