rookie1ja Posted March 30, 2007 Report Share Posted March 30, 2007 Why 1 = 2 - Back to the Cool Math Games Find the mistake in these mathematical equations. x = 2 x(x-1) = 2(x-1) x2-x = 2x-2 x2-2x = x-2 x(x-2) = x-2 x = 1 This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Why 1 = 2 - solution The equation is solved the right way, apart from one little detail. There must be stated that x does not equal y, because there would be dividing by zero, which is not defined in maths. Link to comment Share on other sites More sharing options...

Guest Posted April 20, 2007 Report Share Posted April 20, 2007 x(x-2) = x-2 implies that either x = 1 or x = 2. Since x = 2, x = 1 does not have to be true. Therefore, concluding x = 1 from above equation is wrong. Link to comment Share on other sites More sharing options...

rookie1ja Posted April 20, 2007 Author Report Share Posted April 20, 2007 x(x-2) = x-2 implies that either x = 1 or x = 2. Since x = 2, x = 1 does not have to be true. Therefore, concluding x = 1 from above equation is wrong. x(x-2) = x-2 ... then dividing the equation by (x-2) ... this is what can't be done ... we can't divide by zero x(x-2)/(x-2) = (x-2)/(x-2) ... which can be simplified to x = 1 Link to comment Share on other sites More sharing options...

Guest Posted April 28, 2007 Report Share Posted April 28, 2007 Technically there is no error. If anything is wrong, it was introducing x into the multiple in the second line. By including x, a quadratic equation was created, which will always have 2 solutions for x. If you solve the quadratic equation, the 2 solutions are 1 and 2. As we already knew the stated value for x from line 1, applying this to line 2 removes x as an unknown variable and hence removes the possibility of multiple potential values for x. Link to comment Share on other sites More sharing options...

Guest Posted May 7, 2007 Report Share Posted May 7, 2007 There are two solutions. Both can be gotten with the quadratic equation (I don't remember how anymore), or they can be gotten the way you did it. There's your solution which gets x = 1, then there's this solution which gets x = 2: x(x - 1) = 2(x - 1) x(x - 1) / (x - 1) = 2(x - 1) / (x - 1) x = 2 Notice that I did the exact same thing you did, but I didn't re-arrange it to get the other solution. Now, we can verify that both x = 1 and x = 2 are correct: Substitute for x = 2: x(x - 1) = 2(x - 1) 2(x - 1) = 2(x - 1) which is true just because both sides are the same 2(2 - 1) = 2(2 - 1) 2(1) = 2(1) 2 = 2 Since the result is true, X = 2 is true. Substitute 1 for x = 1: x(x - 1) = 2(x - 1) 1(1 - 1) = 2(1 - 1) 1(0) = 2(0) 0 = 0 Since the result is true, X = 1 is true. Of course, since we were already given that x = 2, there was no need to do anything but substitute to verify that x = 2 is correct. So the obvious mistake is that you were trying to solve for x when you already knew its value. Link to comment Share on other sites More sharing options...

Guest Posted May 19, 2007 Report Share Posted May 19, 2007 This has nothing to do with why 1=2 it just shows how you can input a one or a two into and equation and it comes out equal on both sides, and since plugging in one gave you zero=zero and plugging in two gave you two=two they are not equal. Link to comment Share on other sites More sharing options...

Guest Posted May 21, 2007 Report Share Posted May 21, 2007 Fellows, that's not what the puzzle is about at all. Roockie is introducing a proof that shows that 1 = 2. There is obviously something wrong with the proof. AND that what is wrong is that you can not divide both sides of an equation with (x-2) when x=2 because you'll get ZERO (0) and you are not allowed to divide by "0". { x/0 = undefined } Tangents: the equation x(x-1)=2(x-1) has 2 solutions but that is unrelated to anything. Because from the get go the host announced that x=2. And this equation was a manipulation of x=2. And you can not get a new value for x because x only = to 2 and nothing else. Link to comment Share on other sites More sharing options...

Guest Posted June 23, 2007 Report Share Posted June 23, 2007 Alright, try this one on for size: X=.9999999... 10x=9.99999... (10x)-x=9 (9x)/9=1 x=1 .999999...=1 Link to comment Share on other sites More sharing options...

