Results 1 to 4 of 4

Thread: Isomorphic

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    70

    Isomorphic

    Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2009
    Posts
    29
    I'll give it a shot. Assume $\displaystyle \phi: \mathbb{R} \rightarrow \mathbb{Q}$ is such an isomorphism. Say $\displaystyle \phi(1)=k$ where $\displaystyle k \in \mathbb{Q}$. $\displaystyle \phi$ must satisfy
    $\displaystyle \phi(ab) = \phi(a)\phi(b)$
    so
    $\displaystyle kab = k^2 ab \Rightarrow k = 1$

    Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist.
    Last edited by eeyore; Oct 13th 2009 at 10:48 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    125
    Hi,
    here's my try:
    sentence
    $\displaystyle \forall x \exists y \; x=y\cdot y\cdot y$
    holds in $\displaystyle \mathbb{R}$ but not in $\displaystyle \mathbb{Q}$. So they are not elementarily equivalent, therefore they can't be isomorphic.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Another one: assume $\displaystyle f:\mathbb{R}\rightarrow\mathbb{Q}$ is a ring isomorphism, then $\displaystyle f(\sqrt{2})^2=f(\sqrt{2}^2)=f(2)=f(1+1)=f(1)+f(1)= 1+1=2$.

    But no rational is a squared root of 2, so $\displaystyle f$ cannot exist.

    (assume $\displaystyle p,q$ belong to $\displaystyle \mathbb{N}^*, gcd(p,q)=1$ and $\displaystyle (\frac{p}{q})^2=2$ then $\displaystyle p^2=2q^2$.
    But this is impossible since $\displaystyle \nu_2(p^2)$ is even and $\displaystyle \nu_2(2q^2)$ is odd (where $\displaystyle \nu_2(n)$ is the greatest $\displaystyle k$ such that $\displaystyle 2^k$ divides $\displaystyle n$))
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. is U14 isomorphic to U18?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Aug 21st 2010, 09:52 AM
  2. Isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Mar 8th 2010, 11:13 AM
  3. Isomorphic
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Oct 13th 2008, 08:55 PM
  4. Isomorphic or not?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 13th 2008, 04:41 PM
  5. Isomorphic?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 15th 2007, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum