11th std chapter 2 exercise 2.1

JK MATHS

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

227∈Q

Question 2.

Prove that 3–√ is an irrational number.

(Hint: Follow the method that we have used to prove 2–√ ∉ Q.

Solution:

Suppose that 3–√ is rational

⇒ 3 is a factor of q also

so 3 is a factor ofp and q which is a contradiction.

⇒ 3–√ is not a rational number

⇒ 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 + 2–√ and 1 + 2–√

Their difference is a rational number. But if we take two irrational numbers as 2 – 3–√ and 4 + 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.

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 + 3–√ and 3 – 3–√

Their sum is 2 + 3–√ + 3 – 33–√ which is a rational number.

But the sum of 3 + 5–√ and 4 – 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 3π their product is 3–√ and 2–√ = 3–√ × 2–√ which is irrational, So the product of two irrational numbers is either rational or irrational.

Question 5.

Find a positive number smaller than 121000. Justify.

Solution:

There will not be a positive number smaller than 0.

So there will not be a +ve number smaller than 121000

227∈Q