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ashah Elite user 474 Posts 
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Here is another one.
Let x = -0.5 2x = -1 2x + 1 = 0 x^2 + 2x + 1 = x^2 (x + 1)^2 = x^2 (factor) x + 1 = x (take square root of both sides) 1 = 0 This doesn't have a divide-by-zero problem, but there is another problem which I'll leave you to figure out. 
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jessicashurtz New user 86 Posts
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Quote:
On 2005-01-14 12:29, daffy wrote: that looked like a bit of abstract algebra to me. so what do you call something purple that commutes? ... an abelian grape. 
-j
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RealDeal JU Veteran user New Jersey 375 Posts
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E=MC^2
Thought I would get into all this math. I understand all that is being said but I agree with Bill Palmer, "why?" Jim
"Challenge yourself to come up with your own
material, rather than buying into the idea that you have to do the same thing that everyone else is doing to be a good magician." |
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trevcmagicman New user 52 Posts
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OMG, I'm confused!!! I thought I was good at math....
The magic of Trevor Crandall and Luke Vlassis. The magic men of Nipomo, California.
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SteveFowkes New user Oxon, UK 35 Posts
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E=MC^2 proves that light is merely a figment of everyones imagination.
Energy = Mass x Speed of light in a vaccuum ^2 But light has no mass, hence the answer to the question, "If light travels so fast, how come it doesn't hurt when it hits us?" Anything multiplied by zero = zero. So Energy = zero x 12 million miles a minute (roughly) Energy = 0 So, light doesn't exist as it has no energy. By the way, if you type dark sucker theory into any decent search engine, it's an interesting, if not a bit abstract read.
The world is a wonderful place.
Let's make it more so. Steve |
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Mishel New user Israel 100 Posts
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I have not seen this here yet:
M = 1 + 2 + 4 + 8 + 16 + 32 + 64 + ... M = 1 + (2 + 4 + 8 + 16 + 32 + 64 + ...) M = 1 + 2*(1 + 2 + 4 + 8 + 16 + 32 + 64 + ...) M = 1 + 2M M - 2M = 1 -M = 1 M = -1 hence 1 + 2 + 4 + 8 + 16 + 32 + 64 + ... = -1
Mishel.
Don't let the same dog bite you twice. |
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Eric Leclerc Inner circle Ottawa Ontario 1185 Posts
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On 2004-04-24 21:13, Bill Palmer wrote: eeeexactly.... Imagine dropping this on spectators... the 21 card trick would look first class next to this!!! hehe |
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stanalger Special user St. Louis, MO 998 Posts
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...and 5 x 14 = 25.
Let Ma and Pa Kettle prove it to you: http://www.youtube.com/watch?v=du4idjyMyng |
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Carlos the Great Inner circle California 1234 Posts
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Quote:
On 2004-04-29 22:05, landmark wrote: Hi Jack, A couple (or perhaps, a few) millenia ago, there was a similar type of argument put forth in China. We spent a considerable amount of time on it in my Chinese Philosophy class (to illustrate the different approaches taken versus "Western" philosophy). I wrote a 20-page paper on it but I still never really "got it". In any case, the argument is that "a white horse is not a horse". I'll do some internet searches to see if I can find the whole argument but the first premise has to do with the fact that if you ask for a white horse and you get a black horse, you did not get what you asked for (that is, a "horse" does not fit what you asked for). Anyway, there is a brief (very) excerpt of it here ( http://www.yorkshire-divers.com/forums/n......rse.html ). -Carlos
Cognite tute
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Noel D Regular user 197 Posts
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Quote:
On 2005-07-23 12:08, SteveFowkes wrote: I could be wrong, but e=mc^2 squared refers to the mass of the object you are converting to energy. For example, a 1 ounce paper clip would be 1 ounce x (3x(10^8)^2) And you just kind of go from there, no? |
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Darkfrog Regular user 137 Posts
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On 2006-12-08 18:02, Noel D wrote: No, you are correct. The value c (speed of light) is merely the multiplier to calculate the energy that a certain amount of mass has or how much energy is needed to create that mass. The fact that light has no mass has nothing to do with the equation, it is merely a constant, no different than the Plank constant or Avogadro's constant. |
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Angelo the Magician Loyal user Vienna(Austria/Europe) 217 Posts
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Hi,
one more proof for 1 = 2: a^2 - a^2 = a^2 - a^2 which is obviously true. left side I take a and use distributive rule (I hope this is correct english), and on the right side the formula: a^2 - b^2 = (a+b)*(a-b) so: a*(a-a) = (a+a)*(a-a) divide both sides by (a-a) so: a = a+a and: a = 2a divide both sides by a so 1 = 2 |
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Jonathan Townsend Eternal Order Ossining, NY 27300 Posts
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Lots of things happen when you divide by zero.
