1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Irrational Numbers

  1. Sep 3, 2006 #1
  2. jcsd
  3. Sep 3, 2006 #2

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    the link for me says

    "You do not have the permission to view this blog"

    I guess you have to register to see it?

    Can you type the problem here?
     
  4. Sep 3, 2006 #3
    you should be able to see it now. changed the settings
     
  5. Sep 3, 2006 #4

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    I'm getting a page that says it's your blog, but it also says "total entries: 0" and there's not much tehre besides a calender.

    Is it that long of a problem that you couldn't just type it in here?
     
  6. Sep 3, 2006 #5
    yeah, now it should work.
     
    Last edited: Sep 3, 2006
  7. Sep 3, 2006 #6

    StatusX

    User Avatar
    Homework Helper

  8. Sep 3, 2006 #7

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    Alright. You get to p^2=n*q^2 and then assume that n=2*t? That's completely unjustified.

    If n is not a perfect square then you can write it as n=r*m^2 where r has the property that if s is a prime and s divides r then s^2 does not divide r, r>1 obviously as well (you should prove this). Proceed from p^2=n*q^2 and see what happens.

    You could also just look at the prime factorization of n if you have unique factorization at this point.
     
  9. Sep 3, 2006 #8
    Ok, so [tex] p^{2} = rm^{2}q^{2} [/tex] and [tex] p^{2} [/tex] is divisible by [tex] rm^{2} [/tex]. So then I have to show that [tex] q^{2} [/tex] is also divisible by [tex] rm^{2} [/tex]. Is [tex] m [/tex] just any positive number?

    Thanks
     
  10. Sep 3, 2006 #9

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    It's probably easiest at this point to focus on some prime that divides r, say s. Then it will look very much like the usual proof sqrt(2) is irrational.

    However we have a little snag. You are looking at p^2=r*(mq)^2 and you don't necessarily have mq and p relatively prime, but you should be able to fix this without much trouble.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Irrational Numbers
  1. Irrational Numbers (Replies: 15)

  2. Irrational numbers (Replies: 10)

Loading...