Why do irrational numbers have unique decimal expansions?

[tringo]

Hello,
I am looking for an example of two input injective function, f(x1,x2), R x R ->R.
I am very grateful if you can find one. Thanks

 

[tringo]

it seems that there is no two input injective formula, RxR->R because the size of domain is R x R then the size of codomain must be > R x R. But here, the size of codomain is R.

 

[Tom Gilroy]

it seems that there is no two input injective formula, RxR->R because the size of domain is R x R then the size of codomain must be > R x R. But here, the size of codomain is R.

R and R x R have the same cardinality, there's certainly an injective function f:R x R->R. I don't however, have an example of one offhand. I'll try to think of one.

 

[tringo]

For example, in a plane xOy. each couple (x,y) determine a point in the plane. If injective function f exists. it means it will map each point in this plane to a point in x (or y)axis. I think it is a contradiction.

 

[discrete*]

As Tom Gilroy said, RxR and R both have the same cardinality. I think you could use the mapping (0,1)x(0,1)->(0,1). I don't have a proof; it's just intuition, though.

 

[losiu99]

There exists an explicit definition(s, actually) of surjective mapping of [0, 1] onto the unit square (so called space-filling curves, they're even continuous). We can discard the boundary, and define the inverse (perhaps there is a way to do it explicitly, if not, it can be done by axiom of choice - moderately satisfactory, though). We then stretch the $$(0,1)\times(0,1)$$ to the whole plane by tangent or whatever. This provides us with the desired injection.
I know it's just a sketch, but perhaps it will prove helpful. I may try to finish it later.

 

[tringo]

I do not really understand it, then the 2 input injective function exists? Anybody knows one?

 

[losiu99]

Yes, it has been said already. Injective function $$\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$$ does exist. If an explicit formula is what you are asking for, then it may be quite troublesome. It is, however, a matter of fact that such functions exist.
Edit: you may want to learn a bit about set theory, cardinal numbers in particular.

 

[tringo]

If we fix y, for example y=1, then for all x in R, f(x,1) is unique, e.g. f(x,1)=x.
Then if y=2, how can we map f(x.2) to a value different from f(x,1)?
If f(x,y) exists, then f(x1,x2,x3,...,xn), RxRx...xR->R, exists. And then we can use it to reduce the data?

 

[losiu99]

1. Hmm, what makes you think f(x, 1) = x necessarily?
2. Yes, there exist injections from arbitrary finite cartesian product of reals into reals. Once we have RxR --> R case, it follows easily by induction. By "reducing th data", do you mean representing tuples of reals as single numbers? It's possible in theory, but it's not going to work in any practical considerations, since we can't really "represent" arbitrary real numbers as, for example, binary data.

 

[disregardthat]

We can make an injective function from (0,1)x(0,1) to (0,1) as such:

Let x1=0.a1a2a3a... and x2=0.b1b2b3... be the decimal representations of x1 and x2 in (0,1). We always pick the one which doesn't have an infinite sequence of successive 9's. Define f(x1,x2)=0.a1b1a2b2a3b3a4... This function is clearly injective (actually bijective), since two different decimal expansions result in two different numbers (unless it "ends" in 9's, but our restriction on x1 and x2 eliminates this possibility).

Take any bijective function from (0,1) to R, say $$g(x) = \tan(\pi x-\frac{\pi}{2})$$. Thus we have an explicit injective (bijective) function from RxR to R, namely $$h=g \circ f(g^{-1}(x_1),g^{-1}(x_2))$$.

We have $$g^{-1}(x) = \frac{1}{2}+\frac{\arctan(x)}{\pi}$$ by calculation.


Now this can be generalized to a bijective function from RxRx...xR to R by taking $$H(x_1,...,x_n)=h(x_1,h(x_2,h(...,h(x_{n-1},x_n))...)$$

EDIT: Just for fun: $$h:R \times R \to R$$ can be calculated as such:

$$h(x,y)=tan\left( \pi \left( \sum^{\infty}_{n=0}10^n(\lfloor 10^n(\frac{1}{2}+\frac{\arctan(x)}{\pi}) \rfloor+10\lfloor 10^n(\frac{1}{2}+\frac{\arctan(y)}{\pi}) \rfloor) \right)-\frac{\pi}{2} \right)$$

 

[tringo]

thanks a lot. Can you explain more why x1 and x2 doesn't have an infinite sequence of successive 9's? I am slow :)
Thank you for all comments. I learn a lot from them.

 

[disregardthat]

thanks a lot. Can you explain more why x1 and x2 doesn't have an infinite sequence of successive 9's? I am slow :)
Thank you for all comments. I learn a lot from them.

The same reason why 1.0000 = 0.9999... If we shift this back some steps you will see what I mean. For example 0.1439999...=0.1440000... Hence decimal expansions are not unique. It is obvious that you can compare two numbers by comparing the corresponding digits if they do not end in 9's, so if you disallow infinite sequences of 9's, the expansions will be unique. And, of course, if a decimal expansion does end in 9's, we can also write it in another way such that it doesn't.

An exercise for you: show that all irrational numbers have unique decimal expansions.