Fun math paradoxes, problems, and stuff

Level 16
Joined
Feb 4, 2009
Messages
1,315
It depends on your definition of the word "possible" ;)

EDIT:
LOL It reminds me of string theory :D
According to string theory, that's impossible. At one point, you'd have reached the "string" level where you'll
find 1 Dimensional strings forming a 2D and 3D universe =p You can't go further than that .. or can you ;)

how often did humanity believe you can't go any further?

I really want to rename "atoms"
the name does not make sense
 
how often did humanity believe you can't go any further?

I really want to rename "atoms"
the name does not make sense

It's all just theory my friend :p
About atoms, let's call them macros ^^ (the macro prefix playing a major role :p)

EDIT:
More math: ^^

8/0 > 6/0
Aren't I correct?
If you take a system of axes, the line representing the division of 8 would reach higher values for the same denominator compared to the division of 6.
Thus, the line representing the division of 8 for ANY value would reach infinity before the line representing the division of 6 would ^^

Correct me if I'm mistaken.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
@Alain.Mark: Yay! http://en.wikipedia.org/wiki/Taxicab_geometry

And more here from the xkcd geniuses.

Magtheridon96 said:
8/0 > 6/0
Aren't I correct?
If you take a system of axes, the line representing the division of 8 would reach higher values for the same denominator compared to the division of 6.
Thus, the line representing the division of 8 for ANY value would reach infinity before the line representing the division of 6 would ^^

Correct me if I'm mistaken.
I don't get what you're saying. D:
 
Level 15
Joined
Jun 24, 2009
Messages
1,410
Well maybe these were posted before. But it's time for TROLL MATH!
An easy one.
a=x
2a=a+x
2a-2x=a+x-2x
2(a-x)=a+x-2x
2(a-x)=a-x
2=1

Well that one is "true" but it was astonishing for me because I always thought that math is an exact science.
x=0.9* __________ / x10
10x=9.9* __________ / -x
9x=9 __________ / /9
x=1
0.9*=1

And a harder one :)
16-36=25-45 ____________________ Both giving -20 as result
16-36+81/4=25-45+81/4 ____________________ Adding 81/4 to both sides to get two squares
42-2*4*(9/2)+(9/2)2=52-2*5*(9/2)+(9/2)2 ____________________ ----->a2-2*a*b+b2=(a-b)2
(4-9/2)2=(5-9/2)2 ____________________ Non expanded form of the squares
4-9/2=5-9/2
4=5
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
Well maybe these were posted before. But it's time for TROLL MATH!
An easy one.
a=x
2a=a+x
2a-2x=a+x-2x
2(a-x)=a+x-2x
2(a-x)=a-x
2=1
Yeah, that's just divison by zero. I was reluctant to add it to the OP since everyone knows it anyway.

Well that one is "true" but it was astonishing for me because I always thought that math is an exact science.
x=0.9* __________ / x10
10x=9.9* __________ / -x
9x=9 __________ / /9
x=1
0.9*=1
This was covered earlier in the thread, and I posted some nicer proofs (at least, in my opinion they illustrate the solution better). Why does this make math not an "exact science?"

And a harder one :)
16-36=25-45 ____________________ Both giving -20 as result
16-36+81/4=25-45+81/4 ____________________ Adding 81/4 to both sides to get two squares
42-2*4*(9/2)+(9/2)2=52-2*5*(9/2)+(9/2)2 ____________________ ----->a2-2*a*b+b2=(a-b)2
(4-9/2)2=(5-9/2)2 ____________________ Non expanded form of the squares
4-9/2=5-9/2
4=5
(4 - 9/2)2 = (5 - 9/2)2 iff either 4 - 9/2 = 5 - 9/2 or 4 - 9/2 = 9/2 - 5.
 
Level 15
Joined
Jun 24, 2009
Messages
1,410
This was covered earlier in the thread, and I posted some nicer proofs (at least, in my opinion they illustrate the solution better). Why does this make math not an "exact science?".

Beacause 0.9* will never be 1, it will always need a 0.0000000...1 to be 1. The only reason people write it as "1" that noone can write infinite 9-s.(And no, Chuck Norris can't do that.)
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
Beacause 0.9* will never be 1, it will always need a 0.0000000...1 to be 1. The only reason people write it as "1" that noone can write infinite 9-s.(And no, Chuck Norris can't do that.)

