Question
Show that root of 3 is an irrational number
Answers
GPT 3.5
To show that the square root of 3 is an irrational number, we can use a proof by contradiction.
Assume that √3 is a rational number. This means that there exist two integers a and b (where b is not equal to 0) that are relatively prime, such that:
√3 = a/b
Squaring both sides of the equation, we get:
3 = a^2 / b^2
3b^2 = a^2
This implies that a^2 is a multiple of 3. Therefore, a must be a multiple of 3 as well. Let's denote a as 3k, where k is an integer.
Substitute a = 3k back into the equation:
3b^2 = (3k)^2
3b^2 = 9k^2
b^2 = 3k^2
This implies that b^2 is also a multiple of 3, and hence b must also be a multiple of 3. However, this contradicts the initial assumption that a and b are relatively prime.
Since our assumption led to a contradiction, we can conclude that the square root of 3 is an irrational number.
Assume that √3 is a rational number. This means that there exist two integers a and b (where b is not equal to 0) that are relatively prime, such that:
√3 = a/b
Squaring both sides of the equation, we get:
3 = a^2 / b^2
3b^2 = a^2
This implies that a^2 is a multiple of 3. Therefore, a must be a multiple of 3 as well. Let's denote a as 3k, where k is an integer.
Substitute a = 3k back into the equation:
3b^2 = (3k)^2
3b^2 = 9k^2
b^2 = 3k^2
This implies that b^2 is also a multiple of 3, and hence b must also be a multiple of 3. However, this contradicts the initial assumption that a and b are relatively prime.
Since our assumption led to a contradiction, we can conclude that the square root of 3 is an irrational number.