wild_goat_14 Posted May 15, 2009 Share Posted May 15, 2009 This is just something I have began to notice happening notice on the forums. That's right, people trying to prove things mathematically with proofs that have simple algebraic errors in them. Why am I posting this? To hopefully avoid people making themselves look like idiots by posting things like this. EDIT: Oh, I nearly forgot, one of the most annoying things is when they assume in their "equation" that something that is not true is, such as in this example, 1/0 = infinity.(I am using this example because I think it was used in jest, sorry if it wasn't. [hide=2 equals 1]1/0 = infinity thus infinity x 0 = 1 We also know 2 x 0 = 0 So infinity x 0 x 2 = 1 (infinity x 0) x 2 = 1 (1) x 2 = 1 2 = 1[/hide] I am using this equation because it uses all three of the common "proof" errors that I see. 1) You can't divide by zero. Period. 1/0 = infinity. 5478/0 = infinity. Therefore 1 = 5478. NO. 2) Infinity is not a tangible number. Don't use it in a proof. 1 x infinity = infinity. 27463 x infinity = infinity. Therefore 1 = 27463. NOO! 3) Parentheses with no reason to them. 2 + 7 x 4 = 30. Order of operations. (2 + 7) x 4 = 36. Brackets need a REASON to be there. Now I will end this rant with a funny little equation I've heard. [hide=FAIL]A line contains an infinite amount of points. A plane contains an infinate amount of points. There are more points on a plane than a line(you can have 3 noncolinnear points). So with that logic... Points on a line < Points in a Plane infinity < infinity infinity/infinity < infinity/infinity 1 < 1[/hide] I shall take my flock underneath my own wing, and kick them right the [bleep] out of the tree. If they were meant to fly, they won't break their necks on the concrete.So, what is 1.111... equal to?10/9. Please don't continue. Link to comment Share on other sites More sharing options...

superson Posted May 15, 2009 Share Posted May 15, 2009 I think you may be wrong, I am reasonably sure that every person in OT has a PhD in mathematics. In Soviet Russia, glass eats OTers. Alansson Alansson, woo woo woo! Pink owns yes, just like you!GOOOOOOOOOO ALAN! WOO! Link to comment Share on other sites More sharing options...

Riku3220 Posted May 15, 2009 Share Posted May 15, 2009 I think you may be wrong, I am reasonably sure that every person in OT has a PhD in mathematics. We're also heart surgeons, astronomers, astrologers, astronauts, and monarchs of our own planet. Contributing: 1/3 = .3333..... 1/3*3 = 3/3 3/3 = 1 .3333.....*3 = 1 x = .99... 10x = 9.99.... 10x - x = 9 x = 1 Link to comment Share on other sites More sharing options...

Zierro Posted May 15, 2009 Share Posted May 15, 2009 Do you believe 0.999... = 1? If so, then you're doing the same exact thing - using "proofs" to get an "answer". I can't really speak for that poster, but if you ask me I think his point was to show you that infinity can't be applied to regular equations because there are always going to be inevitable flaws. Everyone knows that 1 =/= 2. Link to comment Share on other sites More sharing options...

Wizz Posted May 15, 2009 Share Posted May 15, 2009 2=1 So 2+2=2? Everything I learned is wrong! Wongton is better than me in anyway~~ Link to comment Share on other sites More sharing options...

ElkNight Posted May 15, 2009 Share Posted May 15, 2009 I divided by zero. 8,180WONGTONG IS THE BEST AND IS MORE SUPERIOR THAN ME#1 Wongtong stalker.Im looking for some No Limit soldiers! Link to comment Share on other sites More sharing options...

walka92 Posted May 15, 2009 Share Posted May 15, 2009 I divided by zero. error *explodes* ... as for the brackets thing, i see no problem with them. many people, me included, use the more than required, to help ourself keep track of what we are doing I'm gonna be walking down an alley in varrock, and walka is going to walk up to me in a trench coat and say "psst.. hey man, wanna buy some sara brew"walka92- retired with 99 in attack, strength, defence, health, magic, ranged, prayer and herblore and 137 combat. some day i may return to claim 138 combat, but alas, that time has not yet come Link to comment Share on other sites More sharing options...

