Is it possible to generate a unique real number for each fixed length sequence of real numbers?

Let A be the set of all sequences of real numbers of size $n$. Does there exist an injection from A to R?

I know this is possible if we are only considering integers instead of real numbers; But I am not sure if it is possible if we consider real numbers instead.

For integers, we can generate a unique integer using the following method: Let S be a sequence of integers of size n. $S = s_1,s_2,\ldots,s_n$. Let $P = p_1,p_2,\ldots,p_n$ be the sequence of $n$ primes. Then $f(S) = (p_1^{s_1})(p_2^{s_2})\cdots(p_n^{s_n})$ creates a unique integer for each sequence $S$.

If each $s_i$ was a real number instead, would $f(S)$ still be an injection? If not, is there an alternative invective function from A to R?

edit:
I fixed some of my poor wording.

I am trying to find an injection function from A to R. Such a function does exist and the function I proposed clearly does not work (From the comments).

If possible, I would like to find an injective function that does not involve directly manipulating the decimal expansions.

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(i) Your wording is bad; "Would $f(S)$ create a unique number?" is not what you want to ask (it would, because the expression is defined and is a real number); what you really want to ask is whether the resulting function $f$ is one-to-one (i.e., if different sequences yield different real numbers). (ii) Such an injection necessarily exists by simple set-theoretic arguments, since the set $A$ and the set $\mathbb{R}$ have the same cardinality; you can even find a bijection! So perhaps you are asking for a special kind of function, perhaps an explicit or constructive embedding? Say so – Arturo Magidin Feb 6 '12 at 17:22
Think about $(.a_1a_2a_3,\ldots,.b_1b_2b_3,\ldots )\rightarrow .a_1b_1a_2b_2\ldots$ – David Mitra Feb 6 '12 at 17:22
Your $f$ does not permit unique recovery of an ordered pair. Take $n=2$, $p_1=2$, $p_2=3$. We are asking, if $2^x3^y=2^u3^v$, do we have $x=u$, $y=v$? No, let $x=0$, $u=1$. Can we solve $3^y=2\cdot 3^v$? Sure. For a moderately explicit function that does the job, interleave the decimal expansions of $x$ and $y$. There are some technical difficulties, because the decimal expansion is not unique, $1=0.999\dots$. The difficulties can be resolved, but it is easier to interleave continued fraction expansions. The method goes back to Cantor. – André Nicolas Feb 6 '12 at 17:24
Ah,classical analysis. So beautiful,so intuitive-yet so frustrating sometimes to make precise. – Mathemagician1234 Feb 6 '12 at 19:37

I wasn't sure whether we were considering all finite sequences of reals, or just sequences of a predetermined length, so I answered for all finite sequences. Of course, this also works for a predetermined length sequence.

Suppose we have a sequence $\{x_k\}_{k=1}^n$ of real numbers. Generate an integer based on the signs of those reals as follows: $$\sigma(\{x_k\})=2^{n-1}-1+\sum_{k=1}^{n-1}2^{k-1}\frac{1+\rm{sign}(x_k)}{2}$$ taking $\rm{sign}(0)=1$. Interlace the digits of the absolute values of the $x_k$ (not ending in repeating $9$s) to get a unique non-negative real number $\mathscr{I}(\{x_k\})$ and map it to $[0,1)$ with $$\phi(x)=\frac{x}{\sqrt{x^2+1}}$$ Then we get a unique real using $$\rm{sign}(x_n)\left(\phi(\mathscr{I}(\{x_k\}))+\sigma(\{x_k\})\right)$$

Simpler method:

Get a unique real using the seqence of positive reals $\left\{e^{x_k}\right\}$ with $$n+\phi\left(\mathscr{I}\left(\left\{e^{x_k}\right\}\right)\right)$$

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Pick a bijection from $\mathbb{R}$ to the set of all infinite sequences of, say, non-negative integers (exercise if you don't already know that this is possible; you can build such a bijection out of continued fraction exansions). Then an element of $\mathbb{R}^n$ (the set of ordered $n$-tuples of real numbers) can be identified with an ordered $n$-tuple of infinite sequences of non-negative integers $a_{i, j}, 0 \le i \le n-1, j \in \mathbb{N}$. Now one can simply interweave these sequences to get a single infinite sequence by setting $$b_{jn + i} = a_{i, j}$$

and this is clearly a bijection.

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To handle negative elements in the sequence, you can just use the continued fractions of $e^{x_k}$ for each $x_k$. – robjohn Feb 6 '12 at 20:09