Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.1

Question 1. Classify each element of \(\{\sqrt{7}, -\frac{1}{4}, 0, 3.14, 4, \frac{22}{7}\}\) as a member of N, Q, R – Q or Z.
Answer:
\( \sqrt{7} \) is an irrational number because it cannot be written as a simple fraction. Therefore, \( \sqrt{7} \in R - Q \).
\( -\frac{1}{4} \) is a negative rational number as it is a fraction. So, \( -\frac{1}{4} \in Q \).
\( 0 \) is an integer, and it can also be written as \( \frac{0}{1} \), making it a rational number. So, \( 0 \in Z, Q \).
\( 3.14 \) is a rational number because it is a terminating decimal. Hence, \( 3.14 \in Q \).
\( 4 \) is an integer and a natural number, and it can also be written as \( \frac{4}{1} \), making it rational. So, \( 4 \in Z, N, Q \).
\( \frac{22}{7} \) is a rational number because it is expressed as a fraction, although it's often used as an approximation for pi. Therefore, \( \frac{22}{7} \in Q \).
In simple words: We sort each number into groups like natural numbers (N), integers (Z), rational numbers (Q), or irrational numbers (R-Q). This helps us understand what kind of number each one is.

🎯 Exam Tip: Remember that irrational numbers are non-repeating and non-terminating decimals, while rational numbers can be expressed as a fraction \( \frac{p}{q} \) where \( q \neq 0 \).

Question 2. Prove that \(\sqrt{3}\) is an irrational number.
Answer:Let us assume, for the sake of contradiction, that \(\sqrt{3}\) is a rational number.
If \(\sqrt{3}\) is rational, we can write it as \( \sqrt{3} = \frac{m}{n} \), where \(m\) and \(n\) are integers, \(n \neq 0\), and \(m, n\) have no common factors greater than 1 (meaning the fraction is in its simplest form).
Squaring both sides gives:
\( (\sqrt{3})^2 = (\frac{m}{n})^2 \)
\( 3 = \frac{m^2}{n^2} \)
\( 3n^2 = m^2 \) (Equation 1)
This equation shows that \(m^2\) is a multiple of 3.
If \(m^2\) is a multiple of 3, then \(m\) must also be a multiple of 3.
We can write \(m\) as \(3k\), where \(k\) is some integer.
Substitute \(m = 3k\) into Equation (1):
\( 3n^2 = (3k)^2 \)
\( 3n^2 = 9k^2 \)
Divide both sides by 3:
\( n^2 = 3k^2 \)
This means \(n^2\) is a multiple of 3.
If \(n^2\) is a multiple of 3, then \(n\) must also be a multiple of 3.
So, we have found that both \(m\) and \(n\) are multiples of 3.
This contradicts our initial assumption that \(m\) and \(n\) have no common factors other than 1. Since our assumption led to a contradiction, our initial assumption must be false.
Therefore, \(\sqrt{3}\) cannot be a rational number. It must be an irrational number.
In simple words: We pretended that \(\sqrt{3}\) could be written as a simple fraction. But when we worked it out, it showed that the top and bottom numbers of the fraction would always share a common factor, which means it couldn't be a simple fraction after all. So, \(\sqrt{3}\) is not rational; it is irrational.

🎯 Exam Tip: The proof by contradiction is key here. State the assumption clearly, follow the logical steps, and point out the contradiction that proves the number is irrational.

Question 3. Are there two distinct irrational numbers such that their difference is a rational number? Justify.
Answer:Yes, there are two distinct irrational numbers such that their difference is a rational number.
Consider two irrational numbers: \( (2 + \sqrt{2}) \) and \( (1 + \sqrt{2}) \).
Both \( (2 + \sqrt{2}) \) and \( (1 + \sqrt{2}) \) are irrational numbers because adding a rational number to an irrational number results in an irrational number.
Now, let's find their difference:
\( (2 + \sqrt{2}) - (1 + \sqrt{2}) = 2 + \sqrt{2} - 1 - \sqrt{2} = 2 - 1 = 1 \)
Since \(1\) is a rational number, the difference between these two distinct irrational numbers is indeed rational.
Another example: Let the two irrational numbers be \( (4 + \sqrt{7}) \) and \( \sqrt{7} \).
Their difference is \( (4 + \sqrt{7}) - \sqrt{7} = 4 \), which is a rational number. This demonstrates that it is possible to find such numbers.
In simple words: Yes, you can find two numbers that are both irrational, but when you subtract one from the other, the answer is a simple, rational number. It happens when the "irrational part" cancels itself out.

🎯 Exam Tip: To show the difference between two irrational numbers is rational, choose numbers where the irrational parts are identical and will cancel out upon subtraction.

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) Two irrational numbers whose sum is a rational number:
Let the two irrational numbers be \( (2 + \sqrt{3}) \) and \( (3 - \sqrt{3}) \).
Both are irrational because they contain \( \sqrt{3} \).
Their sum is:
\( (2 + \sqrt{3}) + (3 - \sqrt{3}) = 2 + \sqrt{3} + 3 - \sqrt{3} = 2 + 3 = 5 \)
Since \(5\) is a rational number, their sum is rational.
(ii) Two irrational numbers whose product is a rational number:
Yes, you can find two irrational numbers whose product is a rational number.
Consider \( \sqrt{3} \) and \( \sqrt{12} \). Both are irrational.
Their product is:
\( \sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36} = 6 \)
Since \(6\) is a rational number, their product is rational.
Another example: \( \sqrt{5} \) and \( \sqrt{5} \). Their product is \( \sqrt{5} \times \sqrt{5} = 5 \), which is rational.
In simple words: Yes, it's possible for two irrational numbers to add up to a rational number, and also for them to multiply together to make a rational number. This happens when the messy, non-repeating parts cancel out during the sum or product.

🎯 Exam Tip: For sums, choose irrational numbers that have a common irrational part with opposite signs. For products, choose numbers that multiply to form a perfect square or a simple integer.

Question 5. Find a positive number smaller than \( \frac{1}{2^{1000}} \). Justify.
Answer:We need to find a positive number that is smaller than \( \frac{1}{2^{1000}} \).
We know that \( 1000 < 1001 \).
When we use these numbers as exponents for a base greater than 1, the inequality holds:
\( 2^{1000} < 2^{1001} \)
Now, if we take the reciprocal of both sides, the inequality sign flips:
\( \frac{1}{2^{1000}} > \frac{1}{2^{1001}} \)
Therefore, a positive number smaller than \( \frac{1}{2^{1000}} \) is \( \frac{1}{2^{1001}} \). This number is positive because \(2^{1001}\) is positive.
In simple words: To find a smaller number, we can just make the bottom part of the fraction bigger. Since \(2^{1001}\) is bigger than \(2^{1000}\), the fraction \( \frac{1}{2^{1001}} \) will be smaller than \( \frac{1}{2^{1000}} \).

🎯 Exam Tip: When dealing with fractions, remember that making the denominator (bottom number) larger makes the overall fraction smaller, assuming the numerator (top number) stays the same and both are positive.