Results 1 to 2 of 2

Thread: question on group of real numbers modulo 1

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    question on group of real numbers modulo 1

    let $\displaystyle x \in \mathbb{R} $ and let [x] be the greatest integer less than or equal to x. define $\displaystyle (x) = x - [x] $ to be the fractional part of x. On $\displaystyle [0, 1) $ we define addition modulo 1 by a + b mod 1 = (a + b) = fractional part of a + b.

    my book then says that $\displaystyle \phi_{1}: \mathbb{R} \rightarrow [0, 1) $ defined by $\displaystyle \phi_{1}(x) = (x) $ is a homomorphism between the additive group of the real numbers and the additive group of the reals modulo 1 since $\displaystyle \phi_{1}(x + y) = (x + y) = (x) + (y) = \phi_{1}(x) + \phi_{1}(y) $.

    what i don't understand is how (x + y) = (x) + (y). after experimenting with numbers such as 0.5 and 0.8, i find that (0.5 + 0.8) = (1.3) = 1.3 - [1.3] = 0.3 while (0.5) + (0.8) = 0.5 - [0.5] + 0.8 - [0.8] = 1.3. so is my book wrong? is this map is not a homomorphism? help clearing this up will be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: question on group of real numbers modulo 1

    Quote Originally Posted by oblixps View Post
    let $\displaystyle x \in \mathbb{R} $ and let [x] be the greatest integer less than or equal to x. define $\displaystyle (x) = x - [x] $ to be the fractional part of x. On $\displaystyle [0, 1) $ we define addition modulo 1 by a + b mod 1 = (a + b) = fractional part of a + b.

    my book then says that $\displaystyle \phi_{1}: \mathbb{R} \rightarrow [0, 1) $ defined by $\displaystyle \phi_{1}(x) = (x) $ is a homomorphism between the additive group of the real numbers and the additive group of the reals modulo 1 since $\displaystyle \phi_{1}(x + y) = (x + y) = (x) + (y) = \phi_{1}(x) + \phi_{1}(y) $.

    what i don't understand is how (x + y) = (x) + (y). after experimenting with numbers such as 0.5 and 0.8, i find that (0.5 + 0.8) = (1.3) = 1.3 - [1.3] = 0.3 while (0.5) + (0.8) = 0.5 - [0.5] + 0.8 - [0.8] = 1.3. so is my book wrong? is this map is not a homomorphism? help clearing this up will be greatly appreciated.
    You forgot to mod out once again. The way you can think about this is to define, for each $\displaystyle x\in\mathbb{R}$, the set $\displaystyle [x]=\left\{y\in\mathbb{R}:x-y\in\mathbb{Z}\right\}$ (where $\displaystyle [x]$ denotes a set NOT the lower integer function $\displaystyle \lfloor\cdot\rfloor$ as you have used $\displaystyle [\cdot]$) and then your group is defined by $\displaystyle [x]+[y]$ is DEFINED to mean $\displaystyle [x+y]$. So then, $\displaystyle [1.3]=[.3]$ since $\displaystyle 1.3-.3=1\in\mathbb{Z}$.

    In fact, it should be easy to see if you have done quotient groups that your group is isomorphic to $\displaystyle \mathbb{R}/\mathbb{Z}$ (in fact, the way I tried to help you understand it I actually used elements of $\displaystyle \mathbb{R}/\mathbb{Z}$. If this confuses you, maybe this would help: your question in analogous to the following question "Somebody told me that the function $\displaystyle f:\mathbb{Z}\to\mathbb{Z}_n$ given by reduction $\displaystyle x\mapsto x\text{ mod }n$ is a homomorphism. But, for $\displaystyle n=4$ one can see that $\displaystyle f(3+1)=f(4)=4\text{ mod }4=0$ but $\displaystyle f(3)+f(1)=(3\text{ mod }4)+(1\text{ mod }4=3+1=4$ and $\displaystyle 0\ne 4$. What's happening?!"
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: Dec 10th 2010, 03:42 AM
  2. Real Numbers question.
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Nov 1st 2010, 09:26 AM
  3. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  4. Replies: 1
    Last Post: May 14th 2010, 01:53 AM
  5. Real Numbers - Real Anaylsis
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Sep 3rd 2008, 10:54 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum