Question
Show work for square root of 5 + square root of 36 as examples of irrational numbers
Answers
GPT 3.5
To show that the square root of 5 and the square root of 36 are examples of irrational numbers, we need to prove that they cannot be expressed as fractions or terminating decimals.
1. Square root of 5 (√5):
Assume √5 can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1. Squaring both sides, we get:
(√5)^2 = (a/b)^2
5 = (a^2)/(b^2)
Cross-multiplying, we have:
5b^2 = a^2
This means that a^2 is divisible by 5, so a must also be divisible by 5. Let a = 5k for some integer k. Substituting this into the equation, we get:
5b^2 = (5k)^2
5b^2 = 25k^2
Dividing both sides by 5, we have:
b^2 = 5k^2
This shows that b^2 is also divisible by 5, implying that b must also be divisible by 5. However, this contradicts our assumption that a and b have no common factors other than 1 (since they are integers). Hence, √5 cannot be expressed as a fraction and is thus irrational.
2. Square root of 36 (√36):
Assume √36 can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1. Squaring both sides, we get:
(√36)^2 = (a/b)^2
36 = (a^2)/(b^2)
Cross-multiplying, we have:
36b^2 = a^2
This shows that a^2 is divisible by 36, so a must also be divisible by 6. Let a = 6k for some integer k. Substituting this into the equation, we get:
36b^2 = (6k)^2
36b^2 = 36k^2
Dividing both sides by 36, we have:
b^2 = k^2
This implies that b^2 and k^2 have a common factor of 1, which contradicts our assumption that a and b have no common factors other than 1. Therefore, √36 cannot be expressed as a fraction and is thus irrational.
In both cases, we have shown that √5 and √36 cannot be expressed as fractions, proving that they are examples of irrational numbers.
1. Square root of 5 (√5):
Assume √5 can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1. Squaring both sides, we get:
(√5)^2 = (a/b)^2
5 = (a^2)/(b^2)
Cross-multiplying, we have:
5b^2 = a^2
This means that a^2 is divisible by 5, so a must also be divisible by 5. Let a = 5k for some integer k. Substituting this into the equation, we get:
5b^2 = (5k)^2
5b^2 = 25k^2
Dividing both sides by 5, we have:
b^2 = 5k^2
This shows that b^2 is also divisible by 5, implying that b must also be divisible by 5. However, this contradicts our assumption that a and b have no common factors other than 1 (since they are integers). Hence, √5 cannot be expressed as a fraction and is thus irrational.
2. Square root of 36 (√36):
Assume √36 can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1. Squaring both sides, we get:
(√36)^2 = (a/b)^2
36 = (a^2)/(b^2)
Cross-multiplying, we have:
36b^2 = a^2
This shows that a^2 is divisible by 36, so a must also be divisible by 6. Let a = 6k for some integer k. Substituting this into the equation, we get:
36b^2 = (6k)^2
36b^2 = 36k^2
Dividing both sides by 36, we have:
b^2 = k^2
This implies that b^2 and k^2 have a common factor of 1, which contradicts our assumption that a and b have no common factors other than 1. Therefore, √36 cannot be expressed as a fraction and is thus irrational.
In both cases, we have shown that √5 and √36 cannot be expressed as fractions, proving that they are examples of irrational numbers.