Real Numbers 1 mark Solutions
1. Insert 4 rational numbers between 3/4 and 1 without using formula a + b/2 (M’15)
We convert to a common denominator:
3/4 = 15/20, 1 = 20/20
Choose four numbers between 15/20 and 20/20:
16/20, 17/20, 18/20, 19/20
Simplifying: 4/5, 17/20, 9/10, 19/20
3/4 = 15/20, 1 = 20/20
Choose four numbers between 15/20 and 20/20:
16/20, 17/20, 18/20, 19/20
Simplifying: 4/5, 17/20, 9/10, 19/20
2. The prime factorization of a natural number(n) is 2³ × 3² × 5² × 7. How many consecutive zeroes will it have at the end of it? Justify your answer. (J’15)
Trailing zeros are determined by the number of pairs of factors 2 × 5.
Here, exponent of 2 is 3, exponent of 5 is 2.
So, number of trailing zeros = min(3, 2) = 2
Here, exponent of 2 is 3, exponent of 5 is 2.
So, number of trailing zeros = min(3, 2) = 2
3. Find the value of log₅ 125 (M’16)
125 = 5³ ⇒ log₅ 125 = 3
4. Write any two irrational numbers lying between 3 and 4 (J’16)
Examples:
√10 (≈ 3.162) and π (≈ 3.1416)
5. Find the value of log√₂ 256 (M’17)
√2 = 2¹ᐟ², 256 = 2⁸ ⇒ log√₂ 256 = 8/(1/2) = 16
6. Find the HCF and LCM of 90, 144 by prime factorization method (J’17)
90 = 2 × 3² × 5, 144 = 2⁴ × 3²
HCF = common primes with lowest exponents: 2¹ × 3² = 18
LCM = all primes with highest exponents: 2⁴ × 3² × 5 = 720
HCF = common primes with lowest exponents: 2¹ × 3² = 18
LCM = all primes with highest exponents: 2⁴ × 3² × 5 = 720
7. Is log₃ 81 rational or irrational? Justify your answer. (J’17)
81 = 3⁴ ⇒ log₃ 81 = 4, which is a whole number.
Therefore, it is rational.
Therefore, it is rational.
8. Expand log₁₀ 385 (M’18)
385 = 5 × 7 × 11 ⇒ log₁₀ 385 = log₁₀ 5 + log₁₀ 7 + log₁₀ 11
9. Find the value of log√₂ 128 (J’18)
√2 = 2¹ᐟ², 128 = 2⁷ ⇒ log√₂ 128 = 7/(1/2) = 14
10. Find the HCF of 24 and 33 by using division algorithm (M’19)
Using Euclidean algorithm:
33 ÷ 24 → remainder 9
24 ÷ 9 → remainder 6
9 ÷ 6 → remainder 3
6 ÷ 3 → remainder 0
HCF = 3
33 ÷ 24 → remainder 9
24 ÷ 9 → remainder 6
9 ÷ 6 → remainder 3
6 ÷ 3 → remainder 0
HCF = 3
11. Ramu says, “If log₁₀ x = 0, value of x = 0”. Do you agree with him? Give reason. (J’19)
log₁₀ x = 0 ⇒ x = 10⁰ = 1 ≠ 0
Therefore, I disagree with Ramu. The correct value is 1.
Therefore, I disagree with Ramu. The correct value is 1.
12. Expand log(a³b²c⁵) (May 2022)
log(a³b²c⁵) = 3 log a + 2 log b + 5 log c
13. Expand log(32/81) (Aug. 2022)
log(32/81) = log(32) – log(81) = log(2⁵) – log(3⁴) = 5 log 2 – 4 log 3
14. Find the mean of the factors of 24 (Jun’23)
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60
Number of factors = 8
Mean = 60/8 = 7.5
Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60
Number of factors = 8
Mean = 60/8 = 7.5
Math Problems – 2 Mark Questions
1
Write any three numbers of two digits. Find the L.C.M. and H.C.F. for the above numbers by the “Prime factorization method”.
