# Thread: Decimal Expansion, 1-1 onto

1. ## Decimal Expansion, 1-1 onto

Use the fact that every real number has a decimal expansion to produce a 1-1 function that maps S into (0,1). Discuss whether the formulated function is onto.

S={(0,1):0<x, y<1}

I don't even know where to begin. The whole decimal expansion business has me confused.

2. Originally Posted by kathrynmath
Use the fact that every real number has a decimal expansion to produce a 1-1 function that maps S into (0,1). Discuss whether the formulated function is onto.

S={(0,1):0<x, y<1}

I don't even know where to begin. The whole decimal expansion business has me confused.
I don't understand the notation you used to define S (x and y appear on the right side of the colon but not the left side.. ???), thus I don't know exactly what you're asking, but it looks related to this

Cantor's diagonal argument - Wikipedia, the free encyclopedia

3. oh, the (0,1) in S should have been (x,y)

4. I'm sorry I can read the link, but don't really understand it. We haven't studied Cantor's Theorem yet.

5. Originally Posted by kathrynmath
oh, the (0,1) in S should have been (x,y)
Ah, that makes more sense.

I think we can just express x as its decimal expansion $\,0.x_1x_2x_3\dots$ where $\,x_1$ is the first digit after the decimal point, similarly with y, then define $\,z = x_1y_1x_2y_2\dots$, that is f(x,y) = z. Note that with this function it is not possible to get, for instance, 0.09090909...

6. Ahh, that makes more sense.