Now what's all this about plankton and avocado numbers above?
...to all the coins I've dropped here
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Jonathan_Miller Loyal user CT 211 Posts
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Quote:
On 2005-07-26 11:39, Mishel wrote: The problem here is when you make the claim that (1+2+4+...)=M in the third line you are reindexing the series but when you do that you must also reindex the original series which will make everything work out fine. I like the following proof that 0 = 1 0 = 0 + 0 + 0 + ... = (1-1) + (1-1) + (1-1) + ... = 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 + (-1 + 1) + (-1 + 1) + ... = 1 + 0 + 0 + 0 + ... so 0 = 1. |
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Jonathan_Miller Loyal user CT 211 Posts
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On 2005-04-29 01:42, ashah wrote: When you take the square root of both sides you will technically end up with absolute values. so abs(x+1)=abs(x) but since abs(x+1)<= abs(x) + 1 by the Triangle Inequality you don't end up with the bad result of 1 = 0 but instead with the true result that 0 <= 1. |
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stanalger Special user St. Louis, MO 998 Posts
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Quote:
On 2007-05-21 04:20, Jonathan_Miller wrote: Jonathan_Miller, In your critique of the fallacious proof that 1 + 2 + 4 + 8 + ...= -1, you claim to see a problem with "reindexing." This strikes me as very odd... since there's not an index in sight. Sounds like a fallacious explanation of a fallacious "proof". Would you please elaborate? |
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Jonathan Townsend Eternal Order Ossining, NY 27300 Posts
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Quote:
On 2004-05-02 10:19, MRReed wrote: And the person would be? And the principle would be?
...to all the coins I've dropped here
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stanalger Special user St. Louis, MO 998 Posts
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MRReed hasn't posted at the Café in quite a while.
We'll probably never know what he was talking about, but I'll take a guess at what "more primary math principle" he was referring to. Some algebra teachers do not allow their students to divide both sides of an equation by any expression that involves variables. This prevents "division by zero" errors...but it's a bit of overkill. It's perfectly acceptable to divide both sides of an equation by a variable expression if you know that the variable expression never takes on a value of zero. (If A*B = A*C then either A = 0 or B = C. If you know that A doesn't equal zero, then you can conclude that B = C.) MRReed, are you still visiting the Café? |
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Jonathan_Miller Loyal user CT 211 Posts
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Quote:
On 2007-05-21 09:55, stanalger wrote: Sure I will try. We start off with the series 2^n from n=0 to infinity. Then in the second step we rewrite it as 1+ Sum(2^n) from n=1 to infinity. From this we factor out a 2 and rewrite it as 1+ 2*Sum(2^n-1) from n=1 to infinity. The claim is then made that Sum(2^n-1) from n=1 to infinity is equal to our original series so we just substitute in what we called it, in this case M. However in order to make it M we weould need to make it be Sum(2^n) so we reindex it to be Sum(2^n) from 0 to infinity. Formally what we are doing however is the following substitution, k=n-1, So we get Sum(2^k) from 0 to infinity. In isolation we could just have used the index n again because it doesn't matter to the behavior of the series (i.e. it still diverges and still has the same partial sums) but when we then choose to make the subtraction M-2M we cant have it both ways. We can keep M the same, in which case "2M" is really more like "2K" or we can make 2M be what we want but then our "M" loses the first term. Either way the subtraction gives us 1. I hope that made sense. It's not easy writing math on Magic Café, lol. Basically you can't call the new series M without paying some price which will result in the subtraction working out the way it should. The 0=1 "proof" I gave above is similar in the fact that it also uses infinite series only there what is important are the partial sums. Posted: May 21, 2007 1:16pm --------------------------------- Quote:
On 2007-05-21 10:44, stanalger wrote: Just a fun division by zero story: I was grading some Calc 3 papers (I am a grader at my university) and someone wrote the following (x/0)=0. You would think by Calc 3 you would know that's not true. I see my fair share of 1/(a+b) = 1/a + 1/b but an engineer dividing by 0 is bad. This is why bridges fall down? Just kidding. |
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Moxahalla Special user Los Angeles 751 Posts
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Abbott & Costello at their best. Greg ______________ Sure, Greg...how else could Bud & Lou pay Mr. Fields $28 rent - for 7 weeks, at $13 per week?!:-) |
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Posted: Apr 29, 2005 05:42 am 
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