Answer :
Let's tackle each part of the question step by step:
### Part 1: Find the LCM and HCF of 15, 18, and 45 by the prime factorization method.
#### Step-by-Step Solution:
1. Prime Factorization of the Numbers:
- 15 = 3 × 5
- 18 = 2 × 3^2
- 45 = 3^2 × 5
2. Finding the Least Common Multiple (LCM):
- Identify the highest power of each prime number that appears in the factorization of any of the numbers:
- The highest power of 2 is 2^1 (from 18).
- The highest power of 3 is 3^2 (from 18 and 45).
- The highest power of 5 is 5^1 (from 15 and 45).
- The LCM is therefore 2^1 × 3^2 × 5^1 = 2 × 9 × 5 = 90.
3. Finding the Highest Common Factor (HCF):
- Identify the lowest power of each prime number common to all the numbers:
- The only prime common to all factorizations is 3.
- The lowest power of 3 present in all the factorizations is 3^1.
- The HCF is therefore 3.
Hence, the LCM of 15, 18, and 45 is 90, and the HCF of 15, 18, and 45 is 3.
### Part 2: Maximum number of mixed sections for boys and girls
#### Step-by-Step Solution:
To determine the maximum number of mixed sections such that every section has an equal number of boys and an equal number of girls, we must find the HCF of 372 and 444.
1. Calculate the HCF using the Euclidean algorithm:
- The Euclidean algorithm is a method to find the greatest common divisor (GCD) of two numbers by repeatedly applying the division algorithm.
- Let's find the HCF/GCD of 372 and 444:
- Step 1: 444 % 372 = 72 (remainder)
- Step 2: 372 % 72 = 12 (remainder)
- Step 3: 72 % 12 = 0 (remainder)
2. Since the remainder is now 0, the last non-zero remainder is the HCF.
Hence, the HCF of 372 and 444 is 12.
So, the number of such sections is 12.
### Part 3: Prove that √2 is an irrational number and show that 5 + 3√2 is also an irrational number
#### Step-by-Step Solution:
1. Prove that √2 is irrational:
- Assume that √2 is rational. Therefore, it can be expressed as a fraction of two integers p and q (where p and q are coprime):
[tex]\( \sqrt{2} = \frac{p}{q} \)[/tex]
- Squaring both sides, we get:
[tex]\( 2 = \frac{p^2}{q^2} \)[/tex]
- Rearrange to:
[tex]\( p^2 = 2q^2 \)[/tex]
- This implies that [tex]\( p^2 \)[/tex] is even (since it is 2 times some integer).
- If [tex]\( p^2 \)[/tex] is even, then [tex]\( p \)[/tex] must also be even (since only the square of an even number is even).
- Let [tex]\( p = 2k \)[/tex] for some integer [tex]\( k \)[/tex]. Substituting back, we get:
[tex]\( (2k)^2 = 2q^2 \)[/tex]
- Simplifying, we get:
[tex]\( 4k^2 = 2q^2 \)[/tex]
- Dividing both sides by 2, we obtain:
[tex]\( 2k^2 = q^2\)[/tex]
- This implies that [tex]\( q^2 \)[/tex] is also even, and so [tex]\( q \)[/tex] too must be even.
- However, if both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are even, they have a common factor of 2, which contradicts the assumption that they are coprime. Therefore, √2 cannot be rational and is irrational.
2. Show that 5 + 3√2 is irrational:
- Assume that 5 + 3√2 is rational. Therefore, it can be expressed as a fraction of two integers:
[tex]\( 5 + 3\sqrt{2} = \frac{a}{b} \)[/tex] where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers and [tex]\( b \neq 0 \)[/tex].
- Rearranging, we get:
[tex]\( 3\sqrt{2} = \frac{a}{b} - 5 \)[/tex]
[tex]\( \sqrt{2} = \frac{\frac{a}{b} - 5}{3} = \frac{a - 5b}{3b} \)[/tex]
- Since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers, [tex]\( \frac{a - 5b}{3b} \)[/tex] is rational.
- This implies that √2 is rational, which contradicts the fact that √2 is irrational (as previously proven).
- Hence, our initial assumption is false and therefore, 5 + 3√2 must be irrational.
