RBSE Solutions Class 10 Maths Chapter 2 Real Numbers More Ques

Question 1. Sum of the power of prime factors of 196
(a) 1
(b) 2
(c) 4
(d) 6
Answer: (c) 4
\( 196 = 2 \times 98 = 2 \times 2 \times 49 = 2 \times 2 \times 7 \times 7 = 2^2 \times 7^2 \)
The prime factors are 2 and 7. The power of 2 is 2, and the power of 7 is 2.
Sum of the powers = \( 2 + 2 = 4 \). This method helps break down larger numbers into their simplest building blocks.
In simple words: First, find the prime numbers that multiply to make 196. Then, add together the small numbers (exponents) that show how many times each prime number is used.

🎯 Exam Tip: Remember to only use prime numbers when finding prime factors. Common errors include stopping at composite numbers like 4 or 6.

Question 2. If two numbers are written in the form m = \( pq^3 \) and n = \( p^3q^2 \) then HCF of m, n whereas p, q prime numbers are:
(a) pq
(b) \( pq^2 \)
(c) \( p^2q^2 \)
(d) \( p^3q^3 \)
Answer: (b) \( pq^2 \)
To find the HCF of two numbers given in prime factor form, we take the lowest power of each common prime factor.
For m = \( p^1 q^3 \) and n = \( p^3 q^2 \):
The common prime factor p has powers 1 and 3. The lowest power is \( p^1 \).
The common prime factor q has powers 3 and 2. The lowest power is \( q^2 \).
So, the HCF = \( p^1 \times q^2 = pq^2 \). This method ensures we find the largest factor that divides both numbers evenly.
In simple words: When numbers are shown as \( p \) and \( q \) multiplied together with small powers, find the smallest power for each \( p \) and \( q \) that is common to both numbers. Multiply these smallest powers together to get the HCF.

🎯 Exam Tip: For HCF, always choose the lowest power of each common prime factor. For LCM, choose the highest power of all prime factors present in either number.

Question 3. HCF of 95 and 152 is:
(a) 1
(b) 19
(c) 57
(d) 38
Answer: (b) 19
First, we find the prime factors of each number:
\( 95 = 5 \times 19 \)
\( 152 = 2 \times 76 = 2 \times 2 \times 38 = 2 \times 2 \times 2 \times 19 = 2^3 \times 19 \)
The highest common factor (HCF) is the product of the common prime factors raised to their lowest powers. In this case, the only common prime factor is 19.
So, H.C.F. = 19. Finding the HCF helps simplify fractions and solve problems involving sharing items into equal groups.
In simple words: Break down 95 and 152 into their smallest prime number parts. The number that appears in both lists of prime parts is their HCF.

🎯 Exam Tip: Always list all prime factors completely for both numbers before identifying the common ones to avoid missing any.

Question 4. Product of two is 1080 and their HCF is 30 then their LCM is:
(a) 5
(b) 16
(c) 36
(d) 108
Answer: (c) 36
We know the important relationship that states: Product of two numbers = H.C.F. \( \times \) L.C.M.
We are given:
Product of both numbers = 1080
H.C.F. = 30
Let L.C.M. = x
So, \( 1080 = 30 \times x \)
To find x, we divide 1080 by 30:
\( x = \frac {1080}{30} \)
\( x = 36 \)
Thus, the L.C.M. is 36. This formula is a quick way to find either HCF or LCM if the other values are known.
In simple words: When you multiply two numbers, the answer is the same as multiplying their HCF and LCM. So, if you have the product and the HCF, you can divide to find the LCM.

🎯 Exam Tip: This formula (Product = HCF × LCM) is fundamental for number theory questions and saves time on calculations.

Question 5. Decimal expansion of number \( \frac {441}{{ 2 }^{2}\times { 5 }^{7}\times {7}^{2}} \) will be:
(a) Terminating
(b) Non-terminating repeating
(c) Non-terminating and repeating
(d) Non-rational
Answer: (a) Terminating
To determine the nature of the decimal expansion, we first simplify the fraction:
The numerator is \( 441 = 21^2 = (3 \times 7)^2 = 3^2 \times 7^2 \).
The denominator is \( 2^2 \times 5^7 \times 7^2 \).
So, the fraction is \( \frac{3^2 \times 7^2}{2^2 \times 5^7 \times 7^2} \).
We can cancel out \( 7^2 \) from both the numerator and the denominator:
\( \frac{3^2}{2^2 \times 5^7} = \frac{9}{2^2 \times 5^7} \)
A rational number has a terminating decimal expansion if its denominator, in simplest form, has only 2 and 5 as prime factors. In this simplified fraction, the denominator \( 2^2 \times 5^7 \) only contains prime factors 2 and 5. Therefore, its decimal expansion will be terminating. This property is a key characteristic of rational numbers that can be expressed as finite decimals.
In simple words: First, simplify the fraction as much as you can. Then, look at the bottom number (denominator). If its prime factors are only 2s and 5s, the decimal will stop (terminate).