Guest Posted June 24, 2007 Report Share Posted June 24, 2007 Alright, try this one on for size: X=.9999999... 10x=9.99999... (10x)-x=9 (9x)/9=1 x=1 .999999...=1 I see where the problem could lie. This one sure is an interesting one. I was starting to think that some of the arithmetical principals we accept are dogmatic. ^^^^^^^^^^^^^^^^^^^^^^^^^^__ The difference between 1 and 0.9 we can not represent in a number format. At least the level of math that I have reached I have never encountered a way of putting a number like that down on paper. (unless you accept 1 * 10^-infinity) ^^^^__ 1 - 0.9 = is a theoretical number. I could use some feedback here. Am i correct here or no? Therefore: 10 * that theoretical number can not be the same as the "theoretical number" unless that number is 0. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__ So, 10x - x /= (does not equal) 9x if and only if x = 0.9. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__ 10x = 10 - 10 * "theoretical number (1 -0.9)" and ^__^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__ 9.9 = 10 - 1 * "theoretical number (1 -0.9)" and since those 2 things are not equal that's where I think the problem lies. Link to comment Share on other sites More sharing options...

Guest Posted June 25, 2007 Report Share Posted June 25, 2007 9.9 = 10 - 1 * "theoretical number (1 -0.9)" and since those 2 things are not equal that's where I think the problem lies. Sorry dude, In mathematics, the recurring decimal 0.999â€¦ denotes a real number equal to 1. In other words, "0.999â€¦" represents the same number as the symbol "1". The equality has long been accepted by professional mathematicians and taught in textbooks. Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience. I do not understand your objection, and I think you're attempting to be contrarywise. However, here it is in proof form to overcome your objections. Proof: 0.9999... = Sum 9/10^n (n=1 -> Infinity) = lim sum 9/10^n (m -> Infinity) (n=1 -> m) = lim .9(1-10^-(m+1))/(1-1/10) (m -> Infinity) = lim .9(1-10^-(m+1))/(9/10) (m -> Infinity) = .9/(9/10) = 1 Perhapse you'd like to add two 0.99999 and get 2? 0.999... + 0.999... = Sum[n = 1 to infinity](9*10^-n) + Sum[n = 1 to infinity](9*10^-n) = Sum[n = 1 to infinity](18*10^-n) = Sum[n = 1 to infinity](10*10^-n + 8*10^-n) = Sum[n = 1 to infinity](10^(-n+1)) + Sum[n = 1 to infinity](8*10^-n) = Sum[n = 0 to infinity](10^-n) + Sum[n = 1 to infinity](8*10^-n) = 1 + Sum[n = 1 to infinity](10^-n) + Sum[n = 1 to infinity](8*10^-n) = 1 + Sum[n = 1 to infinity]((8+1)*10^-n) = 1 + Sum[n = 1 to infinity](9*10^-n) = 1.999... But you can't, you get 1.9999... interesting eh? Perhapse if that does not satisfy, you could see it this way. 1.000000000... -0.999999999... =0.000000000... You may ask about the "1" in the last decimal place, and I'll write that when I'm done writing an infanate string of "0"s. So the 4th decimal place has a "0", and the 4 billionth has a "0", and the decimal place corrisponding to any number you could or could not name or imagine has a "0." It can be seen that a number with a "0" in every decimal place must be equal to 0. Or there is a quite technical and compleate work up of many difrent proofs and systems to arrive at such conclusions at Wikipedia http://en.wikipedia.org/wiki/0.999... Link to comment Share on other sites More sharing options...

Guest Posted June 28, 2007 Report Share Posted June 28, 2007 I am not arguing about whether or not the limit of 0.99999999999 is 1. That is obvious due to the definition of the word "limit". And, if you are going to use limits in the proof of .999999 = 1 then its the same as just saying that the limit of .99999 = 1. Because, inherently we accept that (for ie) any number to the power of negative infinity equals to 0. That's probably the first thing a teacher will tell you when he/she introduces limits and such. But we both know that 0 and x^-inf are two different things conceptually. Just as duality exists in the way scientists accept light, wave or particle, there is duality in this, no more no less. I stand firm on the belief that .9999999 = 1 practically but not conceptually. And the trick with the equation is that you can't always treat irrational numbers like natural numbers. I looked up the wikepedia link and here is what I copied from it: "A 19th-century reaction against such liberal summation methods resulted in the definition that still dominates today: the sum of a series is defined to be the limit of the sequence of its partial sums." So, it is simply agreed that to refer to a function like that one can express it in terms of its limit. And the limit of 0.9999.... is 1 but they are not the same. Although, after reading the wikepedia page on 0.99999 I retract everything that I have said. It's just mind numbing how deep that rabbit hole goes. People far more able than me have messed with it and other similar, complicated dudu. There is a lot to learn out there. I am not disenchanted though, just humbled. Link to comment Share on other sites More sharing options...