There is no 0.000...1 to be added for it to be 1, since, well suppose there is. Say such a number has k 0's in its decimal expansion.
Then 0.999... + 0.000...1 = 1
so 0.999... = 1 - 0.000...1 = 0.999...9
Where 0.999...9 has k 9's in its decimal expansion. This is impossible by definition of 0.999...

And of course, if you say that 0.000...1 has infinitely many 0's in its decimal expansion, then that number is 0, so 0.999... + 0 = 0.999... = 1
 
Level 15
Joined
Jun 24, 2009
Messages
1,410
Well, my mind is full of millions and millions of fuuuu-
I can't even understand that thing in math. I accept it but erm...
hte8ev.png
 
Level 11
Joined
Aug 1, 2009
Messages
963
Consider the expression: the smallest positive integer not representable in under twelve English words.

Does such an integer exist? (this problem is tricky as shit)
No, because if it did it would be representable by the sentence given above, which has eleven words. That didn't seem very tricky. :\

What you said reminds me of the above. Imagine if you have a powerful magnification device (too bad i have non but my imagination) and you magnify a thread you will see > /\/\/\/\/\/\/\/\/\/\/\/\ and for every "/" you see you can zoom in again and see smaller > /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ . But is that even possible?
You can't have anything shorter than a planck length (which is like 10^-30something meters), so eventually you wouldn't be able to zoom in any more.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
mrzwach said:
No, because if it did it would be representable by the sentence given above, which has eleven words. That didn't seem very tricky. :\
Yes. Now, since no such integer exists, then every postitive integer must be representable in under twelve English words. But this is impossible since there are finitely many English words and finitely many permutations on under twelve words.

But I think it was also established earlier in the thread that the problem statement doesn't make sense in the first place haha.
 
Level 5
Joined
Aug 12, 2010
Messages
149
Imagine if you have a powerful magnification device (too bad i have non but my imagination) and you magnify a thread you will see > /\/\/\/\/\/\/\/\/\/\/\/\ and for every "/" you see you can zoom in again and see smaller > /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ . But is that even possible?


You can't have anything shorter than a planck length (which is like 10^-30something meters), so eventually you wouldn't be able to zoom in any more.

Long before you hit the Planck length, you won't be able to describe the string classically. Most straight-chain alkanes are molecular string that look like ".../\/\/..." if you leave out the hydrogen atoms. The verticies are carbon atoms and the lines are C-C bonds. The system is already so small that you have to use quantum mechanics to completely describe it accurately. For example, you don't have a "string" but instead string-like 3-d regions where the probability of finding an electron is greater than some value. And remember - that picture is on a scale about 24 orders of magnitude larger than the Planck length. Things get immensely weirder when you shrink the scale a few more orders of magnitude to that of the nucleus. And the Planck length? That's where even the most audacious physicists will admit, "all bets are off".


That doesn't matter, though... it's a thought experiment so let's suspend disbelief. The string is basically a fractal. Which is cool.
 
Level 7
Joined
May 13, 2011
Messages
310
Well it's not. The notation 0.000...1 implies finitely many 0's in its decimal expansion. I was just saying, if you said there are infinitely many 0's in 0.000...1's decimal expansion, then it would be equal to 0.

Guys, you are messing with infinity here. It doesn't even exist (at least not yet), so don't go too hard out or your heads might explode...

explode.jpg
 
Level 7
Joined
Apr 27, 2011
Messages
272
Gentlemen, the blue parts are the mirrors. I placed this here for stimulation, what do to some how end those bouncing reflections w/ out removing any mirror? There are no wrong answers but be sure they are logical to keep rolling :D (ive got no interesting stuff, "yet")
Mirror.jpg
 
Last edited:
Level 7
Joined
May 13, 2011
Messages
310
Logical solution: Create a black hole. Set it close enough to the mirrors to bend a bit of the light, but not to suck in everything (including you). That little bend will break the bouncing of the reflections.

Illogical solution: Kill infininty.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
http://en.wikipedia.org/wiki/Grandi's_series

In particular:
Wikipedia said:
In modern mathematics, the sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. The sequence of partial sums of Grandi's series is 1, 0, 1, 0, …, which clearly does not approach any number (although it does have two accumulation points at 0 and 1). Therefore, Grandi's series is divergent.

