Any Diophantic-like surjection from Z to R?

[Pion]

Is there any example, or theorem, or know-how on this problem?
f(x1,x2...xm, k, n): Z1 x Z2 x...x Zm x N x N* -> R,
with form:
f(x1,x2...xm, k, n)=k-root[(x1^k+x2^k+...+xm^k)/n],
where: k-root[...] is the k-th root of [...]. Not k minus SQR[...] !

Equally important here are the surjectivity on R and its form.

P.S. of course I do not mind if the answer is available only for the "trivial" Pithagora cases: k=2 and n=1.
Same, if the answer comes from any of mathematics specialty.

This is no home-work of any kind. It is simply a long time present curiosity.
Thank you all for any (relevant) answers.

 

[Pion]

I do not mind if N* is replaced by Q*.

 

[pka]

To Pion, This is is a theorem you must consider.
Theorem: [imath]\bf{F:A\mapsto B}[/imath] is an injection if and only if [imath]\bf{\exists\text{ a surjection }G:B\mapsto A}[/imath].
Think about the cardinalities involved in your question.

[imath][/imath][imath][/imath]

 

[Pion]

Thank you.
I understand the theorem, also in a few other different but equivalent ways. I will be grateful if you have the name, or link or know a book where to find it: please tell me. I want to read it completely.

Thank you more, for hint with cardinalities: it it absolutely clear now that I was wrong. It cannot exist be such a surjection as above, if the definition domain is limited to Q^n, while n-finite or n=Aleph0. It can be a surjection: only if n=Aleph1 (also, practically to eliminate the ordinality of the tuples).

But, anyway, could it be mathematically true that if we organize (QxQ)^n, with n-infinite, then by 2^aleph0=aleph1 and 2=numbers_of_tuple_elements and 1*\aleph1 = \aleph1, results:

[math]\sqrt{\overbrace{\frac{a^2+b^2}{2^1} + \frac{a^2+b^2}{2^2}+\frac{a^2+b^2}{2^3}+...}^{\aleph0 \, \text{of identical a-b pairs}}}=\sqrt{\displaystyle\sum_{n=1}^\infin (a^2+b^2)\frac{1}{2^n}}=\sqrt{(a^2+b^2)\overset{\infin}{\sum_{\substack{n=1}}} \frac{1}{2^n}}=\sqrt{(a^2+b^2)}[/math]
which apparently could fix at least the cardinality problem with R, while maintaining a clear and unique a,b tuple ordinality?
It may not work with Q, but for QxQ tuples (or QxQxQ, etc) apparently is just needed to measure the distance a countable-infinite number of times to obtain any real number.
More, in discrete spaces one can define metrics which do not accept notion of time. If there is no time, then re-measuring distance between two points a countable-infinite number of times is no problem. But the results will map the R. So, distances in a 2D (or 3D, or more) timeless discrete spaces will create an image space which is continuum, and which does have time.
Then, if above is mathematically true, I would conclude that it is the continuum space of R-distances which creates time. The underlying space would have the same number of points, be countable, ie. discrete, having any suitable metric which does not support a definition of time.

This is just a friendly fun over a beer or wine, no career to sacrifice. Please criticize. I promise to learn from critics having a swift and obviously clear debunk. Somewhere above, I surely did a mistake, most probably an elementary (not fully educated) one.