fastortoise Posted May 15, 2009 Share Posted May 15, 2009 hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 4. (a-B)(a+B) = b(a-B) 5. a+b = b 6. b+b = b 7. 2b = b 8. 2 = 1 =D> Link to comment Share on other sites More sharing options...

gonpost Posted May 15, 2009 Share Posted May 15, 2009 I'm a student electrical engineer... As a general rule, keep all of your numbers in fractional form until you want your answer. However, even then you should only change it over to decimal if you want to apply it to something physical. If I say that this screw needs to be 17/21 inches in diameter exactly, you change it to decimal and carry it out as far as your machining equipment can accurately make the screw. But when you're doing math, keep things in fractional form. And yes, .999... repeating really does equal 1. You're essentially asking "what is the limit of x as x -> 1 from the left side?" Believe me, it's 1. Furthermore, .999... is a NUMBER. It is NOT A PROCESS. You are not moving out the 9s. They are already there, they are a number, and they are equal to one. I can prove this with a very basic infinite sum as well. Where a = 9, 0.999... = a0.(a1)(a2)(a3)... = a0 + (a1)(1/10)^1 + (a2)(1/10)^2 + ... = (ar)/(1-r) = 9(1/10)/(1-1/10) = .9/.9 = 1 I can also use a better limit example but it's really a pain to type out on a forum. In a different number system, namely binary, .999... = 0.11111... = 01 Here are some common errors in thought, courtesy of wikipedia, -Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.[33] -Some students interpret "0.999" (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".[34] -Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999" as meaning the sequence rather than its limit.[35] -Some students regard 0.999 as having a fixed value which is less than 1 by an infinitesimal but non-zero amount. -Some students believe that the value of a convergent series is at best an approximation, that .999... is approximately equal to 1. But alas, you poor students are WRONG. :D hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 > You messed up going from step 1 to step 2. If you go backwards from 2 to 1, you cancelled out an a. That's a big NO NO. You essentially got rid of an entire solution. a^2 = aa, so just rewriting step 2 in another manner, aa-ab = 0 a(a-B)=0 therefore, a = 0 OR a = b In the case of your proof, a = 0 would indeed work, and that is the only solution (which is trivial)! Runescape Name: "unbug07"Expand your mind. Link to comment Share on other sites More sharing options...

fastortoise Posted May 15, 2009 Share Posted May 15, 2009 hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 > You messed up going from step 1 to step 2. If you go backwards from 2 to 1, you cancelled out an a. That's a big NO NO. You essentially got rid of an entire solution. a^2 = aa, so just rewriting step 2 in another manner, aa-ab = 0 a(a-B)=0 therefore, a = 0 OR a = b In the case of your proof, a = 0 would indeed work, and that is the only solution (which is trivial)! ? That step is perfectly legitimate. Notice how I said "assume a=b" before my 'proof'. so lets say a=b=4 4=4 4^2 = 4x4 16 = 16 "going backwards" just involves dividing by 4 on each side. You didn't find the correct incorrect step mr. electrical engineer :-$ Link to comment Share on other sites More sharing options...

andylax38 Posted May 15, 2009 Share Posted May 15, 2009 hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 > You messed up going from step 1 to step 2. If you go backwards from 2 to 1, you cancelled out an a. That's a big NO NO. You essentially got rid of an entire solution. a^2 = aa, so just rewriting step 2 in another manner, aa-ab = 0 a(a-B)=0 therefore, a = 0 OR a = b In the case of your proof, a = 0 would indeed work, and that is the only solution (which is trivial)! ? That step is perfectly legitimate. Notice how I said "assume a=b" before my 'proof'. so lets say a=b=4 4=4 4^2 = 4x4 16 = 16 "going backwards" just involves dividing by 4 on each side. You didn't find the correct incorrect step mr. electrical engineer :-$ The problem is going from steps 4 to 5, in step 4 you have (a-B)(a+B) = b(a-B) you simplify this by dividing each side by (a-B) which according to the proof you wrote gives us 5 which is a+b = b. For 5 to be right (a-B)/(a-B) = 1 would have to be true. In step 1 you say a=b so (a-B) = 0. When you divide each side by (a-B) obtaining (a-B)(a+B)/(a-B) = b(a-B)/(a-B) your essentially dividing 0 by 0 which stated by the OP cannot be done. I'm a mechanical engineering student btw :P. ~>~Check Out My Bursting Guide~<~Current goals - 99 Slayer, 99 Summoning, 85 DungeoneeringAchieved goals - Quest Cape, 99 Strength, 85 Slayer, 99 Range, 88 Summoning, 90 Slayer, 99 Magic, 99 Farming Link to comment Share on other sites More sharing options...

