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Sagot :
Certainly! Let's address each part of the question in detail:
### 1. Prove √2, √3, √5, √7, and √11 are irrational numbers.
Proof by contradiction:
- For √2:
- Assume √2 is rational. Then √2 = p/q, where p and q are integers with no common factors and q ≠ 0.
- Then [tex]\(2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2\)[/tex].
- This means p^2 is even, so p must also be even (since the square of an odd number is odd).
- Let p = 2k. Then [tex]\( (2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2 \)[/tex].
- Hence, q^2 is even, so q must also be even.
- If both p and q are even, they have a common factor of 2, contradicting our original assumption that [tex]\( p \)[/tex] and [tex]\( q \)[/tex] have no common factors.
- Therefore, √2 is irrational.
- For √3, √5, √7, √11:
- The proofs for √3, √5, √7, and √11 follow a similar structure to that for √2.
- Assume each is rational, express it as p/q, and follow the same steps. You will find a contradiction in each case, proving the irrationality of each.
### 2. Solve an example of factorization.
Example: Factorize [tex]\( x^2 - 5x + 6 \)[/tex].
- Given the quadratic expression [tex]\( x^2 - 5x + 6 \)[/tex].
- To factorize, we look for two numbers that multiply to +6 and add to -5.
- These numbers are -2 and -3.
- Therefore, [tex]\( x^2 - 5x + 6 = (x - 2)(x - 3) \)[/tex].
### 3. Solve an example of LCM and HCF.
Example: Find the LCM and HCF of 12 and 15.
- Finding the HCF (Highest Common Factor):
- Factors of 12: 1, 2, 3, 4, 6, 12
- Factors of 15: 1, 3, 5, 15
- Common factors: 1, 3
- Highest common factor: 3
- Finding the LCM (Least Common Multiple):
- Multiples of 12: 12, 24, 36, 48, 60, ...
- Multiples of 15: 15, 30, 45, 60, ...
- Smallest common multiple: 60
- Therefore, LCM of 12 and 15 is 60.
Combining all the details, here are the confirmed results for the examples provided:
- Irrational numbers: √2, √3, √5, √7, and √11.
- Factorization example: [tex]\( x^2 - 5x + 6 \)[/tex] factorizes to [tex]\( (x - 3)(x - 2) \)[/tex].
- LCM and HCF example: For 12 and 15, LCM is 60 and HCF is 3.
### 1. Prove √2, √3, √5, √7, and √11 are irrational numbers.
Proof by contradiction:
- For √2:
- Assume √2 is rational. Then √2 = p/q, where p and q are integers with no common factors and q ≠ 0.
- Then [tex]\(2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2\)[/tex].
- This means p^2 is even, so p must also be even (since the square of an odd number is odd).
- Let p = 2k. Then [tex]\( (2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2 \)[/tex].
- Hence, q^2 is even, so q must also be even.
- If both p and q are even, they have a common factor of 2, contradicting our original assumption that [tex]\( p \)[/tex] and [tex]\( q \)[/tex] have no common factors.
- Therefore, √2 is irrational.
- For √3, √5, √7, √11:
- The proofs for √3, √5, √7, and √11 follow a similar structure to that for √2.
- Assume each is rational, express it as p/q, and follow the same steps. You will find a contradiction in each case, proving the irrationality of each.
### 2. Solve an example of factorization.
Example: Factorize [tex]\( x^2 - 5x + 6 \)[/tex].
- Given the quadratic expression [tex]\( x^2 - 5x + 6 \)[/tex].
- To factorize, we look for two numbers that multiply to +6 and add to -5.
- These numbers are -2 and -3.
- Therefore, [tex]\( x^2 - 5x + 6 = (x - 2)(x - 3) \)[/tex].
### 3. Solve an example of LCM and HCF.
Example: Find the LCM and HCF of 12 and 15.
- Finding the HCF (Highest Common Factor):
- Factors of 12: 1, 2, 3, 4, 6, 12
- Factors of 15: 1, 3, 5, 15
- Common factors: 1, 3
- Highest common factor: 3
- Finding the LCM (Least Common Multiple):
- Multiples of 12: 12, 24, 36, 48, 60, ...
- Multiples of 15: 15, 30, 45, 60, ...
- Smallest common multiple: 60
- Therefore, LCM of 12 and 15 is 60.
Combining all the details, here are the confirmed results for the examples provided:
- Irrational numbers: √2, √3, √5, √7, and √11.
- Factorization example: [tex]\( x^2 - 5x + 6 \)[/tex] factorizes to [tex]\( (x - 3)(x - 2) \)[/tex].
- LCM and HCF example: For 12 and 15, LCM is 60 and HCF is 3.
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