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/sci/ - Science & Math

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9107731 No.9107731 [Reply] [Original]

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

>> No.9107734

countable

>> No.9107737

[eqn]\aleph_0[/eqn]

>> No.9107746

>>9107734
>>9107737
Now establish its bijection with the set of integers.

>> No.9107750

>>9107746
Multiplication by a suitable power of 10

>> No.9107756

>>9107750
Won't that require more integers? If 0.1 maps to 1, what does 0.01 map to? What about 0.001?

>> No.9107763

>>9107746
t. wants their homework done for them

>> No.9107764

>>9107746
1.0
0.1
2.0
1.1
0.2
3.0
2.1
1.2
0.3
You keep going like that, seeing the decimal part as a mirrored integer.

>> No.9107771

>>9107763
Not even in school, and *certainly* not in a math program. Just a problem I thought of last week and couldn't solve.
>>9107764
You'll need to explain that in a more brainlet-friendly way.

>> No.9107775

>>9107756
You could map 0.1 to 1, 0.01 to 100, 0.001 to 1000, 0.456 to 456000, etc. That only works for decimals in [-1,1] though.

>> No.9107779

>>9107775
>That only works for decimals in [-1,1] though.
Exactly my conclusion. That's why I'm stumped.

>> No.9107780

>>9107771
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

>> No.9107781

>>9107731
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.

>> No.9107793

>>9107781
I reasoned that out, but I wanted to see a bijection with the integers.
>>9107780
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.

>> No.9107795

>>9107771
the lemma you want to prove is that the set of finite strings over a finite or countable alphabet is countable

>> No.9107799

>>9107795
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 >>9107781 described that it was countably infinite pretty quickly, but then my mind wandered into this problem.

>> No.9107810

>>9107793
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.

>> No.9107815

>>9107810
But doesn't this just map the integers to fractions between 0 and 1?

>> No.9107823

>>9107815
19.0 isn't between 0 and 1...

>> No.9107835

>>9107823
But I thought you were mapping the integer 19 to the fraction 0.91

>> No.9107864

>>9107746
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.

>> No.9107870 [DELETED] 

>>9107731
>aleph null
What kike came up with that? Why not just call it alpha null?

>> No.9107872 [DELETED] 

>>9107737
>aleph null
What Jew came up with that? Why not just call it alpha null?

>> No.9107884 [DELETED] 

>>9107746
The composition of the inverse of the bijection from naturals to rationals and the bijection from terminal decimal numbers to fractions.

>> No.9107889

>>9107737
>aleph null
What Jew came up with that? Why not just call it alpha null?

>>9107746
The composition of the inverse of the bijection from naturals to rationals and the bijection from terminal decimal numbers to fractions.

>> No.9107902
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9107902

>>9107872
[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.

>> No.9108013

>>9107746
Not neccessary, it's an infinite subset of [math]\mathbb{Q}[/math]

>> No.9108021

Could you get a doctorate for an amateur thesis?

>> No.9108024

>>9108013
I believe this argument depends on the axiom of choice.

>> No.9108042
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9108042

>>9108024
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.

>> No.9108176

>>9107779
>>9107775
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.

>> No.9108776

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

Damn with all this "mathematicians" in /sci/.

>> No.9108785

>>9107746
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

>> No.9110139

>>9107731
>>9107746
>>9108785
Correct answer - a terminating decimal rational, and rationals have been proven to be countable.