| Forum Index > General Forum |
|
imBLIND
United States2626 Posts
On October 09 2016 14:08 Epishade wrote: ![[image loading]](http://i.imgur.com/jClSWXD.jpg) What do you guys think of my mathematical proof that square root of 2 equals 2? It'll fool a lot of people if they don't have a basic understanding of trigonometry... it's not 1 + 1 = 2, it's 2*sin(pi/4) = sqrt(2). | ||
|
Thieving Magpie
United States6752 Posts
On October 09 2016 14:08 Epishade wrote: What do you guys think of my mathematical proof that square root of 2 equals 2? Step one cannot happen. | ||
|
Cascade
Australia5405 Posts
On October 09 2016 14:08 Epishade wrote: What do you guys think of my mathematical proof that square root of 2 equals 2? Neat.  It also illustrates how it's not as straight forward as you would think to mathematically define length. Not to mention area or volume. | ||
|
Cascade
Australia5405 Posts
It's the last step that's dodgy. | ||
|
Khalum
Austria831 Posts
I like how 1+1 is already not equal to sqrt(2) but you still claim this to be the case in the end. I'll show this to some people and see how they react. [edit] That's not what I meant to respond with but I'm not quite awake yet and that's what came out.. So yeah... Still amazing "proof". | ||
|
LaughNgamez
Canada525 Posts
| ||
|
Thouhastmail
Korea (North)876 Posts
| ||
|
Deleuze
United Kingdom2102 Posts
On October 09 2016 18:42 Thouhastmail wrote: how do you guys read "imgur"? "Imgur". How do you read it? | ||
|
Simberto
Germany11941 Posts
On October 09 2016 18:11 Khalum wrote: Amazing! I like how 1+1 is already not equal to sqrt(2) but you still claim this to be the case in the end. I'll show this to some people and see how they react. [edit] That's not what I meant to respond with but I'm not quite awake yet and that's what came out.. So yeah... Still amazing "proof". That is not the structure of the fake proof. The structure is that one of the sides of the thing is sqrt 2, and the other is 1+1. Then, the argument makes a limit observation which does not change the length of the edged part of the figure. That "shows" that both of them are equal. And thus, sqrt2 needs to be equal to 2. The problem here is that the limit observation doesn't actually work that way. Which is way more subtle then just starting with a wrong assumption. People are bad with infinites. | ||
|
Khalum
Austria831 Posts
On October 09 2016 20:18 Simberto wrote: That is not the structure of the fake proof. The structure is that one of the sides of the thing is sqrt 2, and the other is 1+1. Then, the argument makes a limit observation which does not change the length of the edged part of the figure. That "shows" that both of them are equal. And thus, sqrt2 needs to be equal to 2. The problem here is that the limit observation doesn't actually work that way. Which is way more subtle then just starting with a wrong assumption. People are bad with infinites. Indeed, as I said, the thing I was trying to say was not what I actually typed. But since it was already posted I just went ahead with an edit to not make myself look more stupid than I actually am I meant to say that I liked the way he constructed his "proof", even if it's nonsense. I found it to be creative and it made me smile. | ||
|
ZigguratOfUr
Iraq16955 Posts
On October 09 2016 20:18 Simberto wrote: That is not the structure of the fake proof. The structure is that one of the sides of the thing is sqrt 2, and the other is 1+1. Then, the argument makes a limit observation which does not change the length of the edged part of the figure. That "shows" that both of them are equal. And thus, sqrt2 needs to be equal to 2. The problem here is that the limit observation doesn't actually work that way. Which is way more subtle then just starting with a wrong assumption. People are bad with infinites. The last step is amusing, with the key observation being that the length of the step function is the sum of its differentiable components, the number of which approach infinity at the same rate as the difference in length is decreasing for each segment. If you replaced the step-wise with a fractal and applied the same reasoning you could probably argue that sqrt(2) = ∞. | ||
|
Cascade
Australia5405 Posts
length(limit(paths)) Is not always equal to limit(length(paths)), even if both converge nicely. So yes, be careful around infinities. | ||
|
SoSexy
Italy3725 Posts
Still, I love to watch them unfold, but without the boring parts walk from here to here etc. Do you guys know if there are any good montages? Like this one: I know it's a trailer but I would like to watch 40-1 hour recap of the game. | ||
|
Thieving Magpie
United States6752 Posts
Neither 1 + 1 = Root 2 nor does .5^2+.5^2+.5^2+.5^2 = 2 as the proof suggests. Step one is impossible to be true. | ||
|
ZigguratOfUr
Iraq16955 Posts
