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Math Help - question on group of real numbers modulo 1

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    question on group of real numbers modulo 1

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

    my book then says that  \phi_{1}: \mathbb{R} \rightarrow [0, 1) defined by  \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  \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.
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    MHF Contributor Drexel28's Avatar
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    Re: question on group of real numbers modulo 1

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

    my book then says that  \phi_{1}: \mathbb{R} \rightarrow [0, 1) defined by  \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  \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 x\in\mathbb{R}, the set [x]=\left\{y\in\mathbb{R}:x-y\in\mathbb{Z}\right\} (where [x] denotes a set NOT the lower integer function \lfloor\cdot\rfloor as you have used [\cdot]) and then your group is defined by [x]+[y] is DEFINED to mean [x+y]. So then, [1.3]=[.3] since 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 \mathbb{R}/\mathbb{Z} (in fact, the way I tried to help you understand it I actually used elements of \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 f:\mathbb{Z}\to\mathbb{Z}_n given by reduction x\mapsto x\text{ mod }n is a homomorphism. But, for n=4 one can see that f(3+1)=f(4)=4\text{ mod }4=0 but f(3)+f(1)=(3\text{ mod }4)+(1\text{ mod }4=3+1=4 and 0\ne 4. What's happening?!"
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