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, \(-\frac{1}{4}\), 0, 3.14 , 4, \(\frac{22}{7}\)} as a member of N, Q, R – Q or Z.

Answer:

√7 is an irrational number. ∴ √7 ∈ R – Q

\(-\frac{1}{4}\) is a negative rational number. ∴ \(-\frac{1}{4}\) ∈ 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

\(\frac{22}{7}\) is an rational number. ∴ \(\frac{22}{7}\) ∈ Q

Question 2.

Prove that √3 is an irrational number.

Answer:

Suppose that √3 is rational. Let √3 = \(\frac{\mathrm{m}}{\mathrm{n}}\) where m and n are positive integers with no common factors greater than 1.

√3 = \(\frac{\mathrm{m}}{\mathrm{n}}\)

⇒ √3n = m

⇒ 3n^{2} = m^{2} ——– (1)

By assumption n is an integer

∴ n^{2} is an integer. Hence 3n^{2} is an integral multiple of 3.

∴ From equation (1) m^{2} is an integral multiple of 3

⇒ m is an intergral multiple of 3

[Here m is an integer and m^{2} is an integral multiple of 3. That m^{2} is cannot take all integral multiples of 3. For example suppose m^{2} = 3 = 1 × 3 which is an integral multiple of 3. In this case m = √3 which is not an integer. Suppose m^{2} = 6 = 2 × 3 which is an integer multiple of 3 , but m = √2 √3 which is an integer. Hence m^{2} is an integral multiple of 3. Such that m is an integer.

Examples: m^{2} = 4 × 9,

m^{2} = 9,

m^{2} = 9 × 9 etc.]

Let m = 3k

where k is an integer

Using equation (1) we have

3n^{2} = (3k)^{2}

⇒ 3n^{2} = 9k^{2}

⇒ n^{2} = 3k^{2
}∴ n^{2} 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 + \(\sqrt{2}\) 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.

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 + \(\sqrt{3}\) 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.

Answer:

The given number is \(\frac{1}{2^{1000}}\)

We have 1000 < 1001

⇒ 2^{1000} < 2^{1001}

⇒ \(\frac{1}{2^{1000}}\) > \(\frac{1}{2^{1001}}\)

∴ A positive number smaller than \(\frac{1}{2^{1000}}\) is \(\frac{1}{2^{1001}}\)