|
Laremere
|
 |
« Reply #140 on: November 22, 2008, 08:51:53 PM » |
|
≠ ≤ ≥  |
|
|
|
|
Logged
|
If a tree falls in the forest and no one is around to hear it, is sound_tree_fall.play() called?
"Everything that is really great and inspiring is created by the individual who can labor in freedom."
-Albert Einstein
|
|
|
|
Gold Cray
|
|
« Reply #141 on: November 22, 2008, 10:04:53 PM » |
|
x^2 + 2x + 1 = (x+1)^2 (-(x+1))^2 = x^2 + 2x + 1
(x+1)^2 = (-(x+1))^2
(x+1) =/= -(x+1)
y^2 = x^2 y = +/-x (x+1)^2 = (-(x+1))^2 (x+1) = (x+1) -(x+1) = -(x+1) |
|
|
|
|
Logged
|
|
|
|
|
PaleFox
Guest
|
|
« Reply #142 on: November 22, 2008, 10:42:42 PM » |
|
a = b
a + b = b + b
(a + b)^2 = (b + b)^2
a^2 + 2ab + b^2 = 4b^2
a^2 + 2a^2 + b^2 = 4b^2
3a^2 + b^2 = 4b^2
lim 3a^2 + b^2 = 4b^2
b -> 0
3a^2 + 0^2 = 4(0)^2
3a^2 + 0 = 0
3a^2 = 0
(3a^2)/(a^2) = 0/(a^2)
3 = 0
|
|
|
|
|
Logged
|
|
|
|
|
Biggerfish
Guest
|
|
« Reply #143 on: November 22, 2008, 10:51:30 PM » |
|
Indeed
|
|
|
|
|
Logged
|
|
|
|
|
Gold Cray
|
|
« Reply #144 on: November 22, 2008, 11:32:57 PM » |
|
(3a^2)/(a^2) = 0/(a^2)
The problem here is that a = b, so if b goes to 0, so does a, and you get 0/(a^2) = 0/0, which means you have to back up to an earlier part of the problem to figure out what 0/0 is in this case. 0/(a^2) was 3b^2/a^2. a = b, so that is 3. 3 = 3 |
|
|
|
|
Logged
|
|
|
|
|
Laremere
|
|
« Reply #145 on: November 23, 2008, 06:30:58 AM » |
|
Ga, you can use stinking right signs:
Alt = is ≠
Alt < is ≤
Alt > is ≥
Alt shift = is ±
Also, I think your dividing by zero PaleFox, but I need to get ready for the day, so I can't double check.
|
|
|
|
|
Logged
|
If a tree falls in the forest and no one is around to hear it, is sound_tree_fall.play() called?
"Everything that is really great and inspiring is created by the individual who can labor in freedom."
-Albert Einstein
|
|
|
|
PaleFox
Guest
|
|
« Reply #146 on: November 23, 2008, 10:46:13 AM » |
|
k k
( Σ n ) / ( Σ 2n) = .5
n=1 n=1
k k
2( Σ n ) = ( Σ 2n)
n=1 n=1
∞
( Σ n ) = ∞
n=1
∞
2( Σ 2n ) = ∞
n=1
k k
2( Σ n ) = ( Σ 2n)
n=1 n=1
2 ∞ = ∞
2 = 1
By the way, doing "alt =" etc does not work for me, I have to copy and paste over and over for that sort of thing.
|
|
|
|
|
Logged
|
|
|
|
|
team_q
|
|
« Reply #147 on: November 23, 2008, 11:16:47 AM » |
|
2 X infinity is infinity, its not incorrect.
|
|
|
|
|
Logged
|
|
|
|
|
Gold Cray
|
|
« Reply #148 on: November 23, 2008, 11:33:35 AM » |
|
∞/∞ is not always 1.
|
|
|
|
|
Logged
|
|
|
|
|
Chris Whitman
|
|
« Reply #149 on: November 23, 2008, 11:40:29 PM » |
|
x^2 + 2x + 1 = (x+1)^2 (-(x+1))^2 = x^2 + 2x + 1
(x+1)^2 = (-(x+1))^2
(x+1) =/= -(x+1)
a^2=b^2 only implies that |a|=|b|, not that a=b, so this is just bad algebra. a = b
a + b = b + b
(a + b)^2 = (b + b)^2
a^2 + 2ab + b^2 = 4b^2
a^2 + 2a^2 + b^2 = 4b^2
3a^2 + b^2 = 4b^2
lim 3a^2 + b^2 = 4b^2
b -> 0
3a^2 + 0^2 = 4(0)^2
3a^2 + 0 = 0
3a^2 = 0
(3a^2)/(a^2) = 0/(a^2)
3 = 0
The limit is totally unnecessary, but the real problem is here: 3a^2 = 0 (3a^2)/(a^2) = 0/(a^2) Since the reals are a field, they have no zero divisors and since 3!=0, 3a^2=0 implies that a^2=0 and a=0. If a=0, you cannot divide by a since division by zero is undefined, so this is also just bad algebra. |
|
|
|
|
Logged
|
Formerly "I Like Cake."
