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

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.1 Text Book Back Questions and Answers, Notes.

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

Question 1.
Classify each element of {√7, −1/4, 0, 3.14 , 4, 22/7} as a member of N, Q, R – Q or Z.
Answer:
√7 is an irrational number. ∴ √7 ∈ R – Q
−1/4 is a negative rational number. ∴ −1/4 ∈ Q
0 is an integer. ∴ 0 ∈ Z , Q
3.14 is a rational number. ∴ 3.14 ∈ Q
4 is a positive integers. ∴ 4 ∈ Z, N, Q
22/7 is an rational number. ∴ 22/7 ∈ Q

Question 2.
Prove that √3 is an irrational number.
Answer:
Suppose that √3 is rational. Let √3 = mn where m and n are positive integers with no common factors greater than 1.
√3 = mn
⇒ √3n = m
⇒ 3n² = m² ——– (1)
By assumption n is an integer
∴ n² is an integer. Hence 3n² is an integral multiple of 3.
∴ From equation (1) m² is an integral multiple of 3
⇒ m is an intergral multiple of 3

[Here m is an integer and m² is an integral multiple of 3. That m² is cannot take all integral multiples of 3. For example suppose m² = 3 = 1 × 3 which is an integral multiple of 3. In this case m = √3 which is not an integer. Suppose m² = 6 = 2 × 3 which is an integer multiple of 3 , but m = √2 √3 which is an integer. Hence m² is an integral multiple of 3. Such that m is an integer.
Examples: m² = 4 × 9,
m² = 9,
m² = 9 × 9 etc.]

Let m = 3k
where k is an integer
Using equation (1) we have
3n² = (3k)²
⇒ 3n² = 9k²
⇒ n² = 3k²
∴ n² is an integral multiple of 3. Since, n is an integer, we have n is also an integral multiple of 3.

Thus we have proved both m and n are integral multiple of 3. Hence both m and n have common factor 3, which is a contradiction to our assumption that m and n are integers with no common factors greater than 1.

Hence our assumption that √3 is a rational number is wrong.
∴ √3 is an irrational number.

Question 3.
Are there two distinct irrational numbers such that their difference is a rational number? Justify.
Answer:
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?
Answer:
(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.