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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.1
Question 1.
Solution:
3.14 ∈ Q
0, 4 are integers and 0 ∈ Z, 4 ∈ N, Z, Q
\(\frac{22}{7} \in \mathrm{Q}\)
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
⇒ 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:
Let the two irrational numbers be 2 + √5 and 4 + √5
Their difference = (2 + √5) – (4 + √5)
= 2 + √5 – 4 – √5
= 2 – 4 = – 2
which is a rational number.
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:
Let the two irrational numbers be 3 + √5 and 7 – √5
Their sum = 3 + √5 + 7 – √5 = 3 + 7 = 10
which is a rational number
Consider the two irrational numbers 2 + √3, 2 – √3
Their product = (2 + √3) (2 – √3)
= 22 – (√3)2 = 4 – 3 = 1
which is a rational number.
Question 5.
Find a positive number smaller than \(\frac{1}{2^{1000}}\). Justify.
Solution:
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
So let p = 5c
substitute p = 5c in (1) we get
(5c)2 = 5q2 ⇒ 25c2 = 5q2
⇒ 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
Question 2.
Prove that 0.33333 = \(\frac{1}{3}\)
Solution:
Let x = 0.33333….
10x = 3.3333 ….
10x – x = 9x = 3