🎯 Exam Tip: Always simplify the fraction to its lowest terms before checking the prime factors of the denominator. If a factor other than 2 or 5 remains, it will be non-terminating repeating.

Question 6. In the decimal expansion of rational number \( \frac { 43 }{ { 2 }^{ 2 }\times { 5 }^{ 3 } } \), after how many digits decimal will end?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
To find how many digits the decimal expansion will have, we need to make the powers of 2 and 5 in the denominator equal to the higher of the two powers.
The given fraction is \( \frac { 43 }{ { 2 }^{ 2 }\times { 5 }^{ 3 } } \).
Here, the power of 2 is 2 and the power of 5 is 3. The higher power is 3.
To make the powers equal, we multiply the numerator and denominator by 2 to get \( 2^3 \times 5^3 \):
\( \frac { 43 \times 2 }{ { 2 }^{ 2 } \times { 5 }^{ 3 } \times 2 } = \frac { 86 }{ { 2 }^{ 3 } \times { 5 }^{ 3 } } = \frac { 86 }{ (2 \times 5)^3 } = \frac { 86 }{ 10^3 } = \frac { 86 }{ 1000 } \)
Now, we can write this as a decimal: \( 0.086 \).
The decimal ends after 3 digits (0, 8, 6). The number of digits in the terminating decimal is determined by the highest power of 2 or 5 in the denominator after simplification.
In simple words: Look at the powers of 2 and 5 in the bottom part of the fraction. The bigger number between these powers tells you how many decimal places the number will have before it stops.

🎯 Exam Tip: The number of digits after the decimal point is equal to the maximum of the powers of 2 and 5 in the denominator (after simplifying the fraction).

Question 7. The least number, which multiplies by \( \sqrt{27} \) gives a natural number, will be :
(a) 3
(b) \( \sqrt{3} \)
(c) 9
(d) \( 3\sqrt{3} \)
Answer: (b) \( \sqrt{3} \)
First, we simplify the square root \( \sqrt{27} \):
\( \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3} \)
Now, we want to multiply \( 3\sqrt{3} \) by some number to get a natural number (a whole positive number without fractions or decimals).
If we multiply \( 3\sqrt{3} \) by \( \sqrt{3} \), we get:
\( 3\sqrt{3} \times \sqrt{3} = 3 \times (\sqrt{3} \times \sqrt{3}) = 3 \times 3 = 9 \)
Since 9 is a natural number, the least number we need to multiply by is \( \sqrt{3} \). This process helps convert irrational numbers into rational ones.
In simple words: First, break down \( \sqrt{27} \) into \( 3\sqrt{3} \). To make it a whole number, you need to multiply it by another \( \sqrt{3} \) so that \( \sqrt{3} \times \sqrt{3} \) becomes 3.

🎯 Exam Tip: To turn a number with a square root (like \( a\sqrt{b} \)) into a natural number, you usually need to multiply it by \( \sqrt{b} \).

Question 8. If HCF = LCM for two rational numbers, the numbers should be
(a) Composite
(b) Equal
(c) Prime
(d) Non-rational
Answer: (b) Equal
We know that for any two positive integers, A and B, the product of the numbers is equal to the product of their HCF and LCM:
\( A \times B = \text{HCF}(A, B) \times \text{LCM}(A, B) \)
If HCF = LCM, let's call this common value X. So, HCF(A, B) = X and LCM(A, B) = X.
Substituting this into the formula:
\( A \times B = X \times X \)
\( A \times B = X^2 \)
Since X is both the HCF and LCM, X must be a common factor and a common multiple. The only way for the HCF and LCM of two numbers to be the same is if the numbers themselves are identical. For example, HCF(5, 5) = 5 and LCM(5, 5) = 5. This property holds true for any pair of identical numbers, regardless of whether they are prime or composite.
In simple words: If the biggest number that divides two numbers (HCF) is the same as the smallest number that both numbers can divide into (LCM), then those two numbers must be exactly the same.