Guest Posted June 29, 2007 Report Share Posted June 29, 2007 Update! I've been inspired. I realized something important yesterday. Here is a laymanâ€™s proof on why 1 and 0.9999999... are distinct: What is an infinitesimal? ( 1-0.99999...= ) It is the difference between every "consecutive" number whether it is rational, real, irrational or any as long as it has its place on the number line. If numbers are the limits of the subset they represent beneath them than its just from the number system that we use that 1 is neatly capped at 1 because otherwise 1 is as unlimited as 0.9999999.... If after-all every number describes and infinitesimal point on the number line then their separations are also infinitesimal and therefore 1 and 0.9999999... are as distinct as any other number on the number line. Link to comment Share on other sites More sharing options...

Guest Posted July 4, 2007 Report Share Posted July 4, 2007 Update! I've been inspired. I realized something important yesterday. Here is a laymanâ€™s proof on why 1 and 0.9999999... are distinct: What is an infinitesimal? ( 1-0.99999...= ) It is the difference between every "consecutive" number whether it is rational, real, irrational or any as long as it has its place on the number line. If numbers are the limits of the subset they represent beneath them than its just from the number system that we use that 1 is neatly capped at 1 because otherwise 1 is as unlimited as 0.9999999.... If after-all every number describes and infinitesimal point on the number line then their separations are also infinitesimal and therefore 1 and 0.9999999... are as distinct as any other number on the number line. For layman's purposes you are right of course; numbers are arbitrary, and by definition nominal. It's only when dealing with them in a 'laboratory' setting as it where, that these kinds of distinctions are put under greater scrutiny. Link to comment Share on other sites More sharing options...

Guest Posted July 8, 2007 Report Share Posted July 8, 2007 I feel like I need to add something to an earlier reply. I keep thinking about this and as most things that aren't studied to their core by me (which is everything) I come to an inconclusive conclusion. hehe Here is a problem: Say 1 and 0.9999999... are separated by an infinitesimal and can somehow be viewed as "consecutive", where or what is the next consecutive down? It doesn't exist on our number line. We can't just right 0.99999... ...8 cause its an infinite series, and 0.9999999... - infinitesimal makes no sense either. It seems the "number line" isn't really a number line. It's like a line with some points marked on the surface and in between these marked points there are bottomless canyons that only God knows what goes on there. I've enjoyed our discussion on this Incog. Link to comment Share on other sites More sharing options...

Guest Posted August 12, 2007 Report Share Posted August 12, 2007 Here are the given equations: x = 2 x(x-1) = 2(x-1) x2-x = 2x-2 x2-2x = x-2 x(x-2) = x-2 x = 1 Take the equation on the second line and solve it: x(x-1) - 2(x-1) = 0 x**2 -3x + 2 = 0 (x-1)(x-2)=0 (Note: the term x**2 is supposed to be "x-squared") Therefore, x=1 AND x=2 are valid solutions for the original equation. This is Algebra I. Nothing wrong at all. Equations defined with a 2nd power term often have two different numerical solutions. Link to comment Share on other sites More sharing options...

Guest Posted August 13, 2007 Report Share Posted August 13, 2007 oldcab, There is something wrong. This proof shows that, from x=2, you can reach x=1. After that you can say x = 2 x = 1 1 = 2 1 = 1+1 1 = 2+2 1 = 4 etc. etc. and you can prove that every number is equal to every other number, which is of course false. As everyone pointed out, the mistake is dividing by x-2 without checking. When writing a real proof, you might write: x(x-2) = x-2 Divide both sides by x-2, which is OK if x-2 != 0 x = 1 And then you would realize right away that the step is invalid, because x=2 from before, and x-2 is 0. Depending on your proof, there may not have been a definition x=2 beforehand, then a step like this is OK. But you still should write the condition, or at least verify any solutions found to make sure a divide by zero did not occur. Even if the solution is not affected, you may be docked points for not doing the check. Therefore, x=1 AND x=2 are valid solutions for the original equation. More explicitly, this means that x=1 OR x=2, given the premise x(x-1) - 2(x-1) = 0. There is absolutely nothing wrong with what you wrote -- when you start from the second line. What if you started from the first line, x=2? Then x=1 is not a valid solution anymore. In fact, these are only possible solutions, and there is nothing to say that both of them must be valid solutions at all, or that any situations exist where either is valid (say, you add the premise x>10). But the original proof starts from the first line, and all 6 equations must be true at the same time, because no conditions were stated. So the conclusion is x=1 AND x=2 AND (all the other equations are true), given the premise x=1. And this certainly can not be true. This riddle shows why it's important to write proofs the proper way, doing all the necessary checks, annoying as it may seem. And it might also explain why math teachers stress that you should always use 'OR' and write concluding sentences, and freak out if you just leave x=1, x=2, x=7, etc. scattered all over the place Link to comment Share on other sites More sharing options...

