Go through the proof, but remember to replace 2's with 3's.Begin by supposing √3 = a/b. Note we'll have a^2 = 3b^2 and b^2 = 3k^2 following the lines of the proof.
Since a and b are necessarily multiples of 3, a/b can never be in reduced form, which contradicts the premise of our supposition.
Assume by contradiction that sqrt(3) is rational. Then sqrt(3) = a/b for integers a and b with gcd(a, b) = 1 and b =/= 0. Then 3 = a^2 / b^2. Then a^2 = 3b^2. Thus, 3 | a^2. Since 3 is prime, 3 | a. Then a = 3k for some integer k. Then a^2 = 9k^2. Then 9k^2 = 3b^2. Then b^2 = 3k^2. Then 3 | b^2. Since 3 is prime, 3 | b. Thus, 3 | gcd(a, b) since 3 | a and 3 | b. But then 3 | 1 which is impossible. Thus, sqrt(3) is irrational.
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use the standard proof for the irrationality of √2
http://www.homeschoolmath.net/teaching/proof_squar...
Go through the proof, but remember to replace 2's with 3's.Begin by supposing √3 = a/b. Note we'll have a^2 = 3b^2 and b^2 = 3k^2 following the lines of the proof.
Since a and b are necessarily multiples of 3, a/b can never be in reduced form, which contradicts the premise of our supposition.
Assume by contradiction that sqrt(3) is rational. Then sqrt(3) = a/b for integers a and b with gcd(a, b) = 1 and b =/= 0. Then 3 = a^2 / b^2. Then a^2 = 3b^2. Thus, 3 | a^2. Since 3 is prime, 3 | a. Then a = 3k for some integer k. Then a^2 = 9k^2. Then 9k^2 = 3b^2. Then b^2 = 3k^2. Then 3 | b^2. Since 3 is prime, 3 | b. Thus, 3 | gcd(a, b) since 3 | a and 3 | b. But then 3 | 1 which is impossible. Thus, sqrt(3) is irrational.