Is there a map from real numbers to non integers?

[ssamsymn]

Can you help me to construct a 1-1 mapping from real numbers onto non-integers? thanks

 

[ssamsymn]

I came up with some idea but not sure. I can only map all the real numbers to (0,1) interval. And then try to enlarge it to (n,n+1) where n goes to infinity. Is it valid?

 

[jbunniii]

Let $(z_n)$ be an enumeration of the integers, and let $(r_n)$ be any sequence of distinct non-integers ($n = 0, 1, 2, \ldots$). Define the map $f : \mathbb{R} \rightarrow \mathbb{R}\setminus \mathbb{Z}$ by
$$f(r_n) = r_{2n}$$
$$f(z_n) = r_{2n+1}$$
$$f(x) = x \textrm{ for all other }x$$
It's easy to see that this is a bijection.

 

[micromass]

You want a bijection f:\mathbb{R}\rightarrow \mathbb{R}\setminus \mathbb{Z}??

Now, it is not so hard to see that such a bijection has to exist. So if you're only interested in bijection and not existence, then you basically only have to apply the wonderful Cantor-Bernstein-Shroder theorem.

If you want an actual map, then this requires a bit more work. It might be easier to first find a bijection f:\mathbb{R}\rightarrow \mathbb{R}\setminus \{0\} and then apply the same trick on your problem. For the easier problem, the trick is to select a sequence in \mathbb{R} such as x_n=n for n\geq 0. Then we can define

f(x)=x~\text{if}~x\neq x_n~ \text{and}~f(x_n) = x_{n+1}

Do you see that that works?? Can you apply this same idea on your problem?

 

[ssamsymn]

Let $(z_n)$ be an enumeration of the integers, and let $(r_n)$ be any sequence of distinct non-integers ($n = 0, 1, 2, \ldots$). Define the map $f : \mathbb{R} \rightarrow \mathbb{R}\setminus \mathbb{Z}$ by
$$f(r_n) = r_{2n}$$
$$f(z_n) = r_{2n+1}$$
$$f(x) = x \textrm{ for all other }x$$
It's easy to see that this is a bijection.

thank you very much.
I don't fully understant how I can pick real numbers with n indices.
I am not familiar with the notation, I need a valid function to show this map is one to one and onto.

my real numbers set is non countable, non integers set is also.I thought the function can be 1/(1+e^n) for any n picked from real line. after that I try to generalize it.

 

[ssamsymn]

You want a bijection f:\mathbb{R}\rightarrow \mathbb{R}\setminus \mathbb{Z}??

Now, it is not so hard to see that such a bijection has to exist. So if you're only interested in bijection and not existence, then you basically only have to apply the wonderful Cantor-Bernstein-Shroder theorem.

If you want an actual map, then this requires a bit more work. It might be easier to first find a bijection f:\mathbb{R}\rightarrow \mathbb{R}\setminus \{0\} and then apply the same trick on your problem. For the easier problem, the trick is to select a sequence in \mathbb{R} such as x_n=n for n\geq 0. Then we can define

f(x)=x~\text{if}~x\neq x_n~ \text{and}~f(x_n) = x_{n+1}

Do you see that that works?? Can you apply this same idea on your problem?

thank you very much.
I am trying to figure out Cantor's method. I think I should construct disjoint one countable and one uncountable sets and their union should be real numbers.

 

[jbunniii]

thank you very much.
I don't fully understant how I can pick real numbers with n indices.
I am not familiar with the notation, I need a valid function to show this map is one to one and onto.

If you want something specific, just take something like $r_n = (2n+1)/2$, i.e., the sequence $1/2, 3/2, 5/2, \ldots$. Any sequence of real numbers will work as long as they are all distinct and none of them are integers. Drawing a picture might help if it's not clear what the map is doing.

 

[ssamsymn]

If you want something specific, just take something like $r_n = (2n+1)/2$, i.e., the sequence $1/2, 3/2, 5/2, \ldots$. Any sequence of real numbers will work as long as they are all distinct and none of them are integers. Drawing a picture might help if it's not clear what the map is doing.

Yes, thank you very much. I thought that my function such as 2n+1/2 should be valid for any real number but I think the remaining non-integers from half integers are included in f(x)=x otherwise case. thank you