(M’15)
Step 1: Choose three two-digit numbers
Let’s take: 12, 18, and 24
Let’s take: 12, 18, and 24
Step 2: Prime Factorization
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²
24 = 2 × 2 × 2 × 3 = 2³ × 3
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²
24 = 2 × 2 × 2 × 3 = 2³ × 3
Step 3: Find H.C.F. (Highest Common Factor)
H.C.F. = Common prime factors with lowest powers
Common factors: 2 and 3
Lowest power of 2: 2¹
Lowest power of 3: 3¹
∴ H.C.F. = 2¹ × 3¹ = 6
H.C.F. = Common prime factors with lowest powers
Common factors: 2 and 3
Lowest power of 2: 2¹
Lowest power of 3: 3¹
∴ H.C.F. = 2¹ × 3¹ = 6
Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2 and 3
Highest power of 2: 2³
Highest power of 3: 3²
∴ L.C.M. = 2³ × 3² = 8 × 9 = 72
L.C.M. = All prime factors with highest powers
Prime factors: 2 and 3
Highest power of 2: 2³
Highest power of 3: 3²
∴ L.C.M. = 2³ × 3² = 8 × 9 = 72
2
Give an example for each of the following:
i) The product of two rational numbers is a rational number.
ii) The product of two irrational numbers is an irrational number. (M’15)
i) The product of two rational numbers is a rational number.
ii) The product of two irrational numbers is an irrational number. (M’15)
i) Product of two rational numbers is rational:
Let’s take two rational numbers: 2/3 and 4/5
Product = (2/3) × (4/5) = 8/15
8/15 is in the form p/q where p and q are integers and q ≠ 0
∴ 8/15 is a rational number
Let’s take two rational numbers: 2/3 and 4/5
Product = (2/3) × (4/5) = 8/15
8/15 is in the form p/q where p and q are integers and q ≠ 0
∴ 8/15 is a rational number
ii) Product of two irrational numbers is irrational:
Let’s take two irrational numbers: √2 and √3
Product = √2 × √3 = √6
√6 cannot be expressed as a ratio of two integers
∴ √6 is an irrational number
Let’s take two irrational numbers: √2 and √3
Product = √2 × √3 = √6
√6 cannot be expressed as a ratio of two integers
∴ √6 is an irrational number
3
State with reasons which of the following are rational numbers and which are irrational numbers:
(i) √225 × √4
(ii) 6√50 + 8√125 (J’15)
(i) √225 × √4
(ii) 6√50 + 8√125 (J’15)
(i) √225 × √4
√225 = 15 (since 15² = 225)
√4 = 2 (since 2² = 4)
Product = 15 × 2 = 30
30 can be expressed as 30/1 (p/q form where p and q are integers, q ≠ 0)
∴ √225 × √4 = 30 is a rational number
√225 = 15 (since 15² = 225)
√4 = 2 (since 2² = 4)
Product = 15 × 2 = 30
30 can be expressed as 30/1 (p/q form where p and q are integers, q ≠ 0)
∴ √225 × √4 = 30 is a rational number
(ii) 6√50 + 8√125
Simplify each term:
√50 = √(25 × 2) = 5√2
√125 = √(25 × 5) = 5√5
∴ 6√50 + 8√125 = 6(5√2) + 8(5√5) = 30√2 + 40√5
This is a sum of irrational numbers where the irrational parts don’t cancel out
∴ 6√50 + 8√125 is an irrational number
Simplify each term:
√50 = √(25 × 2) = 5√2
√125 = √(25 × 5) = 5√5
∴ 6√50 + 8√125 = 6(5√2) + 8(5√5) = 30√2 + 40√5
This is a sum of irrational numbers where the irrational parts don’t cancel out
∴ 6√50 + 8√125 is an irrational number
4
If x² + y² = 7xy then show that 2 log(x + y) = log x + log y + 2 log 3
(M’16)
Given: x² + y² = 7xy
Step 1: Add 2xy to both sides
x² + y² + 2xy = 7xy + 2xy
(x + y)² = 9xy
x² + y² + 2xy = 7xy + 2xy
(x + y)² = 9xy
Step 2: Take logarithm on both sides
log[(x + y)²] = log(9xy)
2 log(x + y) = log 9 + log x + log y
2 log(x + y) = log(3²) + log x + log y
2 log(x + y) = 2 log 3 + log x + log y
log[(x + y)²] = log(9xy)
2 log(x + y) = log 9 + log x + log y
2 log(x + y) = log(3²) + log x + log y
2 log(x + y) = 2 log 3 + log x + log y
∴ 2 log(x + y) = log x + log y + 2 log 3
Hence proved.