🎯 Exam Tip: This is a conceptual question. Remember the definition of HCF and LCM; if they are equal, it implies the numbers themselves are identical.

Question 9. If LCM of a and 18 is 36 and HCF of a and 18 is 2 then value of a will be:
(a) 1
(b) 2
(c) 5
(d) 4
Answer: (d) 4
We use the fundamental relationship between two numbers, their HCF, and their LCM:
Product of two numbers = H.C.F. \( \times \) L.C.M.
Let the two numbers be 'a' and 18.
Given: L.C.M. (a, 18) = 36
H.C.F. (a, 18) = 2
Substituting these values into the formula:
\( a \times 18 = 2 \times 36 \)
\( a \times 18 = 72 \)
To find 'a', divide 72 by 18:
\( a = \frac{72}{18} \)
\( a = 4 \)
So, the value of 'a' is 4. This principle helps us find missing number components in problems involving HCF and LCM.
In simple words: Multiply the HCF and LCM together. This answer will be the same as multiplying the two original numbers. Use this to find the missing number.

🎯 Exam Tip: Always double-check your calculations, especially multiplication and division, to ensure accuracy in these types of problems.

Question 10. If n is a natural number, then unit digit in \( 6^n - 5^n \) is:
(a) 1
(b) 6
(c) 5
(d) 9
Answer: (a) 1
Let's look at the unit digits of powers of 6 and 5:
For any natural number n, the unit digit of \( 6^n \) is always 6.
\( 6^1 = 6 \)
\( 6^2 = 36 \)
\( 6^3 = 216 \)
For any natural number n, the unit digit of \( 5^n \) is always 5.
\( 5^1 = 5 \)
\( 5^2 = 25 \)
\( 5^3 = 125 \)
Now, we want to find the unit digit of \( 6^n - 5^n \). This will be the unit digit of (Unit digit of \( 6^n \) - Unit digit of \( 5^n \)).
Unit digit = \( 6 - 5 = 1 \). This consistent pattern for unit digits simplifies calculations in number theory.
In simple words: The last digit of any power of 6 is always 6. The last digit of any power of 5 is always 5. So, if you subtract a number ending in 5 from a number ending in 6, the last digit of the answer will be 1.

🎯 Exam Tip: Understanding cyclicity of unit digits for different base numbers is very helpful for these questions. Powers of 0, 1, 5, 6 always end in their respective base digits.

Question 11. If \( \frac {p}{q} \) (q ≠ 0) is a rational number then what condition apply for q whereas \( \frac {p}{q} \) is a terminating decimal ? Is a rational number is terminating decimal.
Answer: For a rational number \( \frac {p}{q} \) (where p and q are co-prime integers and q ≠ 0) to have a terminating decimal expansion, the prime factorization of its denominator, q, must only contain powers of 2 and/or 5. This means q must be in the form \( 2^m \times 5^n \), where m and n are non-negative integers. This rule helps us quickly determine if a fraction will result in a finite decimal without performing the actual division. Not all rational numbers have terminating decimals; for example, \( \frac{1}{3} = 0.333... \) is a rational number but has a non-terminating repeating decimal.
In simple words: If a fraction can be written as a decimal that stops, then the bottom number (after simplifying the fraction) can only have 2s and 5s as its prime factors. If it has any other prime factors, the decimal will go on forever.

🎯 Exam Tip: Remember that p and q must be co-prime (no common factors other than 1) before checking the denominator's prime factors.

Question 12. Simplify \( \frac {2\sqrt{45}+3\sqrt{20} }{ 2\sqrt{5} } \) and clear whether it is rational or irrational number.
Answer: We need to simplify the expression by first simplifying the square roots in the numerator:
\( \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5} \)
\( \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5} \)
Now, substitute these back into the expression:
\( \frac {2(3\sqrt{5}) + 3(2\sqrt{5}) }{ 2\sqrt{5} } \)
\( = \frac {6\sqrt{5} + 6\sqrt{5} }{ 2\sqrt{5} } \)
Combine the terms in the numerator:
\( = \frac { (6+6)\sqrt{5} }{ 2\sqrt{5} } \)
\( = \frac { 12\sqrt{5} }{ 2\sqrt{5} } \)
Now, we can cancel \( \sqrt{5} \) from the numerator and denominator:
\( = \frac {12}{2} \)
\( = 6 \)
Since 6 can be written as \( \frac{6}{1} \), it is a rational number. Simplifying radical expressions helps reduce complex math to its simplest form.
In simple words: First, make the square roots simpler. Then, add the top parts together. After that, divide the top by the bottom. The final answer is a whole number, which means it is a rational number.

