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Exercise on (ir)rational numbers

  1. Oct 12, 2014 #1
    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.

    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??
    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.

    Attached Files:

  2. jcsd
  3. Oct 12, 2014 #2
    For part a) an irrational divided by an irrational is not always an irrational. For example, [itex] \frac {\sqrt{2}}{\sqrt{2}}=1 [/itex]. Rather than using evenness or oddness, I would use the fact the if 2 divides [itex]a^2[/itex], 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 [itex] \sqrt {\frac {p}{q}} \in \mathbb{Q} [/itex].
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