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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)

= 2^{2} – (√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} = 5q^{2} ⇒ 25c^{2} = 5q^{2}

⇒ 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