🎯 Exam Tip: Always simplify square roots as much as possible before combining or dividing to make calculations easier and avoid errors.

Question 13. Prove that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is any integer.
Answer: To prove this, we use Euclid's Division Lemma. This lemma states that for any two positive integers 'a' and 'b', there exist unique integers 'q' and 'r' such that \( a = bq + r \), where \( 0 \le r < b \).
Let 'a' be any positive odd integer.
Let \( b = 4 \).
According to Euclid's Division Lemma, \( a = 4q + r \), where \( 0 \le r < 4 \).
So, the possible values for 'r' are 0, 1, 2, or 3.
This means 'a' can be one of these forms:
If \( r = 0 \), then \( a = 4q + 0 = 4q \).
If \( r = 1 \), then \( a = 4q + 1 \).
If \( r = 2 \), then \( a = 4q + 2 \).
If \( r = 3 \), then \( a = 4q + 3 \).
We are looking for positive odd integers. Let's check which forms fit this condition:
- \( 4q \) is an even number because it is a multiple of 4.
- \( 4q + 1 \) is an odd number because an even number plus 1 is odd.
- \( 4q + 2 = 2(2q + 1) \) is an even number because it is a multiple of 2.
- \( 4q + 3 \) is an odd number because an even number plus 3 (which is odd) is odd.
Since 'a' is a positive odd integer, it cannot be of the form \( 4q \) or \( 4q + 2 \).
Therefore, any positive odd integer must be of the form \( 4q + 1 \) or \( 4q + 3 \). This demonstration shows how number properties can be rigorously proven using basic principles.
In simple words: We can divide any whole number by 4. The remainder can be 0, 1, 2, or 3. If a number is odd, its remainder when divided by 4 can only be 1 or 3, not 0 or 2, because 0 and 2 would make it even.

🎯 Exam Tip: When using Euclid's Division Lemma for proofs, clearly state the values of 'a' and 'b', list all possible remainders 'r', and then analyze each case based on the given conditions (e.g., odd or even).

Question 14. Prove that the product of two consecutive positive integers is divisible by 2.
Answer: Let the two consecutive positive integers be 'n' and 'n + 1'.
Their product is \( n(n + 1) \).
We need to show that this product is always divisible by 2.
There are two possible cases for any positive integer 'n':
Case 1: 'n' is an even number.
If 'n' is even, then by definition, 'n' can be written in the form \( 2k \) for some integer k.
So, the product becomes \( n(n + 1) = (2k)(2k + 1) \).
Since \( 2k \) is a factor in the product, \( n(n + 1) \) is clearly divisible by 2.
Case 2: 'n' is an odd number.
If 'n' is odd, then 'n + 1' must be an even number (because an odd number plus 1 is always even).
If 'n + 1' is even, then 'n + 1' can be written in the form \( 2k \) for some integer k.
So, the product becomes \( n(n + 1) = n(2k) \).
Since \( 2k \) is a factor in the product, \( n(n + 1) \) is clearly divisible by 2.
In both cases, whether 'n' is even or odd, the product \( n(n + 1) \) is always divisible by 2.
Hence proved. This fundamental property highlights the basic arithmetic structure of consecutive numbers.
In simple words: Take any two numbers that come right after each other. One of them will always be an even number (like 2, 4, 6...). When you multiply any number by an even number, the answer will always be even, meaning it can be divided by 2.

🎯 Exam Tip: For proofs involving divisibility, consider breaking the problem into cases based on whether the integer is even or odd (or its remainder when divided by the divisor).

