1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1
Math Poetry - Page 2
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Paljas
Germany6926 Posts
1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1 | ||
DarkPlasmaBall
United States43547 Posts
On November 20 2013 16:18 TheEmulator wrote: Maths too hard for me (I'm an econ major after all), but I still enjoyed the blog Haha well you need a bit of calculus for economics, right? And thank you! On November 20 2013 16:36 Cyx. wrote: Software engineering, heavy on the graphics and physics ^^ so no, not studying it, but I use a SHITLOAD of it lol =) Yeah my course has all STEM (science, technology, engineering, mathematics) majors, so they understand the relevance of some math topics we've reviewed On November 20 2013 16:46 Chairman Ray wrote: Haha amazing job. I wish I had profs like you. On November 20 2013 17:29 flamewheel wrote: This was cute~ On November 20 2013 18:39 Big J wrote: marvelous Thanks! On November 20 2013 17:00 Artisian wrote: Youtube video? with the slides and your ever so wonderful voice? please? If I ever do, I'll definitely post it here! | ||
DarkPlasmaBall
United States43547 Posts
On November 20 2013 18:01 Danglars wrote: Clever. So much good mathematics/engineering science is done in the shower! I'm a surrogate math teacher to 3 struggling trig/algebra students whose public school teachers are uninterested in their student's success. Let me tell you ... we drill trig identities and common 30/60/90 45/45/90 trig functions until the cows come home! Once they have that down, its the unit circle and odd/even followed by 1/2 angles and double angle identities. That sense of pride they have when they can solve elementary proofs ... nobody can take that away, it's so good. Yeah, I agree My students are starting to learn decent strategies for solving trigonometric proofs (e.g., turn all functions into sines and cosines first), and they're having a lot more success than they did in geometry. There are plenty of interesting math proofs out there, but the only two instances where high school students really learn about implementing proof is to either prove that two triangles are congruent in geometry, or proving trigonometric identities in precalculus. It's a pity that we don't reinforce valid mathematical reasoning and proof with more fun or creative arguments (like the fact that every perfect square can be written as either 4k or 8k+1, where k is an integer). On November 20 2013 19:27 jrkirby wrote: I like my math pickup lines: + Show Spoiler + "Hey girl; Are you a matrix? Because you make my vector go through linear transformations." or, "Hey baby. Are you looking for a guy with a large set of vectors to span your null-space?" and, "How's it going? That dress must be a 4x4 matrix, cause you are going through affine transformation." not to mention, "I wanna find your local derivative, cause I'd do gradient decent on that function all night long." it never ends! "Is your name Taylor? Cause if I did my calculations right, you're an infinite series that is converging on my function." I would make a bunch about tensors, but I never took enough math . I could start on the programming, computer science, and computer graphics ones next though. And the classic: I wish I were your derivative so I could lie tangent to your curves Or perhaps: I wish I were your second derivative so I could explore your concavities On November 20 2013 20:00 Recognizable wrote: From step 4 to 5 something has to be wrong... a-b=0 so you can't divide both sides by (a-b) right? Edit: yay i got it right. Cool all that math education is paying off :D Haha well done! Now can you (or anyone else) actually explain why you can't divide by zero in this case? It's a classic mantra in mathematics, that *you can't divide by zero* ( ::cough:: until calculus and limits ::cough:: ), but... why not? On November 20 2013 20:42 Paljas wrote: 1 = -(i²) 1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1 This is also a great false proof I occasionally show this to my students after they cover imaginary numbers, although we don't really do any proofs with i. | ||
Big J
Austria16289 Posts
On November 20 2013 20:42 Paljas wrote: 1 = -(i²) 1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1 I was thinking of that one as well. Also the one in the OP I know in this version: a = b + c |-2a -a = b+c-2a |+(b+c) b+c-a = 2(b+c-a) 1 = 2 And it always reminds me of the time when I took quantum theory 1 and the professor told us that "now we only have to calculate the determinant of this 2x2 matrix which is easy" and wrote det(A)=ad+bc on the blackboard... Physics... | ||
