show that the square root of a prime is irrational.

Discussion in 'Undergraduate Math' started by malonetravis, Oct 9, 2005.

  1. malonetravis

    malonetravis Guest

    I've seen many ways to show that the square root of a prime is
    irrational using a contradiction and saying that p^2=a^2/b^2. Can you
    show that the square root of a prime is irrational using the property
    of primes: if m and n are integers, p is prime and p divides mn then p
    divides m or p divides n.
    Ive tried this but I always get stuck at the same point. I think you
    can say that m and n are relatively prime otherwise there would exist
    a p that would divide m and n.
    mn=pd where d is an integer. implies that p=mn/d but m and n are
    relatively prime so d must be one or p but d cannot equal p because
    then we would have p^2 divides mn but we know that m and n are
    relatively prime so d must equal 1.
    does this mean that any p greater than 1 is irrational? or can anyone
    give me a hint as to what I'm missing.

    Thanks
     
    malonetravis, Oct 9, 2005
    #1
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  2. You mean p = a^2/b^2.
    I don't understand what you are asking. The proof you speak of in the
    first sentence ->uses<- the prime divisor property as one of its
    fundamental parts: from p =a^2/b^2, we deduce that p divides a^2, so p
    divides a*a, and so by the prime divisor property it must divide a;
    hence we can write a = pk, and we have pb^2 = p^2k^2, from which we
    get pk^2 = b^2, hence p divides b... etc.
    Huh?

    Are you asking how to prove the prime divisor property?

    If so, the relativee primeness (or lack thereof) of m and n are
    irrelevant.

    What you need to use is the fact that if p is a prime, and m is any
    integer, then either p divides m or else p and m are relatively
    prime. This follows by the division algorithm: write m = pq + b, with
    0<= b < p. If b=0 then p divides m; if 0<b<q, then gcd(m,p) =
    gcd(m-pq,p) = gcd(b,p) = 1 (since p is prime, hence relatively prime
    to every integer strictly smaller than itself).

    So: suppose p divides mn. We want to show that p|m or p|n. If p
    divides m we are done. If p does not divide m, then
    gcd(p,m)=1. Therefore we can write 1 = ap + bm for some integer a and
    b. So n = anp + bmn. Since p divides anp and divides bmn, it must
    dividee their sum, which is equal to n. Therefore p divides n, QED.
    This extremely long, convoluted, and confused sentence is completely
    opaque to me. I have no idea what it is you are trying to say.
    That's nonsense. Surely you realize that.
    The ability to state clearly what it is you are trying to do?

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Oct 9, 2005
    #2
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  3. malonetravis

    zero Guest

    Your hint was very helpfull and I hope to make full use of it now.

    What I want to do is show that sqrt(prime) is irrational by using the
    property of primes not the definition of a rational number. I do not
    want to prove the prime divisor property.

    I hope this is a little clearer so that a more helpful hint will be
    forth coming.

    If p|mn then p|m or p|n. I think this can be used to show that m and n
    are relatively prime.

    If p|mn
    then mn/p=g; g is an integer
    so mn=pg
    If p|mn then p|m^2n^2
    p|m^2n^2 implies that ((mn)^2)/p=d; d is a (rational?) integer
    so (mn)^2=pd
    mn=sqrt(p)f where f = sqrt(d)

    Doing some rearraging I get pg=mn=sqrt(p)sqrt(d), pg=sqrt(p)sqrt(d)
    and p=sqrt(p)(sqrt(d)/g).

    finally p/sqrt(p)=sqrt(d)/g where sqrt(d)/g is rational. but p
    divided by any number other than one or itself is irrational. so
    sqrt(p) is irrational.

    This is probably another bunch of crap and I will not waste anymore
    time on this here. I will listen to any feedback.
     
    zero, Oct 9, 2005
    #3
  4. malonetravis

    Jim Spriggs Guest

    If that were so, since p|mn => p|m and p|n, you could prove that any two
    integers are relatively prime.
    3/2 is irrational?
     
    Jim Spriggs, Oct 9, 2005
    #4
  5. How are you going to prove something is irrational without using the
    meaning of "rational"? Seems like you are still not clear on what it
    is you want to prove.
    This is FALSE as written. For crying out loud: set p = 2, m=2,
    n=2. Then 2|2*2, but 2 and 2 are not relativiely prime.
    You should avoid division statements. That's the whole point of the
    "|" notation. That you do not do divisions (which are not defined in
    the integers).

    "a|b" means "there exists x such that ax = b."
    Integer. Why "rational?
    Sigh. If mn=pg, then (mn)^2 = (pg)^2, so your d is equal to d=pg^2.

    So sqrt(d) = sqrt(p)g, hence sqrt(p)*sqrt(d)/g = sqrt(p)^2g/g=p.
    So you are assuming that sqrt(p) is rational.
    No. 7 divided by 2 is not irrational. It is rational: 7/2.


    I'm afraid so.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Oct 9, 2005
    #5
  6. Here are a couple of remarks about what the originator of this thread wants to do -- prove irrationality of a square root of a prime.

    1. The general theorem in question here says that for any positive integer n, if n is not the square of an integer, then sqrt(n) is irrational.
    Notice that primes do not play a special role in the result, though they do often play a role in the proofs that are given for the general theorem or special cases of it such as Euclid's proof for the case of 2.

    2. Since the definition of "irrational" is "not rational", it is essentially impossible (in elementary math) to prove a number is irrational except by contradiction -- assume it is rational and then use that to arrive at a contradiction.

    3. I prefer to show my students the proof by Theodore Kestleman (maybe incorrect name) which makes no use at all of divisibility properties or primes.

    Since this proof seems to be not well known, I will give it here. Suppose that n is a positive integer which is not an integer square. Let k be the floor of sqrt(n), so that k < sqrt(n) < (k+1). Lets shift our attention and write the proof using w=sqrt(n)-k. Note that sqrt(n) is irrational if and only if w is irrational. Note also that w*(w+2*k)=n-(k*k) and 0 < w < 1.
    If w is rational, then we can express it as w=p/q where q is the smallest positive integer for which this is possible, i.e. p and q are coprime, p/q is a "reduced" fraction. Notice also that 0 < p < q.
    Now divide the equation above by w to get
    w+2k=[n-k^2]/w =[n-k^2]*(q/p)
    Subtract 2k and you arrive at a contradiction: w is now expressed as an integer+(fraction with denominator p), and p is smaller than q.
     
    Ladnor Geissinger, Oct 9, 2005
    #6
  7. malonetravis

    Jim Spriggs Guest

    Sorry, p|mn => p|m and p|n, should be p|mn => p|m or p|n.
     
    Jim Spriggs, Oct 10, 2005
    #7
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