Exercise on (ir)rational numbers

[PcumP_Ravenclaw]

Dear all,
I have done question 1 of exercise 2.1 from the book Alan F beardon, Abstract Algebra and Geometry. Please answer some of my doubts.

Q1. a) Show that √(2/3) is irrational. b) Use the prime factorization of integers to show that if √p/q is rational, where p and q are positive integers with no common factors, then p = r^2 and q = s^2 for some integers r and s.

ANS:
a) $\sqrt{2}$ is known to be irrational

$\sqrt{3}$ is irrational because 3 is a prime number and its only factors are 1 and 3??
or it can be proved. p/q is the simplest fraction after cancelling common factors. p and q are integers.

$\sqrt{3} = \frac{p}{q} \\ 3 = \frac{p^2}{q^2} \\ 3*q^2 = p^2$

Please check if my observation below is correct??
Observation:
the multiple is 3 on LHS so if q is odd then p will also be odd. if q is even p will also be even. if both p and q were even numbers then p and q are not in their simplest terms therefore it violates $\sqrt{3}$ being a rational number. because $\sqrt{}$ of any number could be written as p/q but in simplest terms. simplest terms means both cannot be even and both cannot be odd. both can be odd if they are prime numbers e.g 7/3 or 9/5 etc..

In the above case, if q is odd (integer or prime number) p will be a odd number also with a factor 3 (because of the multiple 3) therefore it will not be in its simplest form. Therefore, $\sqrt{3}$ is an irrational number??

√2/3 is irrational because √2 and √3 are irrational! but how to prove that an irrational number divided by another irrational number is also irrational?

b) Please check my answer to the second part of the question? I have written in words can anyone translate to math notation?? Thanks..

$\sqrt{PrimeNumber}$ is always irrational because primes only have factors 1 and themself. Any integer number can be broken down into its prime factors. After cancelling all common primes in numerator and denominator. we are left with non-common primes in numerator and denominator. As square root of prime number is always irrational. the irrational numerator and denominator produce an irrational outcome.

so numerator p and denominator q should be squared numbers to produce integer fractions (after square root operation) which are rational.

 

[Mogarrr]

For part a) an irrational divided by an irrational is not always an irrational. For example, $\frac {\sqrt{2}}{\sqrt{2}}=1$. Rather than using evenness or oddness, I would use the fact the if 2 divides $a^2$, then 2 divides a, for a in the integers.

I think you can sharpen up your argument in part b) a little. Just start off with the premise, p and q have no common factors and $\sqrt {\frac {p}{q}} \in \mathbb{Q}$.