Thread: question on group of real numbers modulo 1

1. 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.

2. Re: question on group of real numbers modulo 1

Originally Posted by oblixps
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?!"