|
|
|
|
Chris Whitman
|
|
« Reply #150 on: November 23, 2008, 11:43:50 PM » |
|
k k
( Σ n ) / ( Σ 2n) = .5
n=1 n=1
k k
2( Σ n ) = ( Σ 2n)
n=1 n=1
∞
( Σ n ) = ∞
n=1
∞
2( Σ 2n ) = ∞
n=1
k k
2( Σ n ) = ( Σ 2n)
n=1 n=1
2 ∞ = ∞
2 = 1
By the way, doing "alt =" etc does not work for me, I have to copy and paste over and over for that sort of thing.
Once again, if we're dealing with real numbers in the usual way, a series summing to infinity just means that, given any a in R, there exists some integer N such that the kth partial sum is greater than for all integers k>=N. Adding two divergent series term by term just gives you another divergent series. Additionally, the division in the first step is wrong (Edit: that is to say, applying the result of the division when k goes to infinity is wrong; this is not how infinite series work). It makes no sense to divide divergent series, and if you create a new series by dividing them term by term, the result is a divergent series, not a series which converges to 0.5. None of this is spooky, you're basically just saying that if you do math wrong, you get the wrong answer. Congratulations! |
|
|
|
« Last Edit: November 23, 2008, 11:51:00 PM by I Like Cake »
|
Logged
|
Formerly "I Like Cake."
|
|
|
|
Biggerfish
Guest
|
|
« Reply #151 on: November 24, 2008, 01:42:14 AM » |
|
All this math isn't making the scary models on the frontpage go away.
|
|
|
|
|
Logged
|
|
|
|
shinygerbil
Blew Blow (Loved It)
Level 10
GET off your horse
|
|
« Reply #152 on: November 24, 2008, 01:58:19 AM » |
|
Nor is all the maths.
|
|
|
|
|
Logged
|
olücæbelel
|
|
|
|
Laremere
|
|
« Reply #153 on: November 24, 2008, 02:45:50 AM » |
|
x=.9 repeating
*10
10x=9.9rep
-x
9x=9
/9
x=1
|
|
|
|
|
Logged
|
If a tree falls in the forest and no one is around to hear it, is sound_tree_fall.play() called?
"Everything that is really great and inspiring is created by the individual who can labor in freedom."
-Albert Einstein
|
|
|
|
KennEH!
|
|
« Reply #154 on: November 24, 2008, 04:25:55 AM » |
|
Shit, the is the last place I thought I would be learning.
|
|
|
|
|
Logged
|
Madness takes its toll please have exact change.
|
|
|
|
jstckr
|
|
« Reply #155 on: November 24, 2008, 04:33:05 AM » |
|
Stop trying to break the universe
|
|
|
|
|
Logged
|
|
|
|
|
jeb
|
|
« Reply #156 on: November 24, 2008, 05:08:27 AM » |
|
10x=9.9rep
This isn't true, but well... I guess that's the point, right? I can also break maths: 1 = 2  |
|
|
|
|
Logged
|
|
|
|
|
Laremere
|
|
« Reply #157 on: November 24, 2008, 05:15:21 AM » |
|
Why can you not multiply a repeating decimal?
|
|
|
|
|
Logged
|
If a tree falls in the forest and no one is around to hear it, is sound_tree_fall.play() called?
"Everything that is really great and inspiring is created by the individual who can labor in freedom."
-Albert Einstein
|
|
|
|
jeb
|
|
« Reply #158 on: November 24, 2008, 05:18:56 AM » |
|
There's a 0 at the end of infinity. It's just as interesting as
∞ + ∞ = ∞
->
∞ = ∞ - ∞
->
∞ = 0
Which also is incorrect. And boring.
|
|
|
|
|
Logged
|
|
|
|
|
Hajo
|
|
« Reply #159 on: November 24, 2008, 06:19:53 AM » |
|
There's a 0 at the end of infinity. It's just as interesting as
∞ + ∞ = ∞
->
∞ = ∞ - ∞
This step is not "allowed". There are infinites of different magnitude. Adding still leads to infinity, but subtracting only yields zero if the infinities have the same magnitude. Dunno the English terms though (countable and more-than-countable). Sorry to add more boredom ... couldn't resist. |
|
|
|
|
Logged
|
Per aspera ad astra
|
|
|
|
Player