Question 15. Find the largest number which is divided by 2053 and 967, left the remainder as 5 and 7 respectively.
Answer: We are looking for the largest number that divides 2053 leaving a remainder of 5, and divides 967 leaving a remainder of 7.
If a number divides 2053 with a remainder of 5, it means that \( 2053 - 5 = 2048 \) is exactly divisible by that number.
If a number divides 967 with a remainder of 7, it means that \( 967 - 7 = 960 \) is exactly divisible by that number.
So, we need to find the largest number that divides both 2048 and 960 exactly. This is the definition of the Highest Common Factor (HCF).
Let's find the prime factorization of 2048 and 960:
\( 2048 = 2^{11} \) (since \( 2^{10} = 1024 \), \( 2^{11} = 2048 \))
\( 960 = 96 \times 10 = (2^5 \times 3) \times (2 \times 5) = 2^6 \times 3 \times 5 \)
The HCF is found by taking the lowest power of each common prime factor.
The only common prime factor is 2. The lowest power of 2 common to both is \( 2^6 \).
\( \text{HCF} = 2^6 = 64 \)
So, the largest number is 64. This problem demonstrates a practical application of finding the HCF in remainder-based scenarios.
In simple words: First, subtract the remainders from each big number. This gives you new numbers that can be divided perfectly. Then, find the biggest number that can divide both of these new numbers (their HCF).

🎯 Exam Tip: When a problem states remainders, always subtract the remainders from the given numbers first. The HCF of these 'adjusted' numbers will be the answer.

Question 16. Explain, why \( 7 \times 11 \times 13 + 13 \) and \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \) are composite numbers?
Answer: A composite number is a positive integer that has at least one divisor other than 1 and itself. According to the Fundamental Theorem of Arithmetic, every composite number can be uniquely expressed as a product of prime numbers.
Let's examine the first expression: \( 7 \times 11 \times 13 + 13 \)
We can factor out 13 from both terms:
\( = 13 \times (7 \times 11 + 1) \)
\( = 13 \times (77 + 1) \)
\( = 13 \times 78 \)
Now, we can find the prime factors of 78:
\( 78 = 2 \times 39 = 2 \times 3 \times 13 \)
So, the expression becomes \( 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2 \).
Since \( 7 \times 11 \times 13 + 13 \) can be expressed as a product of prime numbers (2, 3, and 13), and it has factors other than 1 and itself (e.g., 2, 3, 13, 78), it is a composite number.

Now let's examine the second expression: \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \)
We can see that the number 5 appears as a factor. We can rewrite the expression by grouping common factors:
\( = (7 \times 6 \times 4 \times 3 \times 2 \times 1) \times (5 \times 5) \)
\( = (7 \times 6 \times 4 \times 3 \times 2 \times 1) \times 5^2 \)
Since the expression has 5 as a factor (and thus \( 5^2 \) and other numbers like \( 7 \times 6 \times 4 \times 3 \times 2 \times 1 \) are factors), it means it has factors other than 1 and itself. For example, 5 is a factor. Any number that has factors other than 1 and itself is a composite number. The presence of multiple prime factors confirms its composite nature.
In simple words: A composite number can be broken down into smaller whole numbers multiplied together, apart from just 1 and itself. For the first number, we can pull out 13, showing it has other factors. For the second number, it clearly has 5 as a factor, which means it's not a prime number.

🎯 Exam Tip: To prove a number is composite, simply show that it has at least one factor other than 1 and the number itself. Factoring out common terms is often the easiest way.

Question 17. If HCF of two numbers 306 and 657 is 9, then find their LCM.
Answer: We use the key relationship between two numbers, their HCF, and their LCM:
Product of the two numbers = H.C.F. \( \times \) L.C.M.
Given:
First number = 306
Second number = 657
H.C.F. (306, 657) = 9
Let L.C.M. (306, 657) = X
Substitute the values into the formula:
\( 306 \times 657 = 9 \times X \)
To find X, divide the product of the numbers by their HCF:
\( X = \frac{306 \times 657}{9} \)
We can simplify by dividing 306 by 9:
\( \frac{306}{9} = 34 \)
So, \( X = 34 \times 657 \)
\( X = 22338 \)
Therefore, the LCM of 306 and 657 is 22338. This formula is very useful for quickly calculating one value if the others are known.
In simple words: If you multiply the two numbers, the answer is the same as multiplying their HCF and LCM. So, if you know three of these, you can easily find the fourth one by dividing.

🎯 Exam Tip: This formula works only for two positive integers. Always use the numbers given in the question and their HCF/LCM correctly.

