Math paradox

[Nexuapex]

On a side note, our decimal system has the initially-weird consequence that 0.999… is well-defined, but 0.000…001 isn't. The reason for that is that while 0.999… is representable as an infinite series, 0.000…001 isn't—you can't keep adding anything to make the series continue, you have to keep multiplying (by 1/10). That make it unrepresentable as a decimal.

Incidentally, the infinite series 0/1 + 0/10 + 0/100 + 0/1,000 + 0/10,000 + ... isn't equal to 0. It's an infinite series, it doesn't equal anything. It does converge to 0, but it doesn't equal 0.

Originally Posted by Nexuapex Incidentally, the infinite series 0/1 + 0/10 + 0/100 + 0/1,000 + 0/10,000 + ... isn't equal to 0. It's an infinite series, it doesn't equal anything. It does converge to 0, but it doesn't equal 0.

It surprises me that you're willing to accept that 0.000...1 converges on 0 but does not equal zero, yet you cannot accept that 0.[9] converges on 1 but does not equal 1...and I've yet to see a convincing argument mathematically for it aside from "well, nothing breaks" (which isn't true...an if-then statement using n=1 won't fire if n=0.[9]).

I'm tired, brain no werk gud no moar, want...*yawn*...gimmeh teh Grunty, Cairenn...ZzzZzzzzzz....mreh...cookies in the kitchen, Seerah...I'm out of milk...






(Note: Don't assume I'm angry because of what my students like to call my "direct tone" in debate. Things like this are my idea of "fun". It's never personal!

[Nexuapex]

Originally Posted by Amenity It surprises me that you're willing to accept that 0.000...1 converges on 0 but does not equal zero, yet you cannot accept that 0.[9] converges on 1 but does not equal 1...

9/10 + 9/100 + 9/1,000 + 9/10,000 + … ≠ 1.
0/10 + 0/100 + 0/1,000 + 0/10,000 + … ≠ 0.
(I'm not writing 0.000…001 up here because there's no way to express it as an infinite series. You need to write it as an infinite product. That's important.)

0.999… = 1. Well-defined under the decimal numbering system. I suppose one way to explain that is that, when using decimal notation, convergence is implied, primarily because everything that can be expressed with decimal notation is an infinite series that must converge.
0.000…001 doesn't make sense in the decimal numbering system, which doesn't account for multiplicative infinite series. I can't assert anything about it.

and I've yet to see a convincing argument mathematically for it aside from "well, nothing breaks"

Well, I can try. It depends on whether you think 0.[9] is a real number or not. If you hold that it isn't, then we're going to have to enter some weird nether-realm where a decimal number with a finite part and a repeating part doesn't equal a number. If you hold that it is, in fact, a real number, then let me know.

(Also, if you can formally prove that "nothing breaks" when you do something, then you've proven that it's valid.)

(which isn't true...an if-then statement using n=1 won't fire if n=0.[9]).

That isn't a mathematical argument. You have to formally define your terms. If we were in a computer, there'd be no way for a variable to hold an infinite number of nines…

Note: Don't assume I'm angry because of what my students like to call my "direct tone" in debate. Things like this are my idea of "fun". It's never personal!

Good to know.

Meh, called up a buddy in the math department because I knew I wouldn't be able to sleep until I knew.

A tip of the hat to you, sir. I was indeed wrong.

3 * 0.[3] = 3 * (1/3) = (3*1)/3 = 1 <--- Fuuuuuuuuck.

I'm keepin' my ass in the astronomy lab. To hell with all you math guys. I still don't like it. (lol)

[nightcracker]

Ok, very simple:

1/9 = 0.[1]
2/9 = 0.[2]
3/9 = 0.[3]
4/9 = 0.[4]
5/9 = 0.[5]
6/9 = 0.[6]
7/9 = 0.[7]
8/9 = 0.[8]
9/9 = 0.[9]

9/9 = 1

1 = 0.[9]

To answer haste, I've never been on 4chan, I didn't know you meant it's overdiscussed there. All I know is the boxxy thingy.

New paradox:
To not get things overcomplicated, let's assume the chances of getting a boy and girl are even(in reality girls win by 2% or so).
I have 2 children.
One of my children is a girl.
What's the chance that I have 2 girls?

[Polarina]

Originally Posted by nightcracker What's the chance that I have 2 girls?

The chances are 25%.

[Soeters]

2%, your wife must have cheated on you

[Sythalin]

Never too calculus due to failing trig thanks to TWO different instructors basically telling me "well, if you don't get it, sucks to be you" instead trying to help me like I was originally asking for (tutoring, few more examples, etc).

Sad thing is, I understand this more than I did trig.

[Taryble]

Actually, 50%, because the first being a girl is a given - the only chance we're talking here is the chance that the 2nd child is a girl. That's 50%.