Hence proved.
5
Express 2016 as product of prime factors
(J’16)
Step 1: Divide 2016 by smallest prime factor
2016 ÷ 2 = 1008
1008 ÷ 2 = 504
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
2016 ÷ 2 = 1008
1008 ÷ 2 = 504
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
Step 2: Write the prime factorization
2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7
∴ 2016 = 2⁵ × 3² × 7
2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7
∴ 2016 = 2⁵ × 3² × 7
6
Write any two three-digit numbers. Find their L.C.M. and G.C.D. by prime factorization method.
(M’17)
Step 1: Choose two three-digit numbers
Let’s take: 120 and 150
Let’s take: 120 and 150
Step 2: Prime Factorization
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5²
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5²
Step 3: Find G.C.D. (Greatest Common Divisor)
G.C.D. = Common prime factors with lowest powers
Common factors: 2, 3, and 5
Lowest power of 2: 2¹
Lowest power of 3: 3¹
Lowest power of 5: 5¹
∴ G.C.D. = 2 × 3 × 5 = 30
G.C.D. = Common prime factors with lowest powers
Common factors: 2, 3, and 5
Lowest power of 2: 2¹
Lowest power of 3: 3¹
Lowest power of 5: 5¹
∴ G.C.D. = 2 × 3 × 5 = 30
Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2³
Highest power of 3: 3¹
Highest power of 5: 5²
∴ L.C.M. = 2³ × 3 × 5² = 8 × 3 × 25 = 600
L.C.M. = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2³
Highest power of 3: 3¹
Highest power of 5: 5²
∴ L.C.M. = 2³ × 3 × 5² = 8 × 3 × 25 = 600
7
Prove that 2 + √3 is irrational
(J’17)
Proof by Contradiction:
Assume that 2 + √3 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, 2 + √3 = p/q
⇒ √3 = p/q – 2
⇒ √3 = (p – 2q)/q
Since p and q are integers, (p – 2q)/q is a rational number.
This means √3 is rational.
But we know that √3 is irrational (proved separately).
This is a contradiction.
∴ Our assumption that 2 + √3 is rational must be false.
Hence, 2 + √3 is irrational.
Assume that 2 + √3 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, 2 + √3 = p/q
⇒ √3 = p/q – 2
⇒ √3 = (p – 2q)/q
Since p and q are integers, (p – 2q)/q is a rational number.
This means √3 is rational.
But we know that √3 is irrational (proved separately).
This is a contradiction.
∴ Our assumption that 2 + √3 is rational must be false.
Hence, 2 + √3 is irrational.
8
Lalitha says that HCF and LCM of the numbers 80 and 60 are 20 and 120 respectively. Do you agree with her? Justify
(J’18)
Let’s verify Lalitha’s claim by calculating HCF and LCM of 80 and 60
Step 1: Prime Factorization
80 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
80 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
Step 2: Calculate HCF
HCF = Common prime factors with lowest powers
Common factors: 2 and 5
Lowest power of 2: 2²
Lowest power of 5: 5¹
∴ HCF = 2² × 5 = 4 × 5 = 20
HCF = Common prime factors with lowest powers
Common factors: 2 and 5
Lowest power of 2: 2²
Lowest power of 5: 5¹
∴ HCF = 2² × 5 = 4 × 5 = 20
Step 3: Calculate LCM
LCM = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2⁴
Highest power of 3: 3¹
Highest power of 5: 5¹
∴ LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240
LCM = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2⁴
Highest power of 3: 3¹
Highest power of 5: 5¹
∴ LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240
Conclusion:
Lalitha is correct about HCF (20) but incorrect about LCM.
