AGS Steam Games

Forums → The Tavern → Math Thread

386 73057
daleks
offline
daleks
3,766 posts
Chamberlain

This thread is just what the title says it is. It is all about Math. You can post math jokes(the jokes don't have to be good), math questions, what you like about Math, or even why you hate Math.

My math joke: Resistance is not futile. It is voltage divided by amps.

  • 386 Replies
StormWalker
offline
StormWalker
8,231 posts
Jester

you can curse in school! "you're such an Angle-Side-Side hole!"

Gantic
offline
Gantic
11,891 posts
King

[quote]Oh, and another proof that .999...=1: 1/9=.111... 2/9=.222... 3/9=.333... 4/9=.444... 5/9=.555... 6/9=.666... 7/9=.777... 8/9=.888... therefore: 9/9=.999...


But no, that's just like the 1/3. .111.. multiplied by 9 is .999.., just like the one-third EmperorPalpatine pointed out. 1/9 does not exactly equal .111.. [/quote]

Prove 1/3 =/= 0.333...
Gantic
offline
Gantic
11,891 posts
King

And if 1/3 is truly .333.. then .333.. multiplied by three would equal 1, but it does not (.333.. * 3 = .999..). They are just an estimate of the approximate value. That is, unless infinitum caused numbers to increase and decrease, but I don't see how infinite would help in it's case; it doesn't seem to make much sense unless it somehow ended in four, but that would in turn make it too large. Yeah, Iduno.


Then you need to prove that 1 =/= 0.999...

What you're thinking is 0.999... = 0.999...9 with an infinite number of 9's in between the first 9 and the last 9 which is not true because 0.999... does not have a last digit. There is no last 9, because there is no last digit because you can always add another 9 to the end and keep going. 1/3 is equal to 0.333... but that's the same as 0.333...3.
daleks
offline
daleks
3,766 posts
Chamberlain

But infinite is still a value

It is a value but not really a number.
Then you need to prove that 1 =/= 0.999...

Asymptote on 1. It would never reach 1 but could get to .999...
Gantic
offline
Gantic
11,891 posts
King

[quote]Then you need to prove that 1 =/= 0.999...


Asymptote on 1. It would never reach 1 but could get to .999...[/quote]

An asymptote is not a number. It is a line. Saying that a number is not equal to a line to prove that a number is not equal to a number is ludicrous.
daleks
offline
daleks
3,766 posts
Chamberlain

An asymptote is not a number. It is a line. Saying that a number is not equal to a line to prove that a number is not equal to a number is ludicrous.

Have a graph of a curve that has an asymptote on 1. That is what I meant. Probably should have been more clear.
johnmerz
offline
johnmerz
536 posts
Shepherd

I'm know that this doesn't necessary prove anything, but there's even a wiki on it.

aknerd
offline
aknerd
1,416 posts
Peasant

If you had a line that was infinitely long, is it not still a line, or has length?


Infinitely long lines (or just, you know, lines...) are actually circles through the point at infinity. And yes, this does imply that all lines intersect each other at least once, even parallel ones.

.333.. x 3 still equal .999..?


Yes, because .99999.... equals one.

I can prove to you that 1/3 equals exactly 0.3333333333....

This is because 1/3 is a rational number (why? because one and three are both integers). The decimal expression of ALL rational numbers either a) eventually terminates or b) eventually starts repeating some finitely long string of numbers. I could prove this if you really want, but it is very intuitive.

ANYWAY, so, now if we actually go through the calculation (some notation: due to AG's lack of support for math type, b|n.0 = m._ R K means n/b = m.something remainder k)

3|1.0 = 0.3_ R 0.1
=> 3|0.10 = 0.03_ R 0.01
=> 3|0.01 = 0.003_ R 0.001
.
.
. and so on.

Then, it is clear that IF 3|0.(n zeros)1 = 0.(n+1 zeros)3 R 0.(n+1 zeros)1

Then 3|0.(n+1 zeros)1 = 0.(n+2 zeros)3 R 0.(n+2 zeros)1

=> by induction, the sum of 3|0.(n zeros)1 , where n goes from -1 to infinity must = 0.33333333....

Yet this sum is exactly 1 divided by three.

=> 1/3 = exactly 0.3333333....

On the issue of "getting extremely close but never touching".... This doesn't really apply. We say that things get really close to a number without equaling it when we can ALWAYS find a number in between those two numbers.

For instance: The sequence 1/n , where n is the sequence of natural numbers approaches 0 but does not equal zero. This is because we can always find a number 1/(n+1) that is less than 1/n but greater than 0.

