prove that2-root3 is irrational
Answers
Answered by
0
How would you prove that 2-root 3 is an irrational number?
Ad by CMU MSCF
Considering a master's degree?
STEM students can earn more in quant finance. Watch a 1-min video for info on this lucrative career.
Learn More
4 ANSWERS

Marn Rivelle
Answered Jun 14, 2018 · Author has 461answers and 29.1k answer views
We may assume that 2 is rational because 2 = 2/1 and 2 is prime.
So, it is sufficient to prove that sqrt(3) is irrational (because the rationals are closed under multiplication).
Let a/b = sqrt(3) where a, b are relatively prime.
a^2/b^2 = 3
3b^2 = a^2
3b^2 =a^2 means that 3 divides a^2, so 3 divides a (i.e., 3 is one of the prime factors of a)
Then, we can say that a = 3m for some integer.
a^2 = 9m^2
9m^2 = 3b^2, which means that 3m^2 = b^2. Now we may infer that 3 divides b. But that contradicts the assumption that a & b are relatively prime.
Ad by CMU MSCF
Considering a master's degree?
STEM students can earn more in quant finance. Watch a 1-min video for info on this lucrative career.
Learn More
4 ANSWERS

Marn Rivelle
Answered Jun 14, 2018 · Author has 461answers and 29.1k answer views
We may assume that 2 is rational because 2 = 2/1 and 2 is prime.
So, it is sufficient to prove that sqrt(3) is irrational (because the rationals are closed under multiplication).
Let a/b = sqrt(3) where a, b are relatively prime.
a^2/b^2 = 3
3b^2 = a^2
3b^2 =a^2 means that 3 divides a^2, so 3 divides a (i.e., 3 is one of the prime factors of a)
Then, we can say that a = 3m for some integer.
a^2 = 9m^2
9m^2 = 3b^2, which means that 3m^2 = b^2. Now we may infer that 3 divides b. But that contradicts the assumption that a & b are relatively prime.
Similar questions