Guest Posted December 9, 2007 Report Share Posted December 9, 2007 Its not wrong but 1 and 2 work for the last formula. The problem is that the problem was not worked out until the problem had the variable by itself. Link to comment Share on other sites More sharing options...

Guest Posted January 13, 2008 Report Share Posted January 13, 2008 It seems to me that there are two possible problems, depending on how you look at the question. The first line of the proof states that x = 2. Not x = 1, not x = 2 or 1, only x = 2. In the next line of the proof, we multiply both sides of the equation by (x - 1). Well, this is where we go wrong. We've added a solution. That isn't legitimate. When you solve equations, you can do any operation you want, as long as you do the same thing to both sides of the equation. We've done that by multiplying both sides by (x - 1). But you must also go back to the original equation and make sure that all of the solutions you find satisfy the initial conditions. Well, our initial conditions say that x = 2. So when we get that x can be 1 or 2, we go back: do both of these solutions fit the original equation, x = 2? Well, of course x = 1 doesn't work. If this, to some, is evidence that 1 = 2, because x = 1 and x = 2 so of COURSE 1 and 2 must be equal, look at it this way: After we added the factor (x - 1), we have a quadratic. If we write this quadratic in factored form, as (x - a)(x - b) = 0, we get the solution x = a OR x = b. The given proof simply confuses a conjunction and a disjunction, and states that x = a AND x = b rather than or. Link to comment Share on other sites More sharing options...

Guest Posted May 14, 2008 Report Share Posted May 14, 2008 Here is one possible "quick" proof that does not involve the "Sigma" notation (i.e., the notation Sum [n=1 to infinity]). Everyone agrees that 1/3 = 0.333333333....... Multiply both sides by 3 to obtain: 1 = 3 * ( 1/3 ) = 3 * ( 0.33333333........) = 0.99999999................... Thus 1= 0.999999........ And by the way: 0.999...... + 0.9999..... = 1.99999999..... = 2 The reason being the same Link to comment Share on other sites More sharing options...

Guest Posted May 15, 2008 Report Share Posted May 15, 2008 the very original problem in this post is actually perfectly right once you realize that x= is just asking for the zeros of the quadratic equation y=x^2-3x+2 the second one is more intersting but here's my take on it: we assume correctly that .999999999...is a nonterminating decimal with infinite 9s. the problem comes from multiplying that decimal by 10. thinking it through, .9 x10 = 9.0 .99x10 =9.90 etc... when u multiply a decimal by 10 think of it as moving the first zero to the left of the decimal to behind the last non-zero numeral to the right of the decimal. in other words 10x.9999... will actually = 9.9999.....99990. this zero is disputable but if u think of it as being there, then 10x-x will actually equal 8.99999....99999 because 10 x has a zero in the very last place where x has a 9 resulting in infinite carrying over of 10s (reference to how your teacher taught u to subtract 7 from 16 in the first grade). at this point x doesnâ€™t equal 1 because if u divide this number by 9 you simply get .9999999999 and thus this paradox is also solved. As for the next idea that 10/3= 3.33333â€¦. and multiplying that by 3= 9.99999 by arithmetic and 10 by algebra is actually just an exploitation of the fact that math is usually done in base 10. basically the less numerals in a system the less numbers are expressable through decimal. Think about the same situation in binary code. 1010/11 the closest u can get to this is 11.111111â€¦ which when converted back to base 10 = 3+ the summation of 1/(2^n) when n from 1 to infinity which is less accurate than 3.33333 (it will have a mixture of numbers not just 3. but in base 12 10 would be a 1 digit number which when divided by 3 would simply yield 3.4 (because instead of .1 being 1/10 it will be 1/12 and 3.3333 would still be a non-terminating decimal but written differently. but this is just my take on it. Link to comment Share on other sites More sharing options...

