What is the cardinality of the set of all real, positive, terminating decimal numbers?

>That only works for decimals in [-1,1] though.
Exactly my conclusion. That's why I'm stumped.

Transfer 1 from the integer part to the mirrored decimal part (mirrored in that carries are sent to the next decimal). When the integer part becomes null, start over with a bigger one. Here's a little more on the mirror thing:
12.7
11.8
10.9
9.01
8.11
7.21

Every terminating decimal real number has a fractional representation. The set of rationals is countable, any subset of rational numbers is countable. The set is a superset of the naturals and a subset of the rationals. Thus by Schoder Berstein, the set is countably infinite.

I reasoned that out, but I wanted to see a bijection with the integers.
Okay, now I get what your columns of numbers mean, but how does that help? You changed 12.7 to 7.12, neither if which is an integer.

the lemma you want to prove is that the set of finite strings over a finite or countable alphabet is countable

That's actually how I started thinking about it. I was literally trying to figure out how many words you could make with a finite alphabet (ignoring things like readability and pronounceability). I reasoned out the way described that it was countably infinite pretty quickly, but then my mind wandered into this problem.

The previous terms would be
19.0
18.1
17.2
16.3
15.4
14.5
13.6
12.7
Assuming the bijection starts with [math]f\left( 0 \right) \,=\, 1.0[/math], any input in the form [math]\frac{n\, \left( n \,+\, 1 \right)}2 \,-\, 1[/math] will output an integer.

But doesn't this just map the integers to fractions between 0 and 1?

19.0 isn't between 0 and 1...

But I thought you were mapping the integer 19 to the fraction 0.91

The bijection would be weird, since the set of terminating decimals in (0,1), is in easy bijection with the naturals, just take the mirror of the representation. Now do that for every interval of unit length while taking account the integer part of each interval, and the naturals omitted.

>aleph null
What Jew came up with that? Why not just call it alpha null?

The composition of the inverse of the bijection from naturals to rationals and the bijection from terminal decimal numbers to fractions.

[math]\aleph_0[/math] stands out much better in writing, since it's not used for anything else.

Having more symbols is never a bad thing. We should use some nice Cyrillic letters too.

Not neccessary, it's an infinite subset of [math]\mathbb{Q}[/math]

Could you get a doctorate for an amateur thesis?

I believe this argument depends on the axiom of choice.

Yes, the proof that an infinite set has a countably infinite subset relies on AC. When you start really dealing with cardinality AC is indispensable. Got to have those sets well ordered so there is a least ordinal bijective with them.

You already proved that the cardinality is the same. [-1,1] is the same cardinality as R.

Consider the function y = tan(x*pi*0.5)

This function maps every real number in (-1,1) to R. Its trivial to also include the end points.

Any interval of R has equal cardinality to R.

OP just give a grammar that generates them, then prove all languages are countable by enumerating the strings.

Damn with all this "mathematicians" in Veeky Forums.

It's a subset of rationals, so it's at most countable, it contains naturals, because they're positive and have finite decimal expansion, so the set is equinumerous with naturals

Correct answer - a terminating decimal rational, and rationals have been proven to be countable.