Think and Reflect (Section 3.4)
Q1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?
Answer:
Solution:
To add or subtract rational numbers with different denominators, we first find the LCM (Least Common Multiple) of the denominators. Then we convert each fraction to an equivalent fraction with the LCM as the new denominator by multiplying both numerator and denominator by the appropriate factor. Once the denominators are equal, we can add or subtract the numerators.
Answer: Find the LCM of the denominators and convert both fractions to equivalent fractions with that common denominator.
Q2. Verify the distributive law for rational numbers.
Answer:
Solution:
The distributive law states that for any rational numbers a, b, and c, a × (b + c) = a × b + a × c.
Example verification:
Let a = 1/2, b = 3/4, c = 5/6
LHS = (1/2) × (3/4 + 5/6)
First, find common denominator for 3/4 and 5/6 = 12
3/4 = 9/12, 5/6 = 10/12
Sum = 19/12
LHS = (1/2) × (19/12) = 19/24
RHS = (1/2) × (3/4) + (1/2) × (5/6)
= 3/8 + 5/12
Common denominator = 24
= 9/24 + 10/24 = 19/24
LHS = RHS, hence verified.
Answer: The distributive law holds for rational numbers.
Think and Reflect (Section 3.5)
Q1. Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?
Answer:
Solution:
Proof that √3 is irrational:
Assume √3 = p/q in simplest form, where p and q are co-prime integers.
Squaring: 3 = p²/q² → p² = 3q²
This implies p² is divisible by 3, so p is divisible by 3. Let p = 3k.
Substitute: (3k)² = 3q² → 9k² = 3q² → q² = 3k²
This implies q is also divisible by 3.
Thus p and q both have factor 3, contradicting that they are co-prime.
Therefore, √3 is irrational.
Same approach for √5, √7, √10:
- √5 and √7: yes, the same proof works because 5 and 7 are prime.
- √10: yes, the proof works because 10 is not a perfect square and the prime factors (2 and 5) will force both p and q to be divisible by those primes.
Answer: Yes, the same method works for √5, √7, and √10.
Q2. We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
Answer:
Solution:
Lines with irrational lengths can be constructed using geometric methods:
- Using the Pythagorean theorem: For example, in a right triangle with legs of length 1, the hypotenuse is √2, which is irrational.
- Square root spiral method: By constructing right triangles successively, we can obtain lengths √n for any positive integer n.
Answer: Irrational lengths can be obtained using the Pythagorean theorem and geometric constructions like the square root spiral.
Q3. Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
Answer:
Solution:
For √3:
- Construct a right triangle with legs 1 and √2. The hypotenuse will be √(1² + (√2)²) = √3.
- Alternatively, construct a right triangle with legs 1 and 1 (gives √2), then use √2 as one leg with another leg of length 1 to get √3.
For √5:
- Construct a right triangle with legs 1 and 2. The hypotenuse will be √(1² + 2²) = √5.
General method for √n:
To construct √n for any positive integer n:
- If n can be written as a² + b² (using integers), construct a right triangle with legs a and b; the hypotenuse is √n.
- Alternatively, use the square root spiral method: start with a segment of length 1, repeatedly construct right triangles where one leg is 1 and the other leg is the previous hypotenuse.
Answer: By using right triangles and the Pythagorean theorem, we can construct √n for any positive integer n.
Q4. Try to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?
Answer:
Solution:
10/3 = 3.333... = 3.3
The digit 3 repeats indefinitely.
11/12 = 0.91666... = 0.916
The digit 6 repeats indefinitely after the non-repeating part '91'.
Observation:
- 10/3 has a pure repeating decimal (the repeating part starts immediately after the decimal).
- 11/12 has a non-terminating repeating decimal with a non-repeating part before the repeating block.
Answer: Both are non-terminating repeating decimals.
Q5. The decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?
Answer:
Solution:
For a rational number p/q (in simplest form) to have a terminating decimal, the denominator must divide some power of 10, i.e., 10^n = 2^n × 5^n.
If q has prime factors other than 2 and 5, then q cannot divide any power of 10, so the decimal expansion will never terminate (it will repeat).
If q has only prime factors 2 and/or 5, then we can multiply both numerator and denominator by suitable numbers to make the denominator a power of 10, resulting in a terminating decimal.
Answer: Because only powers of 2 and 5 can divide a power of 10 (which is 2^n × 5^n), so denominators with only these prime factors can be converted to a power of 10.
Q6. Consider this puzzle: What is the square root of −1? We know that 1 × 1 = 1 and (−1) × (−1) = 1. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, √−1 cannot exist on number line.
Answer:
Solution:
√−1 does not exist in the real number system because the square of any real number is non-negative. The square root of a negative number cannot be represented on the real number line.
To solve this, mathematicians introduced the imaginary unit i, where i² = −1. This extends the number system to complex numbers.
Answer: √−1 is not a real number; it is an imaginary number denoted by i, where i² = −1.