It can be shown that it is not valid to perform many seemingly innocuous operations on a series, such as reordering individual terms, unless the series is absolutely convergent. Otherwise these operations can alter the result of summation. It's easy to see how terms of Grandi's series can be rearranged to arrive at any integer number, not only 0 or 1.

This last result comes from this badass theorem.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
Sooooo there is nothing wrong on the picture, if I understand it correctly.

Wikipedia said:
Therefore, Grandi's series is divergent.
...
It can be shown that it is not valid to perform many seemingly innocuous operations on a series, such as reordering individual terms, unless the series is absolutely convergent.
...
It is a divergent series, meaning that it lacks a sum in the usual sense. On the other hand, its Cesàro sum is 1/2.
A Cesàro sum being essentially an average of partial sums in a series.

So yes, there is something wrong in the picture because it's not talking about Cesàro sums.

Magtheridon96 said:
maybe a picture would help ^^
All you're saying is that, since 8 > 6, then for strictly positive n, 8/n > 6/n. This is obvious. However, this doesn't make sense when n is 0.
 
Level 15
Joined
Sep 3, 2006
Messages
1,738
A Cesàro sum being essentially an average of partial sums in a series.

So yes, there is something wrong in the picture because it's not talking about Cesàro sums.


All you're saying is that, since 8 > 6, then for strictly positive n, 8/n > 6/n. This is obvious. However, this doesn't make sense when n is 0.

I think he's doing something like 8/8-n, 6/6-n where n =/= 8 or 6 thus resulting in the asymptopes.

Of course, you probably knew that, just explaining it to us Algebra and Geometry students ._.
 
Level 15
Joined
Sep 3, 2006
Messages
1,738
It sounds like you're stating the Poincaré Conjecture, not explaining its solution.

MY BAD.

Perelman (using ideas originally from Hamilton) proved the conjecture by deforming the manifold using something called the Ricci flow (which behaves similarly to the heat equation that describes the diffusion of heat through an object). The Ricci flow usually deforms the manifold towards a rounder shape, except for some cases where it stretches the manifold apart from itself (like hot mozzarella) towards what are known as singularities. Perelman and Hamilton then chop the manifold at the singularities (a process called "surgery") causing the separate pieces to form into ball-like shapes. Major steps in the proof involve showing how manifolds behave when they are deformed by the Ricci flow, examining what sort of singularities develop, determining whether this surgery process can be completed and establishing that the surgery need not be repeated infinitely many times.

That is 134 words. Tack on my previous post and it becomes ~160
 
Level 6
Joined
May 5, 2008
Messages
210
A tricky question:
You are in an TV show.
You have three doors to choose from, one whith a price, 2 whith a sheep
after you pick one, the moderator will open one of the other to and you will see a sheep.
he askes you if you want to change your selection.
will you if you want the price?
 
Level 20
Joined
Jul 6, 2009
Messages
1,886
A tricky question:
You are in an TV show.
You have three doors to choose from, one whith a price, 2 whith a sheep
after you pick one, the moderator will open one of the other to and you will see a sheep.
he askes you if you want to change your selection.
will you if you want the price?
Lol i know that one. I've heard people say that if you choose to change your choice, you have greater chance, but that's not true imo.
You've got choice of 2 doors, both totally equal so chance is 50% and changing the choice doesn't matter.
 

fladdermasken

Off-Topic Moderator
Level 38
Joined
Dec 27, 2006
Messages
3,687
A tricky question:
You are in an TV show.
You have three doors to choose from, one whith a price, 2 whith a sheep
after you pick one, the moderator will open one of the other to and you will see a sheep.
he askes you if you want to change your selection.
will you if you want the price?
Mathematically speaking, the odds are still ~33% regardless. If you're one for gambler routines and superstition though, by all means.
Bottom line, it's false mathematics to assume that you have 50% if you switch and 33 if you stick with your initial door.

...

Sort of like this conundrum :p

3 men go into a motel. The man behind the desk said the room is $30, so each man paid $10 and went to their room.

A while later the man behind the desk realized the room was only $25, so he sent the bellboy to the 3 guys' room with $5. On the way, the bellboy couldn't figure out how to split $5 evenly between 3 men, so he decides to give each man US$1 and kept the other $2 for himself.