DarkPlasmaBall
United States43547 Posts
On November 20 2013 21:26 Big J wrote: I was thinking of that one as well. Also the one in the OP I know in this version: a = b + c |-2a -a = b+c-2a |+(b+c) b+c-a = 2(b+c-a) 1 = 2 And it always reminds me of the time when I took quantum theory 1 and the professor told us that "now we only have to calculate the determinant of this 2x2 matrix which is easy" and wrote det(A)=ad+bc on the blackboard... Physics... What do your bars/ vertical lines ( denoted as | in your false proof) refer to? It doesn't look like division or conditional probability. And he wrote ad+bc instead of ad-bc? Would you mind sharing why you're allowed to change the minus sign? | ||
Big J
Austria16289 Posts
On November 20 2013 21:38 DarkPlasmaBall wrote: What do your bars/ vertical lines ( denoted as | in your false proof) refer to? It doesn't look like division or conditional probability. And he wrote ad+bc instead of ad-bc? Would you mind sharing why you're allowed to change the minus sign? Oh, the bars refer to doing the operation to both sides. E.g.: I have a = b + c and then substract 2a from both sides (so |-2a) which leads to -a on the left and b+c-2a on the right. The joke is that the quantum guy was all like "2x2 determinant, that's so easy". And then calculated it wrong, since it is ad-bc ofc. | ||
DarkPlasmaBall
United States43547 Posts
On November 20 2013 22:06 Big J wrote: Oh, the bars refer to doing the operation to both sides. E.g.: I have a = b + c and then substract 2a from both sides (so |-2a) which leads to -a on the left and b+c-2a on the right. The joke is that the quantum guy was all like "2x2 determinant, that's so easy". And then calculated it wrong, since it is ad-bc ofc. Ah that makes sense I like it ^^ | ||
Release
United States4397 Posts
On November 20 2013 20:42 Paljas wrote: 1 = -(i²) 1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1 I got you 1 = 1 (square root both sides) 1 = -1 + Show Spoiler + Square root aint a one-to-one function yo | ||
kubiks
France1328 Posts
On November 20 2013 21:18 DarkPlasmaBall wrote: [/QUOTE]Haha well done! Now can you (or anyone else) actually explain why you can't divide by zero in this case? It's a classic mantra in mathematics, that *you can't divide by zero* ( ::cough:: until calculus and limits ::cough:: ), but... why not? Because R is a field, and in a field, and 0 is not in the group of units. \box. Well you don't divide by 0 essentially because multipliying sth by 0 is not an bijective function, so taking the inverse doesn't make any sense. On November 20 2013 20:42 Paljas wrote: 1 = -(i²) 1= 1^(1/2) = (-(i²))^(1/2) = (-1)^(1/2)*(i²)^(1/2) =i*i = -1 This is also a great false proof I occasionally show this to my students after they cover imaginary numbers, although we don't really do any proofs with i. [/QUOTE] Nice one. I found one time a paper proving P=NP , and in the middle of the proof a beautifull bijection between Q and R. I didn't had to look further :D | ||
DarkPlasmaBall
United States43547 Posts
On November 21 2013 01:23 Release wrote: I got you 1 = 1 (square root both sides) 1 = -1 + Show Spoiler + Square root aint a one-to-one function yo Haha that's true, although I don't think that's nearly as sneaky a false proof as some other ones (for exactly the reason you presented) | ||
Initiative
United States131 Posts
No general procedure for bug checks will do. Now, I won’t just assert that, I’ll prove it to you. I will prove that although you might work till you drop, you cannot tell if computation will stop. For imagine we have a procedure called P that for specified input permits you to see whether specified source code, with all of its faults, defines a routine that eventually halts. You feed in your program, with suitable data, and P gets to work, and a little while later (in finite compute time) correctly infers whether infinite looping behavior occurs. If there will be no looping, then P prints out ‘Good.’ That means work on this input will halt, as it should. But if it detects an unstoppable loop, then P reports ‘Bad!’ — which means you’re in the soup. Well, the truth is that P cannot possibly be, because if you wrote it and gave it to me, I could use it to set up a logical bind that would shatter your reason and scramble your mind. Here’s the trick that I’ll use — and it’s simple to do. I’ll define a procedure, which I will call Q, that will use P’s predictions of halting success to stir up a terrible logical mess. For a specified program, say A, one supplies, the first step of this program called Q I devise is to find out from P what’s the right thing to say of the looping behavior of A run on A. If P’s answer is ‘Bad!’, Q will suddenly stop. But otherwise, Q will go back to the top, and start off again, looping endlessly back, till the universe dies and turns frozen and black. And this program called Q wouldn’t stay on the shelf; I would ask it to forecast its run on itself. When it reads its own source code, just what will it do? What’s the looping behavior of Q run on Q? If P warns of infinite loops, Q will quit; yet P is supposed to speak truly of it! And if Q’s going to quit, then P should say ‘Good.’ Which makes Q start to loop! (P denied that it would.) No matter how P might perform, Q will scoop it: Q uses P’s output to make P look stupid. Whatever P says, it cannot predict Q: P is right when it’s wrong, and is false when it’s true! I’ve created a paradox, neat as can be — and simply by using your putative P. When you posited P you stepped into a snare; Your assumption has led you right into my lair. So where can this argument possibly go? I don’t have to tell you; I’m sure you must know. A reductio: There cannot possibly be a procedure that acts like the mythical P. You can never find general mechanical means for predicting the acts of computing machines; it’s something that cannot be done. So we users must find our own bugs. Our computers are losers! | ||
docvoc
United States5491 Posts
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DarkPlasmaBall
United States43547 Posts
On November 21 2013 02:30 docvoc wrote: Very cool DPB. I like it haha. I can't verify the proof, but I can marvel at how well it rhymes haha. Haha thanks I tried writing each step out on the side so you can follow along ^^ Here it is again, in its entirety: 1. a = b.............................................Given 2. a^2 = ab........................................Multiplication 3. a^2 - b^2 = ab - b^2........................Subtraction 4. (a + b) (a – b) = b (a – b)................Factoring 5. a + b = b.......................................Division 6. b + b = b......................................Substitution 7. 2b = 1b........................................Addition 8. 2 = 1............................................Division QED. | ||
DarkPlasmaBall
United States43547 Posts
On November 21 2013 02:23 Initiative wrote: Heres a peom for the proof of the Halting problem Ive always liked: + Show Spoiler + No general procedure for bug checks will do. Now, I won’t just assert that, I’ll prove it to you. I will prove that although you might work till you drop, you cannot tell if computation will stop. For imagine we have a procedure called P that for specified input permits you to see whether specified source code, with all of its faults, defines a routine that eventually halts. You feed in your program, with suitable data, and P gets to work, and a little while later (in finite compute time) correctly infers whether infinite looping behavior occurs. If there will be no looping, then P prints out ‘Good.’ That means work on this input will halt, as it should. But if it detects an unstoppable loop, then P reports ‘Bad!’ — which means you’re in the soup. Well, the truth is that P cannot possibly be, because if you wrote it and gave it to me, I could use it to set up a logical bind that would shatter your reason and scramble your mind. Here’s the trick that I’ll use — and it’s simple to do. I’ll define a procedure, which I will call Q, that will use P’s predictions of halting success to stir up a terrible logical mess. For a specified program, say A, one supplies, the first step of this program called Q I devise is to find out from P what’s the right thing to say of the looping behavior of A run on A. If P’s answer is ‘Bad!’, Q will suddenly stop. But otherwise, Q will go back to the top, and start off again, looping endlessly back, till the universe dies and turns frozen and black. And this program called Q wouldn’t stay on the shelf; I would ask it to forecast its run on itself. When it reads its own source code, just what will it do? What’s the looping behavior of Q run on Q? If P warns of infinite loops, Q will quit; yet P is supposed to speak truly of it! And if Q’s going to quit, then P should say ‘Good.’ Which makes Q start to loop! (P denied that it would.) No matter how P might perform, Q will scoop it: Q uses P’s output to make P look stupid. Whatever P says, it cannot predict Q: P is right when it’s wrong, and is false when it’s true! I’ve created a paradox, neat as can be — and simply by using your putative P. When you posited P you stepped into a snare; Your assumption has led you right into my lair. So where can this argument possibly go? I don’t have to tell you; I’m sure you must know. A reductio: There cannot possibly be a procedure that acts like the mythical P. You can never find general mechanical means for predicting the acts of computing machines; it’s something that cannot be done. So we users must find our own bugs. Our computers are losers! Haha that's awesome! I don't have experience with computability theory or computer programming, but I read about the Halting problem here: http://en.wikipedia.org/wiki/Halting_problem Nice rhymes! | ||
DarkPlasmaBall
United States43547 Posts
On November 21 2013 01:34 kubiks wrote: Because R is a field, and in a field, and 0 is not in the group of units. \box. Well you don't divide by 0 essentially because multipliying sth by 0 is not an bijective function, so taking the inverse doesn't make any sense. Agreed. When I explain this to my high school/ early college students, they don't have any knowledge of fields (or even what it means to be bijective), so I just show them a very simplistic version of the errors that can occur by dividing by zero, starting with 0 = 0. For example: 0 = 0..............Given 4(0) = 3(0).......Factoring 4 = 3..............Division of the common zero factor (error here) QED Obviously, 4 and 3 can be replaced with any other two numbers you want lol... | ||
CoughingHydra
177 Posts
The proof is by mathematical induction. For n = 1, it's obvious since there's only one number (a1) so all numbers are the same. Now let's assume that the statement holds for some n natural number, that for every n-tuple (a1,...,an) a1=a2=...=an holds. We choose an arbitrary (n+1)-tuple (a1,...,a(n+1)). Now we notice we can construct two n-tuples (a1,...,an) and (a2,...,a(n+1)) so now by assumption we get a1=a2=..=an and a2=a3=...=a(n+1), thus a1=a2=...=an=a(n+1). PMI QED | ||
kubiks
France1328 Posts
On November 21 2013 05:09 CoughingHydra wrote: Our professor gave us this proof that all numbers in an n-tuple are the same: The proof is by mathematical induction. For n = 1, it's obvious since there's only one number (a1) so all numbers are the same. Now let's assume that the statement holds for some n natural number, that for every n-tuple (a1,...,an) a1=a2=...=an holds. We choose an arbitrary (n+1)-tuple (a1,...,a(n+1)). Now we notice we can construct two n-tuples (a1,...,an) and (a2,...,a(n+1)) so now by assumption we get a1=a2=..=an and a2=a3=...=a(n+1), thus a1=a2=...=an=a(n+1). PMI QED Base case isn't covered because you need at least n=2 for your induction (for infering a1=a2) It's a good exemple for people that don't master inductions | ||
Geiko
France1933 Posts
All triangles are equilateral ! http://www.mathematik.com/Isoscele/index.html Works all the time, especially if you draw it on a black board. | ||
DarkPlasmaBall
United States43547 Posts
On November 21 2013 06:26 Geiko wrote: My favorite false proof in geometry is this one: All triangles are equilateral ! http://www.mathematik.com/Isoscele/index.html Works all the time, especially if you draw it on a black board. Hahahaha. There's a running joke in my math department regarding the mindset of students (especially those who take standardized tests and rely too heavily on triangle diagrams that aren't drawn to scale): Prove that the triangle is equilateral. Well, it looks equilateral in the diagram, so therefore it is! | ||
Yorbon
Netherlands4272 Posts
On November 21 2013 06:50 DarkPlasmaBall wrote: hahahaHahahaha. There's a running joke in my math department regarding the mindset of students (especially those who take standardized tests and rely too heavily on triangle diagrams that aren't drawn to scale): Prove that the triangle is equilateral. Well, it looks equilateral in the diagram, so therefore it is! on the blog: Very nice proof indeed. I missed it the first i read through it, so i'm a bit embarassed. But i smile everytime i see one of these, thanks for the blog^^ | ||
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