# Is there a map from real numbers to non integers?

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

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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
Homework Helper
Gold Member
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.

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?

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.

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
Homework Helper
Gold Member
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.

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