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Proving a number is irrational

  1. May 11, 2012 #1
    I picked up a book by Stephen Abbott called "Understanding Analysis" and it begins talking about rational and irrational numbers then it goes on proving how √2 is irrational. The proof is easy to understand but I wanted to use the same exact proof on a number I knew was rational.

    Let (p/q)2 be a rational number equal to 4 where p and q have no common factors and q ≠0.

    (p/q)2 = 4 which implies p2 = 4*q2. From this we can see that p2 is a multiple of 4 and hence so is p. Let p = 4a where a is an integer.

    16a2 = 4*q2 , 4a2 = q2 which implies q2 is a multiple of 4 and hence q is also.


    Question:
    There for we now know the ratio p/q is not in lowest terms, it has a common factor of 4.

    So how am i suppose to interpret this? This is the same exact proof that proved √2 is irrational by showing p and q have common factors. Well I just showed you that √4 = p/q where p and q have common factors but √4 = 2 is obviously rational.
     
  2. jcsd
  3. May 11, 2012 #2
    This is where you went wrong. While it is true that, for any integer [itex]p[/itex], [itex]2|p[/itex] whenever [itex]2|p^2[/itex], this is due to [itex]2[/itex] being prime. [itex]4[/itex] is not prime.
     
  4. May 11, 2012 #3
    The ratio for the proof is always in lowest terms. To start with 2/4 instead of 1/2 is nonsensical.

    As a caveat, I believe the proof in most books always states ##\text{gcd}(p,q) =1## or says they are relatively prime.
     
  5. May 11, 2012 #4

    jgens

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    You cannot conclude that p is a multiple of 4. You can only conclude that is a multiple of 2. This avoids the contradiction.
     
  6. May 11, 2012 #5
    Revised Solution:

    (p/q)2 = 4 which implies p2 = 4*q2. From this we can see that p2 is a multiple of 4 and p a multiple of 2. Let p = 2a where a is an integer.


    p2 = 4*q2 or 4a2 = 4*q2. From this equation we can see

    Equation 1: (a^2 = q2.)

    Is this correct?
    QUESTIONS: If we know p is a multiple of 2 and from equation 1 we see that a^2 = q2 where a is any integer, from this equation how can we conclude from eq 1 that q is equal to 1?

    or is it the fact that p and q do not have common factors which tells us that the ratio p/q is reduced, assuming p =/= q and that alone implies that there is a rational number when squared that is equal to 4. Is that correct?
     
    Last edited: May 11, 2012
  7. May 11, 2012 #6

    Office_Shredder

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    The fact that p2=4q2 and p and q have no common factors means that the only thing that can divide p are multiples of 2. From there the fact that p and q have no common factors implies that p must be equal to 2 (otherwise q would have to be divisible by 2 as well)

    You're trying to do a proof by contradiction and failing to find a contradiction (since there isn't one to be found), so you shouldn't expect to prove anything like this. You can get around to calculating what the square root of 4 is using what I typed above but it really relies on the fact that 22=4 to begin with, so I don't think there's any value in thinking about it too much. It's better to recognize that trying to prove by contradiction that 4 has no square root won't get you very far because it's not true that 4 has not square root (and figure out where the hole is in your calculation so you can throw it away with a peaceful mind) than to try to figure out how you can salvage the calculations you made and actually prove something
     
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