Question 18. A rectangular veranda is of dimension 18 m 72 cm x 13 m 30 cm. Squared tiles are used to cover them. Find the least number of such tiles.
Answer: To find the least number of square tiles, we need to find the largest possible size for a square tile. The side length of this largest square tile will be the HCF of the length and breadth of the veranda.
First, convert the dimensions of the veranda from meters and centimeters to just centimeters:
Length of veranda = 18 m 72 cm = \( (18 \times 100) \) cm + 72 cm = \( 1800 + 72 \) cm = 1872 cm
Breadth of veranda = 13 m 20 cm = \( (13 \times 100) \) cm + 20 cm = \( 1300 + 20 \) cm = 1320 cm

Next, find the HCF of 1872 and 1320. We use prime factorization:
\( 1872 = 2^4 \times 3^2 \times 13 \)
\( 1320 = 2^3 \times 3^1 \times 5^1 \times 11^1 \)
The HCF is the product of the lowest powers of the common prime factors:
\( \text{HCF}(1872, 1320) = 2^3 \times 3^1 = 8 \times 3 = 24 \)
So, the side length of the largest square tile is 24 cm. This ensures the tiles fit perfectly without any cutting.

Now, calculate the number of tiles needed:
Number of tiles = \( \frac{\text{Area of veranda}}{\text{Area of one tile}} \)
Area of veranda = Length \( \times \) Breadth = \( 1872 \text{ cm} \times 1320 \text{ cm} \)
Area of one tile = Side \( \times \) Side = \( 24 \text{ cm} \times 24 \text{ cm} \)
Number of tiles = \( \frac{1872 \times 1320}{24 \times 24} \)
\( = \frac{1872}{24} \times \frac{1320}{24} \)
\( = 78 \times 55 \)
\( = 4290 \)
Therefore, the least number of tiles required is 4290. This problem is a classic example of using HCF to solve real-world measurement challenges.
In simple words: First, change meters to centimeters for both the length and width of the floor. Then, find the biggest number that can divide both the length and width exactly (this is the HCF). This HCF will be the side length of your square tiles. Finally, divide the total area of the floor by the area of one tile to find how many tiles you need.

🎯 Exam Tip: For problems asking for the "least number of items" to cover a larger area, you typically need to find the HCF of the dimensions to determine the largest possible size of the covering item.

Question 19. Prove that following numbers are irrational numbers
(i) \( 5\sqrt{2} \)
(ii) \( \frac {2}{\sqrt{7}} \)
(iii) \( \frac {3}{2\sqrt{5}} \)
(iv) \( 4 + \sqrt{2} \)
Answer:
(i) Proof for \( 5\sqrt{2} \):
Let's assume, to the contrary, that \( 5\sqrt{2} \) is a rational number.
If \( 5\sqrt{2} \) is rational, then it can be written in the form \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( 5\sqrt{2} = \frac{a}{b} \)
Now, isolate \( \sqrt{2} \):
\( \sqrt{2} = \frac{a}{5b} \)
Since 'a' and 'b' are integers, and 5 is an integer, \( \frac{a}{5b} \) is a rational number (as it's a ratio of integers).
This means that \( \sqrt{2} \) must also be a rational number. However, we already know that \( \sqrt{2} \) is an irrational number (a number that cannot be expressed as a simple fraction).
This creates a contradiction. Our initial assumption that \( 5\sqrt{2} \) is rational must therefore be false.
Hence, \( 5\sqrt{2} \) is an irrational number. This method of proof by contradiction is common for showing numbers are irrational.

(ii) Proof for \( \frac {2}{\sqrt{7}} \):
Let's assume, to the contrary, that \( \frac {2}{\sqrt{7}} \) is a rational number.
If \( \frac {2}{\sqrt{7}} \) is rational, then it can be written in the form \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( \frac {2}{\sqrt{7}} = \frac{a}{b} \)
Now, rearrange the equation to isolate \( \sqrt{7} \):
\( 2b = a\sqrt{7} \)
\( \sqrt{7} = \frac{2b}{a} \)
Since 'a' and 'b' are integers, and 2 is an integer, \( \frac{2b}{a} \) is a rational number (as it's a ratio of integers).
This implies that \( \sqrt{7} \) must be a rational number. But we know that \( \sqrt{7} \) is an irrational number.
This leads to a contradiction. Therefore, our initial assumption that \( \frac {2}{\sqrt{7}} \) is rational must be false.
Hence, \( \frac {2}{\sqrt{7}} \) is an irrational number.

