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
⇒ 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 + \(\sqrt{2}\) 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
⇒ 21000 < 21001
⇒ \(\frac{1}{2^{1000}}\) > \(\frac{1}{2^{1001}}\)
∴ A positive number smaller than \(\frac{1}{2^{1000}}\) is \(\frac{1}{2^{1001}}\)