This implies that the 3 men each paid $9 for the room, which is a total of $27. Adding the $2 that the bellboy kept brings the total to $29.

30=29
 
Last edited:
Level 20
Joined
Jul 6, 2009
Messages
1,886
I've thought of this problem for a bit and i'm kinda not sure, but it seems it is true.
The doors are equal, i couldn't find a reason that one of them has higher chance because...what if you've chosen the other one?

Still, i've got an idea why this could be true even though i couldn't find that explanation on wiki (some of the explanations they gave are false).

Since you've got 2 bad and 1 good choice at start, you've got 2/3 chance to choose the bad one, after the host reveals the second bad choice, switching your choice will give you increased chance since you've most likely picked the wrong one.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
I'd suggest you read and trust the Wikipedia solutions. It really isn't a problem complicated enough to think too much about. If you think you find something specifically wrong with a solution, please post what you think is specifically wrong.
 
I've seen that sequence twice :D
I solved it the first time, but I just can't remember >.<
I think the next one should be:
1
11
21
1211
111221??

1 -> 11
11 -> 21
2 -> 12

I can recall something similar

edit

I just remembered ^^

1 (one 1) ->
11 (two 1s) ->
21 (one 2 one 1) ->
1211 (one 1 one 2 two 1)
111221 (three 1 two 2 one 1)
312211

I love this sequence ^^
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
fladdermasken said:
Turns out this is now a thread to post mathematical conundrums rather than paradoxes.
It has always been about "Fun [...] problems, and stuff!" :D :D

EDIT:
fladdermasken said:
So enough about me and my life, this sequence is fun.

1
11
21
1211

What's the next line in the sequence?
Let's see. Consider the matrix:
gif.latex


I claim that since such a matrix is invertible, I can always find a cubic function which evaluates to any real value from 4 input points in sequence (and more of course). Now let xn be the given sequence. We know that:
x0 = 1
x1 = 11
x2 = 21
x3 = 1211

Consider an explicit definition of xn as follows:
xn = a + b*n + c*n2 + d*n3

We know already that:
1 = a
11 = a + b + c + d
21 = a + 2b + 22c + 23d
1211 = a + 3b + 32c + 33d

This system of equations can be represented in matrix form as follows:
gif.latex


And is easily reduced to the following:
gif.latex


So our final expression for xn is:
xn = 1 + (1210/3)n - 590n2 + (590/3)n3

Which means that
x4 = 4761

Yeah, I was bored.
 
Last edited:
Level 5
Joined
Aug 12, 2010
Messages
149
I've seen that sequence twice :D
I solved it the first time, but I just can't remember >.<
I think the next one should be: ...

Google Aptitute Test FTW!!! Yep, you got it.

Yeah, I was bored.

I noticed that around the time when you started putting images of matricies in your posts lol... you're basically the sort that the XKCD guy was talking about being particularly easy to nerd snipe.

If you have time to blow on fun math-y stuff, go look up Leonard Susskind's lectures on youtube. There are several lecture series (each lecture clocks in at nearly 2 hours) and quite a few cool topics to choose from.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
Zarathustra said:
I noticed that around the time when you started putting images of matricies in your posts lol...
Psh, THW needs LaTeX tags!

Zarathustra said:
If you have time to blow on fun math-y stuff, go look up Leonard Susskind's lectures on youtube. There are several lecture series (each lecture clocks in at nearly 2 hours) and quite a few cool topics to choose from.
Oh yes, I started watching his lectures on General Relativity a while ago, but gave up for some reason. Very interesting stuff. If I could study anything and not have to work in my lifetime, I'd probably study physics before CS, but oh well.
 
Level 7
Joined
May 13, 2011
Messages
310
These conundrums are just getting too much right now. I'm only in Year 11!

...paradoxes were funner.
 
Level 19
Joined
Apr 22, 2007
Messages
1,960
Here is an old problem I liked off a high school math competition I participated in a few years ago. You don't actually need to know very much math at all to solve it.

A piece of software generates random numbers between 0 and 1. It is designed so that for every x between 0 and 1, the probability that it will generate a number smaller than x is three times the probability that it will generate a number smaller than x/4. Moreover, the probability that it will generate a number higher than or equal to x is the same as the probability that it will generate a number smaller than 1 - x. Calculate the probability that this program will generate a number smaller than 1/21.
 
Top