Thus, 5 + 3√2 is also irrational.
### Part 1: Find the LCM and HCF of 15, 18, and 45 by the prime factorization method.
#### Step-by-Step Solution:
1. Prime Factorization of the Numbers:
- 15 = 3 × 5
- 18 = 2 × 3^2
- 45 = 3^2 × 5
2. Finding the Least Common Multiple (LCM):
- Identify the highest power of each prime number that appears in the factorization of any of the numbers:
- The highest power of 2 is 2^1 (from 18).
- The highest power of 3 is 3^2 (from 18 and 45).
- The highest power of 5 is 5^1 (from 15 and 45).
- The LCM is therefore 2^1 × 3^2 × 5^1 = 2 × 9 × 5 = 90.
3. Finding the Highest Common Factor (HCF):
- Identify the lowest power of each prime number common to all the numbers:
- The only prime common to all factorizations is 3.
- The lowest power of 3 present in all the factorizations is 3^1.
- The HCF is therefore 3.
Hence, the LCM of 15, 18, and 45 is 90, and the HCF of 15, 18, and 45 is 3.
### Part 2: Maximum number of mixed sections for boys and girls
#### Step-by-Step Solution:
To determine the maximum number of mixed sections such that every section has an equal number of boys and an equal number of girls, we must find the HCF of 372 and 444.
1. Calculate the HCF using the Euclidean algorithm:
- The Euclidean algorithm is a method to find the greatest common divisor (GCD) of two numbers by repeatedly applying the division algorithm.
- Let's find the HCF/GCD of 372 and 444:
- Step 1: 444 % 372 = 72 (remainder)
- Step 2: 372 % 72 = 12 (remainder)
- Step 3: 72 % 12 = 0 (remainder)
2. Since the remainder is now 0, the last non-zero remainder is the HCF.
Hence, the HCF of 372 and 444 is 12.
So, the number of such sections is 12.
### Part 3: Prove that √2 is an irrational number and show that 5 + 3√2 is also an irrational number
#### Step-by-Step Solution:
1. Prove that √2 is irrational:
- Assume that √2 is rational. Therefore, it can be expressed as a fraction of two integers p and q (where p and q are coprime):
[tex]\( \sqrt{2} = \frac{p}{q} \)[/tex]
- Squaring both sides, we get:
[tex]\( 2 = \frac{p^2}{q^2} \)[/tex]
- Rearrange to:
[tex]\( p^2 = 2q^2 \)[/tex]
- This implies that [tex]\( p^2 \)[/tex] is even (since it is 2 times some integer).
- If [tex]\( p^2 \)[/tex] is even, then [tex]\( p \)[/tex] must also be even (since only the square of an even number is even).
- Let [tex]\( p = 2k \)[/tex] for some integer [tex]\( k \)[/tex]. Substituting back, we get:
[tex]\( (2k)^2 = 2q^2 \)[/tex]
- Simplifying, we get:
[tex]\( 4k^2 = 2q^2 \)[/tex]
- Dividing both sides by 2, we obtain:
[tex]\( 2k^2 = q^2\)[/tex]
- This implies that [tex]\( q^2 \)[/tex] is also even, and so [tex]\( q \)[/tex] too must be even.
- However, if both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are even, they have a common factor of 2, which contradicts the assumption that they are coprime. Therefore, √2 cannot be rational and is irrational.
2. Show that 5 + 3√2 is irrational:
- Assume that 5 + 3√2 is rational. Therefore, it can be expressed as a fraction of two integers:
[tex]\( 5 + 3\sqrt{2} = \frac{a}{b} \)[/tex] where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers and [tex]\( b \neq 0 \)[/tex].
- Rearranging, we get:
[tex]\( 3\sqrt{2} = \frac{a}{b} - 5 \)[/tex]
[tex]\( \sqrt{2} = \frac{\frac{a}{b} - 5}{3} = \frac{a - 5b}{3b} \)[/tex]
- Since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers, [tex]\( \frac{a - 5b}{3b} \)[/tex] is rational.
- This implies that √2 is rational, which contradicts the fact that √2 is irrational (as previously proven).
- Hence, our initial assumption is false and therefore, 5 + 3√2 must be irrational.
Thus, 5 + 3√2 is also irrational.