On October 10 2016 14:04 Thieving Magpie wrote: Neither 1 + 1 = Root 2 nor does .5^2+.5^2+.5^2+.5^2 = 2 as the proof suggests. Step one is impossible to be true. The first step doesn't claim any of that. And why are you squaring the 0.5s? | ||
|
Thieving Magpie
United States6752 Posts
On October 10 2016 14:42 ZigguratOfUr wrote: The first step doesn't claim any of that. And why are you squaring the 0.5s? for 1, 1, Root 2 to be true, it is actually 1^2 + 1^2 = 2^2, and if that is true, then that means he is suggesting that .5^2 + .5^2 + .5^2 + .5^2 = 2^2 + 2^2 If he is trying to say that 1 + 1 = root 2, then that would be false from the start as that is an impossibility. The third option is that he is showing zero relations between multiple pictures and then saying 2=Root 2 at the end--which would be the dumbest of the 3 options. | ||
|
Cascade
Australia5405 Posts
On October 10 2016 14:04 Thieving Magpie wrote: Neither 1 + 1 = Root 2 nor does .5^2+.5^2+.5^2+.5^2 = 2 as the proof suggests. Step one is impossible to be true. The proof doesn't have any text, but they way I read it is that the first few steps are just triangles with sides of different lengths. The length of the hypotenuse calculated from pythagoras theorem. No one is claiming that the (unsquared) lengths of the sides should sum up to the hypotenuse here. At the last step though, taking the limit of the series of zigzag paths, the two become the same. The zigzag paths converge to the hypotenuse. Then the proof is (incorrectly) assuming that because the lengths of all the zigzag paths is 2, then the length of the limit of the paths (the hypotenuse) must also be 2. Which would then show that the lengths of the hypotenuse is both sqrt(2) (from pythagoras theorem in first fig) and 2 (from the incorrect limit argument). I think that's how it is supposed to be read. | ||
|
Thieving Magpie
United States6752 Posts
On October 10 2016 14:59 Cascade wrote: The proof doesn't have any text, but they way I read it is that the first few steps are just triangles with sides of different lengths. The length of the hypotenuse calculated from pythagoras theorem. No one is claiming that the (unsquared) lengths of the sides should sum up to the hypotenuse here. At the last step though, taking the limit of the series of zigzag paths, the two become the same. The zigzag paths converge to the hypotenuse. Then the proof is (incorrectly) assuming that because the lengths of all the zigzag paths is 2, then the length of the limit of the paths (the hypotenuse) must also be 2. Which would then show that the lengths of the hypotenuse is both sqrt(2) (from pythagoras theorem in first fig) and 2 (from the incorrect limit argument). I think that's how it is supposed to be read. I understand. For root 2 on the first image to be true, Pythagorean Theorem *has* to be used. The sequence suggests a "likeness" of all the hypotenuses. However, the 2nd image is wrong if the first image is to be trusted as a benchmark. .5^2+.5^2 = C^2 2.5 + 2.5 = C^2 5 = C^2 Root 5 = C As such, the second image should actually have a hypotenuse of 2root5 and not root2 The 2nd image is also problematic as it actually has a side of 4root.125 | ||
|
Blisse
Canada3710 Posts
On October 10 2016 15:12 Thieving Magpie wrote: I understand. For root 2 on the first image to be true, Pythagorean Theorem *has* to be used. The sequence suggests a "likeness" of all the hypotenuses. However, the 2nd image is wrong if the first image is to be trusted as a benchmark. .5^2+.5^2 = C^2 2.5 + 2.5 = C^2 5 = C^2 Root 5 = C As such, the second image should actually have a hypotenuse of 2root5 and not root2 The 2nd image is also problematic as it actually has a side of 4root.125 .5^2 is .25 not 2.5 C is √0. 5 which is equal to √2/2 The math checks out. | ||
|
Cascade
Australia5405 Posts
On October 10 2016 15:12 Thieving Magpie wrote: I understand. For root 2 on the first image to be true, Pythagorean Theorem *has* to be used. The sequence suggests a "likeness" of all the hypotenuses. However, the 2nd image is wrong if the first image is to be trusted as a benchmark. .5^2+.5^2 = C^2 2.5 + 2.5 = C^2 5 = C^2 Root 5 = C As such, the second image should actually have a hypotenuse of 2root5 and not root2 The 2nd image is also problematic as it actually has a side of 4root.125 Hmm, I really don't understand what you mean. Pythagoras theorem is taken as given, and is valid through all the images. First image: 1^2 + 1^2 = sqrt(2)^2 = 2 fits. Second image, each little triangle has sides of .5, and hypotenuse of sqrt(2)/2 = 1/sqrt(2) .5^2 + .5^2 = (1/sqrt(2))^2 perform the squares... 0.25 + 0.25 = 1/2 both sides are 1/2 so again things fit. The total length of the diagonal line is the sum of the two smaller ones, which is still sqrt(2). I don't see the problem. Maybe you slipped with the squares and square roots? | ||
| ||
|
|
WardiTV Weekly
Sparkling Tuna Cup
INu's Battles
Cure vs herO
ByuN vs Rogue
PiGosaur Cup
The PondCast
Kung Fu Cup
Patches Events
Replay Cast
CrankTV Team League
OSC
[ Show More ] |
|
|
EPT