The correct LCM is 240, not 120.
∴ I don’t agree with Lalitha.
Lalitha is correct about HCF (20) but incorrect about LCM.
The correct LCM is 240, not 120.
∴ I don’t agree with Lalitha.
9
If x² + y² = 10xy, prove that 2 log (x + y) = log x + log y + 2 log 2 + log 3
(J’19, Apr’23)
Given: x² + y² = 10xy
Step 1: Add 2xy to both sides
x² + y² + 2xy = 10xy + 2xy
(x + y)² = 12xy
x² + y² + 2xy = 10xy + 2xy
(x + y)² = 12xy
Step 2: Take logarithm on both sides
log[(x + y)²] = log(12xy)
2 log(x + y) = log 12 + log x + log y
2 log(x + y) = log(4 × 3) + log x + log y
2 log(x + y) = log 4 + log 3 + log x + log y
2 log(x + y) = log(2²) + log 3 + log x + log y
2 log(x + y) = 2 log 2 + log 3 + log x + log y
log[(x + y)²] = log(12xy)
2 log(x + y) = log 12 + log x + log y
2 log(x + y) = log(4 × 3) + log x + log y
2 log(x + y) = log 4 + log 3 + log x + log y
2 log(x + y) = log(2²) + log 3 + log x + log y
2 log(x + y) = 2 log 2 + log 3 + log x + log y
∴ 2 log(x + y) = log x + log y + 2 log 2 + log 3
Hence proved.
Hence proved.
10
If 2304 = 2ˣ × 3ʸ then find the value of logʸₓ
(Jun’23)
Step 1: Prime factorization of 2304
2304 ÷ 2 = 1152
1152 ÷ 2 = 576
576 ÷ 2 = 288
288 ÷ 2 = 144
144 ÷ 2 = 72
72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
2304 ÷ 2 = 1152
1152 ÷ 2 = 576
576 ÷ 2 = 288
288 ÷ 2 = 144
144 ÷ 2 = 72
72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
Step 2: Write the prime factorization
So, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
∴ 2304 = 2⁸ × 3²
So, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
∴ 2304 = 2⁸ × 3²
Step 3: Compare with given expression
Comparing with 2304 = 2ˣ × 3ʸ:
x = 8, y = 2
Comparing with 2304 = 2ˣ × 3ʸ:
x = 8, y = 2
Step 4: Find logʸₓ
We need to find logᵧx = log₂8
Since 2³ = 8, log₂8 = 3
We need to find logᵧx = log₂8
Since 2³ = 8, log₂8 = 3
∴ logʸₓ = 3
Real Numbers 4 marks Answers
1
Prove that 3 + 2√5 is an irrational number.
(M’15)
Proof by Contradiction:
Assume that 3 + 2√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, 3 + 2√5 = p/q
⇒ 2√5 = p/q – 3
⇒ 2√5 = (p – 3q)/q
⇒ √5 = (p – 3q)/(2q)
Since p and q are integers, (p – 3q)/(2q) is a rational number.
This means √5 is rational.
But we know that √5 is irrational (proved separately).
This is a contradiction.
∴ Our assumption that 3 + 2√5 is rational must be false.
Hence, 3 + 2√5 is irrational.
Assume that 3 + 2√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, 3 + 2√5 = p/q
⇒ 2√5 = p/q – 3
⇒ 2√5 = (p – 3q)/q
⇒ √5 = (p – 3q)/(2q)
Since p and q are integers, (p – 3q)/(2q) is a rational number.
This means √5 is rational.
But we know that √5 is irrational (proved separately).
This is a contradiction.
∴ Our assumption that 3 + 2√5 is rational must be false.
Hence, 3 + 2√5 is irrational.
