Question
Prove that root two is irrational
Answers
GPT 3.5
To prove that the square root of two (√2) is irrational, we need to assume the contrary and show that it leads to a contradiction.
Assume that √2 is rational, which means it can be expressed as a fraction in the form a/b, where a and b are integers and b ≠ 0. We can also assume that a/b is in its simplest form, meaning a and b have no common factors other than 1.
Therefore, we have (√2) = a/b, which can be rearranged as (2) = (a^2)/(b^2), squaring both sides. This implies that 2b^2 = a^2.
Since a^2 is even (since a^2 = 2b^2), we can conclude that a must also be even, since the square of an odd number is odd. Therefore, we can express a as a = 2k, where k is an integer.
Substituting a = 2k into 2b^2 = a^2, we get 2b^2 = (2k)^2. Simplifying this equation gives 2b^2 = 4k^2, which can be further simplified to b^2 = 2k^2.
Following the same logic as before, this implies that b^2 is even, and therefore b must also be even. But if both a and b are even, this contradicts our initial assumption that a/b is in its simplest form, as both a and b have a common factor of 2.
Since assuming that √2 is rational leads to a contradiction, we can conclude that √2 is indeed irrational.
Assume that √2 is rational, which means it can be expressed as a fraction in the form a/b, where a and b are integers and b ≠ 0. We can also assume that a/b is in its simplest form, meaning a and b have no common factors other than 1.
Therefore, we have (√2) = a/b, which can be rearranged as (2) = (a^2)/(b^2), squaring both sides. This implies that 2b^2 = a^2.
Since a^2 is even (since a^2 = 2b^2), we can conclude that a must also be even, since the square of an odd number is odd. Therefore, we can express a as a = 2k, where k is an integer.
Substituting a = 2k into 2b^2 = a^2, we get 2b^2 = (2k)^2. Simplifying this equation gives 2b^2 = 4k^2, which can be further simplified to b^2 = 2k^2.
Following the same logic as before, this implies that b^2 is even, and therefore b must also be even. But if both a and b are even, this contradicts our initial assumption that a/b is in its simplest form, as both a and b have a common factor of 2.
Since assuming that √2 is rational leads to a contradiction, we can conclude that √2 is indeed irrational.