[Sythalin]

Originally Posted by Taryble Actually, 50%, because the first being a girl is a given - the only chance we're talking here is the chance that the 2nd child is a girl. That's 50%.

^^ He's right.

[Nexuapex]

I love/hate probability.

Four possibilities for two children, all equally likely:

Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Drop the Boy - Boy case because at least one child is a girl, and poof:

Boy - Girl: 33.[3]%
Girl - Boy: 33.[3]%
Girl - Girl: 33.[3]%

If we didn't have the information that one child was a girl, then the chance would be 25%.
If we had information that child #1 is a girl, then the chance that child #2 is also a girl would be 50%.
With the given information, the final answer is 33.[3]%.

(It's like the riddle—two coins add up to 30 cents, and one of them is not a nickel. What are the two coins? The answer is that one of them is not a nickel, but the other on is.)

Ooh, ooh, can we do the Monty Hall problem?

[zero-kill]

Definitive nature is impossible to predict, who's to say you won't have a hermaphrodite?

Originally Posted by Nexuapex Ooh, ooh, can we do the Monty Hall problem?

Oh God no...it took me for-eh-verrr to come to terms with that.

[Gandoch]

Originally Posted by nightcracker New paradox: To not get things overcomplicated, let's assume the chances of getting a boy and girl are even(in reality girls win by 2% or so). I have 2 children. One of my children is a girl. What's the chance that I have 2 girls?

You seem to be confused about what a paradox is.

Par-a-dox (par-uh-doks) -noun

1. a statement or proposition that seems self-contradictory or absurd but in reality expresses a possible truth.
2. a self-contradictory and false proposition.
3. any person, thing, or situation exhibiting an apparently contradictory nature.

Originally Posted by Gandoch You seem to be confused about what a paradox is. Par-a-dox (par-uh-doks) -noun a statement or proposition that seems self-contradictory or absurd but in reality expresses a possible truth. a self-contradictory and false proposition. any person, thing, or situation exhibiting an apparently contradictory nature.

Technically true, yes. However the meaning of the word has shifted slightly since it's become more of a part of the everyman's jargon.

My turn now:

There's a group of 23 people standing in front of you. How likely is it that any two of them share the same birthday?

[RPZip]

A bit over 50%, although if you were to use an actual set of real people it'd be even higher as birthday distribution isn't uniform throughout the year.

[nightcracker]

Originally Posted by Taryble Actually, 50%, because the first being a girl is a given - the only chance we're talking here is the chance that the 2nd child is a girl. That's 50%.

Wrong, you don't know wether the first or second child is a girl, and yes that matters.

Originally Posted by Nexuapex I love/hate probability. Four possibilities for two children, all equally likely: Boy - Boy Boy - Girl Girl - Boy Girl - Girl Drop the Boy - Boy case because at least one child is a girl, and poof: Boy - Girl: 33.[3]% Girl - Boy: 33.[3]% Girl - Girl: 33.[3]% If we didn't have the information that one child was a girl, then the chance would be 25%. If we had information that child #1 is a girl, then the chance that child #2 is also a girl would be 50%. With the given information, the final answer is 33.[3]%. (It's like the riddle—two coins add up to 30 cents, and one of them is not a nickel. What are the two coins? The answer is that one of them is not a nickel, but the other on is.) Ooh, ooh, can we do the Monty Hall problem?

For the full 100% right! Here, have a cookie!

Originally Posted by Amenity Technically true, yes. However the meaning of the word has shifted slightly since it's become more of a part of the everyman's jargon. My turn now: There's a group of 23 people standing in front of you. How likely is it that any two of them share the same birthday?

Let's assume leap-years do not exist to not get it overcomplicated.

It's easier to say what is the chance of any of them NOT sharing the same birthday.

The first guy can have his birthday without problems on 365 days of the year, so 365/365 = 1. The second guy 364/365, third 363/365, etc.
So person n has a chance of

(365-(n-1))/365

The chance of all birthdays being on different days is the multiplications of those chances.

So

(365*364*363*...*(365-(n-1)))/365^n

equals the chance of everything being on a different birthday.
Now let's write it easier(if you don't know what ! means, it's the symbol for factorial, google it).


Because

(5*4*3*2*1)/(3*2*1) = (5*4)

and

5*4*3*2*1 = 5!

We can write

365*364*363*...*(365-(n-1))

as

365!/(365-n)!

So it becomes

365!/((365-n)!+365^n)

Since this is the chance of every guy not sharing the same birthday, we must substract 1 by it to get the chance that some are sharing the same birthday:

1-365!/((365-n)!+356^n)

And we fill in 23:

1-365!/((365-23)!+356^23) ≈ 50,7%

[zero-kill]

I don't get it...