2
Expand log(1125/32)
(J’15)
Step 1: Prime factorization of numerator and denominator
1125 = 9 × 125 = 3² × 5³
32 = 2⁵
1125 = 9 × 125 = 3² × 5³
32 = 2⁵
Step 2: Apply logarithm properties
log(1125/32) = log(1125) – log(32)
= log(3² × 5³) – log(2⁵)
= log(3²) + log(5³) – log(2⁵)
= 2 log 3 + 3 log 5 – 5 log 2
log(1125/32) = log(1125) – log(32)
= log(3² × 5³) – log(2⁵)
= log(3²) + log(5³) – log(2⁵)
= 2 log 3 + 3 log 5 – 5 log 2
∴ log(1125/32) = 2 log 3 + 3 log 5 – 5 log 2
3
Express the numbers 6825 and 3825 as a product of its prime factors. Find the HCF and LCM of the above numbers by using their products of prime factors. Justify your answer.
(J’15)
Step 1: Prime factorization of 6825
6825 ÷ 3 = 2275
2275 ÷ 5 = 455
455 ÷ 5 = 91
91 ÷ 7 = 13
13 ÷ 13 = 1
∴ 6825 = 3 × 5² × 7 × 13
6825 ÷ 3 = 2275
2275 ÷ 5 = 455
455 ÷ 5 = 91
91 ÷ 7 = 13
13 ÷ 13 = 1
∴ 6825 = 3 × 5² × 7 × 13
Step 2: Prime factorization of 3825
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 ÷ 17 = 1
∴ 3825 = 3² × 5² × 17
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 ÷ 17 = 1
∴ 3825 = 3² × 5² × 17
Step 3: Find HCF (Highest Common Factor)
HCF = Common prime factors with lowest powers
Common factors: 3 and 5
Lowest power of 3: 3¹
Lowest power of 5: 5²
∴ HCF = 3¹ × 5² = 75
HCF = Common prime factors with lowest powers
Common factors: 3 and 5
Lowest power of 3: 3¹
Lowest power of 5: 5²
∴ HCF = 3¹ × 5² = 75
Step 4: Find LCM (Lowest Common Multiple)
LCM = All prime factors with highest powers
Prime factors: 3, 5, 7, 13, 17
Highest power of 3: 3²
Highest power of 5: 5²
Highest power of 7: 7¹
Highest power of 13: 13¹
Highest power of 17: 17¹
∴ LCM = 3² × 5² × 7 × 13 × 17 = 348075
LCM = All prime factors with highest powers
Prime factors: 3, 5, 7, 13, 17
Highest power of 3: 3²
Highest power of 5: 5²
Highest power of 7: 7¹
Highest power of 13: 13¹
Highest power of 17: 17¹
∴ LCM = 3² × 5² × 7 × 13 × 17 = 348075
4
Use Euclid’s division Lemma to show that the cube of any positive integer is of the form 7m or 7m + 1 or 7m + 6.
(M’16)
Proof:
Let a be any positive integer.
By Euclid’s division lemma, a = 7q + r, where 0 ≤ r < 7
Case 1: r = 0
a = 7q
a³ = (7q)³ = 343q³ = 7(49q³) = 7m, where m = 49q³
Case 2: r = 1
a = 7q + 1
a³ = (7q + 1)³ = 343q³ + 147q² + 21q + 1 = 7(49q³ + 21q² + 3q) + 1 = 7m + 1
Case 3: r = 2
a = 7q + 2
a³ = (7q + 2)³ = 343q³ + 294q² + 84q + 8 = 7(49q³ + 42q² + 12q + 1) + 1 = 7m + 1
Case 4: r = 3
a = 7q + 3
a³ = (7q + 3)³ = 343q³ + 441q² + 189q + 27 = 7(49q³ + 63q² + 27q + 3) + 6 = 7m + 6
Case 5: r = 4
a = 7q + 4
a³ = (7q + 4)³ = 343q³ + 588q² + 336q + 64 = 7(49q³ + 84q² + 48q + 9) + 1 = 7m + 1
Case 6: r = 5
a = 7q + 5
a³ = (7q + 5)³ = 343q³ + 735q² + 525q + 125 = 7(49q³ + 105q² + 75q + 17) + 6 = 7m + 6
Case 7: r = 6
a = 7q + 6
a³ = (7q + 6)³ = 343q³ + 882q² + 756q + 216 = 7(49q³ + 126q² + 108q + 30) + 6 = 7m + 6
In all cases, a³ is of the form 7m, 7m + 1, or 7m + 6.