0.333.... and 0.999.... are not sequences.
0.3333... =/= the sequence 0.3, 0.33, 0.333, 0.3333 etc. It is the limit of this sequence for sure, but this sequence does not ever equal its own limit. 0.33333.... is ALWAYS greater than 0.33...3. It is ALWAYS less than 0.33...4.
daleks
offline
daleks
3,766 posts
Chamberlain

And yes, this does imply that all lines intersect each other at least once, even parallel ones.

I thought only parallel lines on different planes would intersect.
nichodemus
offline
nichodemus
14,990 posts
Grand Duke

Am I still Asian if I really can't be bothered with the intricacies of such math?

Ghostofelements
offline
Ghostofelements
99 posts
Nomad

What's the benefit of attesting the equality of two triangles?

aknerd
offline
aknerd
1,416 posts
Peasant

Am I still Asian if I really can't be bothered with the intricacies of such math?

Yes, but this means you definitely are not Eastern European.

I thought only parallel lines on different planes would intersect.


Well, no. I explained it well I some other thread...

So, this is something I learned this year which is pretty nifty. As many of you probably know, we run into problems when we try to take square roots of negative numbers. So, as a nice little workaround, we just define the square root of -1 as the number "i"*, an imaginary number. This gives us a whole domain of complex numbers (we call them z's, by convention) of the form z = x + iy, where both x and y are real numbers (so iy is the imaginary part of z, and x is the real part).

This is where things get interesting: in order to keep track of complex numbers, we don't use a number line like we do for real numbers, but a plane (called the complex plane). We represent each point z=x+iy on the plane by its x and y coordinates, just like we would if we were graphing an equation of one variable onto a real graph.

However, something cool we can do with this complex plane is "map" it onto a sphere (in this case called a Riemann sphere). We create a basic bijective, conformal function that takes each point on the complex plane, and relates it to a 3D point on the sphere. Similarly, we take the inverse of the function to map every point on the sphere back onto the plane.

When we do this, we see that all lines through the origin (the point 0,0) on the plane are mapped to circles on the sphere that intersect the north and south poles, IE longitudes.

THIS is the interesting part (I SWEAR). So, the circles intersect each other twice (once at each pole) but the lines only appear to intersect each other once. Because the function we are using is conformal, we know that every intersection on the sphere must be mapped from an intersection on the plane. So, where is the missing intersection on the plane? I'll tell you! It is at the POINT at infinity! You see, in the complex plane, there is no difference between "negative" infinity and positive infinity. I mean, what would infinitely imaginary mean, anyway? So, we consider all the different kinds of complex infinity to just be one point on the plane that all lines must pass through.

So, all non-parallel, non-identical lines in the complex plane intersect twice, and all parallel lines intersect once. Crazy!

Ahh there it is. The first two paragraphs have already been discussed in this thread, but I think they might be necessary for context. If you are really interested, I can give you the bijective function that maps the complex plane onto the Riemann sphere.
Poizaz00
offline
Poizaz00
605 posts
Peasant

What does a smart turret say? Tangent aquired. *audience imitates dying camel*

daleks
offline
daleks
3,766 posts
Chamberlain

Well, no. I explained it well I some other thread...

So only when you are using a plane with imaginary numbers parallel line will intersect, but on a normal Cartesian plane they wouldn't?
aknerd
offline
aknerd
1,416 posts
Peasant

So only when you are using a plane with imaginary numbers parallel line will intersect, but on a normal Cartesian plane they wouldn't?


With complex numbers, we have the complex plane, and the extended complex plane. The extended plane is the one that includes the point at infinity and can be mapped onto a Riemann sphere. The extended complex plane, though, is basically a Cartesian coordinate system (though we don't call it that to avoid confusion).

A Cartesian coordinate system has more to do with how the axis are set up than with the nature of the numbers being described. So, it should work with real numbers as well, though I don't know the name of the surface it would be mapped onto. There is an analog of the point at infinity on the real number line, however. It is called the "real projective line", and it looks like this:

http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Real_projective_line.svg/150px-Real_projective_line.svg.png

There is also the extended real number line, which has both negative and positive infinity. Which number line you choose depends on what kind of work you are doing.

And, just for fun, this is a Riemann sphere:
http://media-2.web.britannica.com/eb-media/23/70823-004-2AD75C17.jpg
The lines show how points are mapped between the sphere and plane.
Showing 151-165 of 386