Let's discuss Mathematics

:lol:

I can give an example that takes at least 50 iterations before reaching a multiple of 4, so writing out all those possibilities may take awhile.

I also have plenty more interesting problems where that came from, but typing out some of the proofs in the forum will be problematic.
 
I'm pretty sure there's a way to embed LaTex using a website... I'll have a look.

EDIT: Yep, check out www.texify.com

EDIT2: Does the proof involve Fermat's Little Theorem?
 
From XKCD:

collatz_conjecture.png
 
Hmm, I'll have another think about it, where to start is always hardest step ;)

Does it involve counting set bits of the binary representation of the initial number, by any chance?
 
Everything I did was in decimals. I'd have to think how it'd look with binary.

The stuff before with 4k + 3 is sort of on the right track, or at least I can see something there to look at which should lead towards the way I did it.
 
It's fairly easy in binary, even case (bit 0 not set) is just a shift right 1 bit.

Odd case is add a 1 to the end of the binary representation (maps x to 2x+1) then add the original number.

I was doing some calculations on numbers of the form ...11 in binary but wasn't getting very far (which is the 4k+3 case). As long as you reach a case where the 2 least significant bits are 0 you prove the theorem.

The reason I mentioned set bits count is that you discount the case where only 1 bit is set in the binary, because that would mean it's a power of 2 => just repeatedly div by 2 until reaching 1, 4, 2.
 
That would be pointless though DMOC ;)

I'm too drunk to investigate proving that Collatz thingamajig reaches an ineteger divisible by 4 though.

Hope you get in MIT!
 
I've been known to come up with maths stuff while drunk. Anyway, a hint about one of the two things to think about in order to go from what you had to what I did. If 4k + 3 goes to 4m + 3 after two iterations, what's the ratio of m:k?
 
I'm too drunk ;)

Interested in the proof though.

EDIT: Especially cos it wasn't mentioned on mathworld or wiki, so it's either homework or a new discovery :lol:
 
Stop trolling Quackers.

0.999... = 1 is elementary, as I said before, what would you add to get 1?

There is one interesting answer though, so Ill spoiler it.

Spoiler :

LOL do your own research! It begins with a lowercase e though.


2+2 = 4 has not yet been proved under ZFC as far as I'm aware, I'm not that brilliant.
 
0.999... = 1 is elementary, as I said before, what would you add to get 1?
I would call such a number "1 - 0.999999...". It's not like you've proven that all real numbers can be expressed in decimal notation.

If 0.999... = 1 is a question of notation. To be useful, we can define it as a limit, in which case it is true. However, if we do not define it as a limit, then the equality is false. For instance we can define it as "the sum of {for all N|9*10^-N}". Since this definition is not a limit, this infinite sum does not quite reach 1. Each part in the sum is one step closer to 1, but at no point does it quite reach 1. You can't even say that at infinity the sum reaches 1, because all talk about something happening at infinity implicitly means it's limiting behavior; infinity isn't a real number and it is only for convenience that we sometimes treat it like one.

Note that this isn't special to 0.999..., but applies to all repeating decimals. 0.111..., 0.010101...., 0.0055555.... ext.

But 0.9999... not defined as a limit is never useful, therefore we may as well define it as a limit and therefore equal to 1. This allows us to do useful things like write any rational number in the form:
Code:
[FONT=System][FONT="Lucida Console"]   _
a.bc[/FONT][/font]
On the other hand, not defining 0.99999... as a limit, means that 0.999... is it's own distinct real number. Though not rational, it does have a place on the number line. Like any real number except 1 itself, there are an infinite number of other numbers between it and 1. The number comes more naturally from the notation, since nothing in the notation speaks "limit", and since doing so makes every decimal notation number unique.
 
Let x = 0.999...

10x = 9.999...

10x - x = 9.

10x = 9 + x
=> 9x = 9

=> x=1

QED
 
I told you to stop trolling in my thread. Last warning ;)

Explain what you mean by "2" "+" and "=".
 
I would call such a number "1 - 0.999999...". It's not like you've proven that all real numbers can be expressed in decimal notation.

If 0.999... = 1 is a question of notation. To be useful, we can define it as a limit, in which case it is true. However, if we do not define it as a limit, then the equality is false. For instance we can define it as "the sum of {for all N|9*10^-N}". Since this definition is not a limit, this infinite sum does not quite reach 1. Each part in the sum is one step closer to 1, but at no point does it quite reach 1. You can't even say that at infinity the sum reaches 1, because all talk about something happening at infinity implicitly means it's limiting behavior; infinity isn't a real number and it is only for convenience that we sometimes treat it like one.

Note that this isn't special to 0.999..., but applies to all repeating decimals. 0.111..., 0.010101...., 0.0055555.... ext.

But 0.9999... not defined as a limit is never useful, therefore we may as well define it as a limit and therefore equal to 1. This allows us to do useful things like write any rational number in the form:
Code:
[FONT=System][FONT="Lucida Console"]   _
a.bc[/FONT][/font]
On the other hand, not defining 0.99999... as a limit, means that 0.999... is it's own distinct real number. Though not rational, it does have a place on the number line. Like any real number except 1 itself, there are an infinite number of other numbers between it and 1. The number comes more naturally from the notation, since nothing in the notation speaks "limit", and since doing so makes every decimal notation number unique.

Last sentence is wrong. 0.999... is rational. It's equal to 1.

Any recurring decimal is a rational number. Or are you gonna argue 0.3333... is not 1/3?

EDIT: Fun challenge - give me a recurring decimal, I'll tell you the rational number it represents. If you beat me, you win the internets.
 
Last sentence is wrong. 0.999... is rational. It's equal to 1.

Any recurring decimal is a rational number. Or are you gonna argue 0.3333... is not 1/3?

Precisely.

That's what I meant by: Note that this isn't special to 0.999..., but applies to all repeating decimals. 0.111..., 0.010101...., 0.0055555.... ext.

1/3 > 0.3333...

EDIT: This holds unless of we specially define it as such; unless we define a recurring decimal as a limit.
 
You're both wrong, sorry.

1/3 is not > than 0.333... [EDIT: Cos they're the same]

By 10x I mean multiplying by 10 which shifts the radix.

You do realise this argument has been fought and won many times over?
 
Back
Top Bottom