Hence proved.
Let a be any positive integer.
By Euclid’s division lemma, a = 7q + r, where 0 ≤ r < 7
Case 1: r = 0
a = 7q
a³ = (7q)³ = 343q³ = 7(49q³) = 7m, where m = 49q³
Case 2: r = 1
a = 7q + 1
a³ = (7q + 1)³ = 343q³ + 147q² + 21q + 1 = 7(49q³ + 21q² + 3q) + 1 = 7m + 1
Case 3: r = 2
a = 7q + 2
a³ = (7q + 2)³ = 343q³ + 294q² + 84q + 8 = 7(49q³ + 42q² + 12q + 1) + 1 = 7m + 1
Case 4: r = 3
a = 7q + 3
a³ = (7q + 3)³ = 343q³ + 441q² + 189q + 27 = 7(49q³ + 63q² + 27q + 3) + 6 = 7m + 6
Case 5: r = 4
a = 7q + 4
a³ = (7q + 4)³ = 343q³ + 588q² + 336q + 64 = 7(49q³ + 84q² + 48q + 9) + 1 = 7m + 1
Case 6: r = 5
a = 7q + 5
a³ = (7q + 5)³ = 343q³ + 735q² + 525q + 125 = 7(49q³ + 105q² + 75q + 17) + 6 = 7m + 6
Case 7: r = 6
a = 7q + 6
a³ = (7q + 6)³ = 343q³ + 882q² + 756q + 216 = 7(49q³ + 126q² + 108q + 30) + 6 = 7m + 6
In all cases, a³ is of the form 7m, 7m + 1, or 7m + 6.
Hence proved.
5
Prove that √2 − 3√5 is an irrational number.
(M’16)
Proof by Contradiction:
Assume that √2 − 3√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √2 − 3√5 = p/q
⇒ √2 = p/q + 3√5
⇒ √2 = (p + 3q√5)/q
⇒ q√2 = p + 3q√5
⇒ q√2 – 3q√5 = p
⇒ q(√2 – 3√5) = p
Since p and q are integers, the left side must be rational.
But √2 and √5 are irrational, so their combination √2 – 3√5 is irrational.
The product of a rational number (q) and an irrational number (√2 – 3√5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √2 − 3√5 is rational must be false.
Hence, √2 − 3√5 is irrational.
Assume that √2 − 3√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √2 − 3√5 = p/q
⇒ √2 = p/q + 3√5
⇒ √2 = (p + 3q√5)/q
⇒ q√2 = p + 3q√5
⇒ q√2 – 3q√5 = p
⇒ q(√2 – 3√5) = p
Since p and q are integers, the left side must be rational.
But √2 and √5 are irrational, so their combination √2 – 3√5 is irrational.
The product of a rational number (q) and an irrational number (√2 – 3√5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √2 − 3√5 is rational must be false.
Hence, √2 − 3√5 is irrational.
6
Use Euclid’s division lemma, show that the cube of any positive integer is of the form 3p or 3p + 1 or 3p + 2 for any integer ‘p’.
(J’16)
Proof:
Let a be any positive integer.
By Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3
Case 1: r = 0
a = 3q
a³ = (3q)³ = 27q³ = 3(9q³) = 3p, where p = 9q³
Case 2: r = 1
a = 3q + 1
a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 3(9q³ + 9q² + 3q) + 1 = 3p + 1
Case 3: r = 2
a = 3q + 2
a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 3(9q³ + 18q² + 12q + 2) + 2 = 3p + 2
In all cases, a³ is of the form 3p, 3p + 1, or 3p + 2.
Hence proved.
Let a be any positive integer.
By Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3
Case 1: r = 0
a = 3q
a³ = (3q)³ = 27q³ = 3(9q³) = 3p, where p = 9q³
Case 2: r = 1
a = 3q + 1
a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 3(9q³ + 9q² + 3q) + 1 = 3p + 1
Case 3: r = 2
a = 3q + 2
a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 3(9q³ + 18q² + 12q + 2) + 2 = 3p + 2
In all cases, a³ is of the form 3p, 3p + 1, or 3p + 2.