Guest Posted May 28, 2008 Report Share Posted May 28, 2008 Don't you realize that 2=1 is indeed true?? It's known that all (real) numbers fulfill a/a=1, and 0/a=0. So, for a=0, we have 0/0=1 and 0/0=0. Since 0/0 must be equal to itself, then 1 must be equal to 0. From here we can take that 1+0 = 1. But since 1=0, we can follow that 1+1=1 (we just replaced the 0). But we also know that 1+1=2. If 1+1=2 and 1+1=1, then it's obvious that 2=1, because 1+1=1+1 Of course, that's not true at all. But if someone says that there is an error in the "proof" above, they will be wrong... There are actually three gaps (a/a=1 and 0/a=0 are not true for all reals, but for all except zero; and 0/0 is an indetermination so it doesn't need to be equal to itself). I just hope you had some laugh with it See you some other day, with my proof of PI=1 (that is among my favorites) ------ There are 10 kinds of people in the world: those who know binary, those who don't, and those who also know ternary. Link to comment Share on other sites More sharing options...

Guest Posted June 4, 2008 Report Share Posted June 4, 2008 For the second one, my friend explained it to me simply like this: In order to find the end of 0.999999..., you have to put an infinite number of nines and then a zero at the end. But where is the end to infinity? There is none. Therefore, it is impossible to end 0.99999... and so it must be the same as 1. Link to comment Share on other sites More sharing options...

Guest Posted June 26, 2008 Report Share Posted June 26, 2008 Assume x = y which means 1 = 1 x = y then multply both sides by x so... x^{2} = xy then subtract y^{2} so... x^{2} - y^{2} = xy - y^{2} then expand these quandratric equations to... ( x + y ) ( x - y ) = x ( x - y ) which is just another representation of the previous line. Now, we currently have ( x - y ) on both sides, so lets divide both sides by ( x - y ) to get what is left... (x+y)(x-y) = x(x-y) . (x-y) . . . . (x-y) That leaves us with the following x + y = x if we go back to the beginning and say that x equals y so that x = 1 and y = 1 then 1 + 1 = 1 OR 2 = 1 Logical except for the fact that you can not divide by x-y because if x=y then x-y=0 and you can not divide a number by nothing - ie, nothing goes into one an infinite number of times plus one. Link to comment Share on other sites More sharing options...

Guest Posted July 29, 2008 Report Share Posted July 29, 2008 where it says 'x=2' if you take that and plug it in to 'x=1'you get'2=1' which is inequal so there is your mistakeSorry dude, I do not understand your objection, and I think you're attempting to be contrarywise. However, here it is in proof form to overcome your objections. Proof: 0.9999... = Sum 9/10^n (n=1 -> Infinity) = lim sum 9/10^n (m -> Infinity) (n=1 -> m) = lim .9(1-10^-(m+1))/(1-1/10) (m -> Infinity) = lim .9(1-10^-(m+1))/(9/10) (m -> Infinity) = .9/(9/10) = 1 Perhapse you'd like to add two 0.99999 and get 2? 0.999... + 0.999... = Sum[n = 1 to infinity](9*10^-n) + Sum[n = 1 to infinity](9*10^-n) = Sum[n = 1 to infinity](18*10^-n) = Sum[n = 1 to infinity](10*10^-n + 8*10^-n) = Sum[n = 1 to infinity](10^(-n+1)) + Sum[n = 1 to infinity](8*10^-n) = Sum[n = 0 to infinity](10^-n) + Sum[n = 1 to infinity](8*10^-n) = 1 + Sum[n = 1 to infinity](10^-n) + Sum[n = 1 to infinity](8*10^-n) = 1 + Sum[n = 1 to infinity]((8+1)*10^-n) = 1 + Sum[n = 1 to infinity](9*10^-n) = 1.999... But you can't, you get 1.9999... interesting eh? Perhapse if that does not satisfy, you could see it this way. 1.000000000... -0.999999999... =0.000000000... You may ask about the "1" in the last decimal place, and I'll write that when I'm done writing an infanate string of "0"s. So the 4th decimal place has a "0", and the 4 billionth has a "0", and the decimal place corrisponding to any number you could or could not name or imagine has a "0." It can be seen that a number with a "0" in every decimal place must be equal to 0. And you don't have to take my word for it, Blizzard Entertainment is weighing in on the discussion with their own proof... http://www.blizzard.com/press/040401.shtml Or there is a quite technical and compleate work up of many difrent proofs and systems to arrive at such conclusions at Wikipedia http://en.wikipedia.org/wiki/0.999... Link to comment Share on other sites More sharing options...

Guest Posted July 29, 2008 Report Share Posted July 29, 2008 where it says 'x=2' if you take that and plug it in to 'x=1'you get'2=1' which is inequal so there is your mistake Link to comment Share on other sites More sharing options...

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