Question
Prove that root three is irrational
Answers
GPT 3.5
To prove that √3 is irrational, we start by assuming the opposite: that √3 is rational. This means that it can be written as a fraction in the form a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.
Now, suppose √3 = a/b, where a and b are integers. Squaring both sides of the equation, we get 3 = (a^2)/b^2. Multiplying both sides by b^2, we have 3b^2 = a^2.
From this equation, we can see that a^2 must be a multiple of 3. This implies that a must also be a multiple of 3 since the square of any integer multiple of 3 is also a multiple of 3.
Let a = 3k, where k is an integer. Substituting this value into the equation, we have 3b^2 = (3k)^2 = 9k^2. Simplifying, we get b^2 = 3k^2.
Now, we can see that b^2 must also be a multiple of 3. This means that b must also be a multiple of 3.
However, we originally assumed that a and b have no common factors other than 1, so it is a contradiction to say that both a and b are multiples of 3. Therefore, our initial assumption that √3 is rational must be false.
Hence, √3 is irrational.
Now, suppose √3 = a/b, where a and b are integers. Squaring both sides of the equation, we get 3 = (a^2)/b^2. Multiplying both sides by b^2, we have 3b^2 = a^2.
From this equation, we can see that a^2 must be a multiple of 3. This implies that a must also be a multiple of 3 since the square of any integer multiple of 3 is also a multiple of 3.
Let a = 3k, where k is an integer. Substituting this value into the equation, we have 3b^2 = (3k)^2 = 9k^2. Simplifying, we get b^2 = 3k^2.
Now, we can see that b^2 must also be a multiple of 3. This means that b must also be a multiple of 3.
However, we originally assumed that a and b have no common factors other than 1, so it is a contradiction to say that both a and b are multiples of 3. Therefore, our initial assumption that √3 is rational must be false.
Hence, √3 is irrational.