Hence proved.
7
Prove that √3 – √5 is an irrational number.
(J’16)
Proof by Contradiction:
Assume that √3 – √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √3 – √5 = p/q
⇒ √3 = p/q + √5
⇒ √3 = (p + q√5)/q
⇒ q√3 = p + q√5
⇒ q√3 – q√5 = p
⇒ q(√3 – √5) = p
Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 – √5 is irrational.
The product of a rational number (q) and an irrational number (√3 – √5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √3 – √5 is rational must be false.
Hence, √3 – √5 is irrational.
Assume that √3 – √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √3 – √5 = p/q
⇒ √3 = p/q + √5
⇒ √3 = (p + q√5)/q
⇒ q√3 = p + q√5
⇒ q√3 – q√5 = p
⇒ q(√3 – √5) = p
Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 – √5 is irrational.
The product of a rational number (q) and an irrational number (√3 – √5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √3 – √5 is rational must be false.
Hence, √3 – √5 is irrational.
8
Use Euclid’s division lemma to show that the square of any positive integer is of the form 5n or 5n + 1 or 5n + 4 where n is a whole number.
(M’17) & (J’19)
Proof:
Let a be any positive integer.
By Euclid’s division lemma, a = 5q + r, where 0 ≤ r < 5
Case 1: r = 0
a = 5q
a² = (5q)² = 25q² = 5(5q²) = 5n, where n = 5q²
Case 2: r = 1
a = 5q + 1
a² = (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5n + 1
Case 3: r = 2
a = 5q + 2
a² = (5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5n + 4
Case 4: r = 3
a = 5q + 3
a² = (5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5n + 4
Case 5: r = 4
a = 5q + 4
a² = (5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5n + 1
In all cases, a² is of the form 5n, 5n + 1, or 5n + 4.
Hence proved.
Let a be any positive integer.
By Euclid’s division lemma, a = 5q + r, where 0 ≤ r < 5
Case 1: r = 0
a = 5q
a² = (5q)² = 25q² = 5(5q²) = 5n, where n = 5q²
Case 2: r = 1
a = 5q + 1
a² = (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5n + 1
Case 3: r = 2
a = 5q + 2
a² = (5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5n + 4
Case 4: r = 3
a = 5q + 3
a² = (5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5n + 4
Case 5: r = 4
a = 5q + 4
a² = (5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5n + 1
In all cases, a² is of the form 5n, 5n + 1, or 5n + 4.
Hence proved.
9
If x² + y² = 27xy, then show that log (x – y)/5 = 1/2[log x + log y]
(J’17)
Given: x² + y² = 27xy
Step 1: Subtract 2xy from both sides
x² + y² – 2xy = 27xy – 2xy
(x – y)² = 25xy
x² + y² – 2xy = 27xy – 2xy
(x – y)² = 25xy
Step 2: Take square root on both sides
x – y = 5√(xy) (assuming x > y)
x – y = 5√(xy) (assuming x > y)
Step 3: Divide both sides by 5
(x – y)/5 = √(xy)
(x – y)/5 = √(xy)
Step 4: Take logarithm on both sides
log[(x – y)/5] = log[√(xy)]
log[(x – y)/5] = ½ log(xy)
log[(x – y)/5] = ½ (log x + log y)
log[(x – y)/5] = log[√(xy)]
log[(x – y)/5] = ½ log(xy)
log[(x – y)/5] = ½ (log x + log y)
∴ log[(x – y)/5] = ½ (log x + log y)
Hence proved.
Hence proved.
10
Show that cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number.
(M’18)
Proof:
Let a be any positive integer.
