# Decimal Expansion, 1-1 onto

• Sep 14th 2010, 05:52 PM
kathrynmath
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.
• Sep 14th 2010, 07:08 PM
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Quote:

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
• Sep 14th 2010, 07:18 PM
kathrynmath
oh, the (0,1) in S should have been (x,y)
• Sep 14th 2010, 07:20 PM
kathrynmath
I'm sorry I can read the link, but don't really understand it. We haven't studied Cantor's Theorem yet.
• Sep 14th 2010, 07:24 PM
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Quote:

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 \$\displaystyle \,0.x_1x_2x_3\dots\$ where \$\displaystyle \,x_1\$ is the first digit after the decimal point, similarly with y, then define \$\displaystyle \,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...
• Sep 14th 2010, 07:26 PM
kathrynmath
Ahh, that makes more sense.