(iii) Proof for \( \frac {3}{2\sqrt{5}} \):
Let's assume, to the contrary, that \( \frac {3}{2\sqrt{5}} \) is a rational number.
If \( \frac {3}{2\sqrt{5}} \) is rational, then it can be written in the form \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( \frac {3}{2\sqrt{5}} = \frac{a}{b} \)
Now, rearrange the equation to isolate \( \sqrt{5} \):
\( 3b = 2a\sqrt{5} \)
\( \sqrt{5} = \frac{3b}{2a} \)
Since 'a' and 'b' are integers, and 2 and 3 are integers, \( \frac{3b}{2a} \) is a rational number (as it's a ratio of integers).
This implies that \( \sqrt{5} \) must be a rational number. However, we know that \( \sqrt{5} \) is an irrational number.
This leads to a contradiction. Therefore, our initial assumption that \( \frac {3}{2\sqrt{5}} \) is rational must be false.
Hence, \( \frac {3}{2\sqrt{5}} \) is an irrational number.

(iv) Proof for \( 4 + \sqrt{2} \):
Let's assume, to the contrary, that \( 4 + \sqrt{2} \) is a rational number.
If \( 4 + \sqrt{2} \) is rational, then it can be written in the form \( \frac{a}{b} \), where 'a' and 'b' are co-prime integers and \( b \neq 0 \).
So, \( 4 + \sqrt{2} = \frac{a}{b} \)
Now, isolate \( \sqrt{2} \):
\( \sqrt{2} = \frac{a}{b} - 4 \)
To combine the terms on the right side, find a common denominator:
\( \sqrt{2} = \frac{a - 4b}{b} \)
Since 'a' and 'b' are integers, \( (a - 4b) \) is also an integer, and 'b' is a non-zero integer. Therefore, \( \frac{a - 4b}{b} \) is a rational number.
This means that \( \sqrt{2} \) must also be a rational number. However, we know that \( \sqrt{2} \) is an irrational number.
This leads to a contradiction. Our initial assumption that \( 4 + \sqrt{2} \) is rational must be false.
Hence, \( 4 + \sqrt{2} \) is an irrational number. The sum or difference of a rational and an irrational number is always irrational.
In simple words: To prove these numbers are irrational, we pretend they are rational (can be written as a fraction). Then we move things around to show that a known irrational number (like \( \sqrt{2} \), \( \sqrt{7} \), or \( \sqrt{5} \)) would have to be rational, which we know is wrong. This contradiction means our first guess was false, and the numbers are indeed irrational.

🎯 Exam Tip: The core idea for proving irrationality is contradiction: assume it's rational, manipulate the equation to show an irrational number equals a rational expression, and then conclude your initial assumption was false.

Question 20. What can you say about the prime factors of the denominator of the following rational numbers.
(i) A rational number with a terminating decimal (e.g., \( \frac{P}{100000} \))
(ii) \( 43.\overline{123456789} \)
Answer:
(i) For a rational number whose decimal expansion terminates, the prime factors of its denominator (when the fraction is in its simplest form) must only be 2s and 5s. For example, if we consider a number like \( \frac{P}{100000} \) which terminates, the denominator is \( 100000 = 10^5 = (2 \times 5)^5 = 2^5 \times 5^5 \). In this case, the prime factors of the denominator are only 2 and 5. This property is fundamental to understanding terminating decimals.

(ii) For the number \( 43.\overline{123456789} \), its decimal expansion is non-terminating and repeating. A rational number has a non-terminating repeating decimal expansion if, and only if, the prime factors of its denominator (when the fraction is in its simplest form) contain at least one prime factor other than 2 or 5. If we were to convert \( 43.\overline{123456789} \) into a fraction \( \frac{p}{q} \), the denominator 'q' would have prime factors in addition to, or instead of, 2 and 5. This is because the repeating block in the decimal indicates the presence of other prime factors in the denominator.
In simple words: (i) If a decimal stops, its fraction's bottom number (denominator) can only have 2 and 5 as its prime parts. (ii) If a decimal goes on forever but repeats, its fraction's bottom number must have other prime parts besides 2 and 5.

🎯 Exam Tip: Distinguish carefully between terminating and non-terminating repeating decimals based on the prime factors of their denominators. Always simplify the fraction first.