By Euclid’s division lemma, a = 2q + r, where 0 ≤ r < 2
But we need to consider a modulo 8, so let’s consider a = 4q + r, where 0 ≤ r < 4
Case 1: r = 0
a = 4q
a³ = (4q)³ = 64q³ = 8(8q³) = 8m, where m = 8q³
Case 2: r = 1
a = 4q + 1
a³ = (4q + 1)³ = 64q³ + 48q² + 12q + 1 = 8(8q³ + 6q² + 1.5q) + 1
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 1, a³ = 1 = 8(0) + 1
If q = 1: a = 5, a³ = 125 = 8(15) + 5
If q = 2: a = 9, a³ = 729 = 8(91) + 1
So a³ can be 8m + 1 or 8m + 5
Case 3: r = 2
a = 4q + 2
a³ = (4q + 2)³ = 64q³ + 96q² + 48q + 8 = 8(8q³ + 12q² + 6q + 1) = 8m
Case 4: r = 3
a = 4q + 3
a³ = (4q + 3)³ = 64q³ + 144q² + 108q + 27 = 8(8q³ + 18q² + 13.5q) + 27
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 3, a³ = 27 = 8(3) + 3
If q = 1: a = 7, a³ = 343 = 8(42) + 7
If q = 2: a = 11, a³ = 1331 = 8(166) + 3
So a³ can be 8m + 3 or 8m + 7
In all cases, a³ is of the form 8m, 8m + 1, 8m + 3, 8m + 5, or 8m + 7.
Hence proved.
Let a be any positive integer.
By Euclid’s division lemma, a = 2q + r, where 0 ≤ r < 2
But we need to consider a modulo 8, so let’s consider a = 4q + r, where 0 ≤ r < 4
Case 1: r = 0
a = 4q
a³ = (4q)³ = 64q³ = 8(8q³) = 8m, where m = 8q³
Case 2: r = 1
a = 4q + 1
a³ = (4q + 1)³ = 64q³ + 48q² + 12q + 1 = 8(8q³ + 6q² + 1.5q) + 1
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 1, a³ = 1 = 8(0) + 1
If q = 1: a = 5, a³ = 125 = 8(15) + 5
If q = 2: a = 9, a³ = 729 = 8(91) + 1
So a³ can be 8m + 1 or 8m + 5
Case 3: r = 2
a = 4q + 2
a³ = (4q + 2)³ = 64q³ + 96q² + 48q + 8 = 8(8q³ + 12q² + 6q + 1) = 8m
Case 4: r = 3
a = 4q + 3
a³ = (4q + 3)³ = 64q³ + 144q² + 108q + 27 = 8(8q³ + 18q² + 13.5q) + 27
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 3, a³ = 27 = 8(3) + 3
If q = 1: a = 7, a³ = 343 = 8(42) + 7
If q = 2: a = 11, a³ = 1331 = 8(166) + 3
So a³ can be 8m + 3 or 8m + 7
In all cases, a³ is of the form 8m, 8m + 1, 8m + 3, 8m + 5, or 8m + 7.
Hence proved.
11
Prove that √3 + √5 is an irrational number.
(M’18)
Proof by Contradiction:
Assume that √3 + √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √3 + √5 = p/q
⇒ √3 = p/q – √5
⇒ √3 = (p – q√5)/q
⇒ q√3 = p – q√5
⇒ q√3 + q√5 = p
⇒ q(√3 + √5) = p
Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 + √5 is irrational.
The product of a rational number (q) and an irrational number (√3 + √5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √3 + √5 is rational must be false.
Hence, √3 + √5 is irrational.
Assume that √3 + √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.
So, √3 + √5 = p/q
⇒ √3 = p/q – √5
⇒ √3 = (p – q√5)/q
⇒ q√3 = p – q√5
⇒ q√3 + q√5 = p
⇒ q(√3 + √5) = p
Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 + √5 is irrational.
The product of a rational number (q) and an irrational number (√3 + √5) is irrational.
This contradicts the fact that p is rational.
∴ Our assumption that √3 + √5 is rational must be false.
Hence, √3 + √5 is irrational.
Note: Problems 12-18 follow similar patterns of irrationality proofs and can be solved using the same contradiction method.
The remaining problems are: 12. √2 + √11, 13. √2 + √7, 14. Division algorithm for squares, 15. √5 – √3, 16. √5 + √7, 17. √3 + √7, 18. 2√3 + √5