fastortoise Posted May 15, 2009 Share Posted May 15, 2009 [hide=]hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 > You messed up going from step 1 to step 2. If you go backwards from 2 to 1, you cancelled out an a. That's a big NO NO. You essentially got rid of an entire solution. a^2 = aa, so just rewriting step 2 in another manner, aa-ab = 0 a(a-B)=0 therefore, a = 0 OR a = b In the case of your proof, a = 0 would indeed work, and that is the only solution (which is trivial)! ? That step is perfectly legitimate. Notice how I said "assume a=b" before my 'proof'. so lets say a=b=4 4=4 4^2 = 4x4 16 = 16 "going backwards" just involves dividing by 4 on each side. You didn't find the correct incorrect step mr. electrical engineer :-$ The problem is going from steps 4 to 5, in step 4 you have (a-B)(a+B) = b(a-B) you simplify this by dividing each side by (a-B) which according to the proof you wrote gives us 5 which is a+b = b. For 5 to be right (a-B)/(a-B) = 1 would have to be true. In step 1 you say a=b so (a-B) = 0. When you divide each side by (a-B) obtaining (a-B)(a+B)/(a-B) = b(a-B)/(a-B) your essentially dividing 0 by 0 which stated by the OP cannot be done. I'm a mechanical engineering student btw :P.[/hide] I'm definitely glad at least mechanical engineers know you can't divide by zero. My cal2 teacher once told us a story of a plane crashing because someone divided 0 in some sort of landing sequence calculation, but he was a nutjob and shouldn't be taken too seriously. Link to comment Share on other sites More sharing options...

jemathonical Posted May 15, 2009 Share Posted May 15, 2009 In the end, does it really matter? I hope you all don't get ulcers over things like this. ^Sir Jem 05-The Bunny Drinking Blog?^ Click it! Link to comment Share on other sites More sharing options...

warri0r45 Posted May 15, 2009 Share Posted May 15, 2009 I don't have a mind for maths, unfortunately. Whenever I think of a mathematical problem I do best to keep it simple and, if possible, think of each unit as a physical object. For example, I rationalise dividing by zero to be nonsensical because you can't divide a number of objects into zero groups. They'll always be in groups, even if it's just groups of one. Link to comment Share on other sites More sharing options...

mmmcannibalism Posted May 15, 2009 Share Posted May 15, 2009 [hide=]hai guyz watch me beat math! Let a and b be any non-zero number 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 > You messed up going from step 1 to step 2. If you go backwards from 2 to 1, you cancelled out an a. That's a big NO NO. You essentially got rid of an entire solution. a^2 = aa, so just rewriting step 2 in another manner, aa-ab = 0 a(a-B)=0 therefore, a = 0 OR a = b In the case of your proof, a = 0 would indeed work, and that is the only solution (which is trivial)! ? That step is perfectly legitimate. Notice how I said "assume a=b" before my 'proof'. so lets say a=b=4 4=4 4^2 = 4x4 16 = 16 "going backwards" just involves dividing by 4 on each side. You didn't find the correct incorrect step mr. electrical engineer :-$ The problem is going from steps 4 to 5, in step 4 you have (a-B)(a+B) = b(a-B) you simplify this by dividing each side by (a-B) which according to the proof you wrote gives us 5 which is a+b = b. For 5 to be right (a-B)/(a-B) = 1 would have to be true. In step 1 you say a=b so (a-B) = 0. When you divide each side by (a-B) obtaining (a-B)(a+B)/(a-B) = b(a-B)/(a-B) your essentially dividing 0 by 0 which stated by the OP cannot be done. I'm a mechanical engineering student btw :P.[/hide] I'm definitely glad at least mechanical engineers know you can't divide by zero. My cal2 teacher once told us a story of a plane crashing because someone divided 0 in some sort of landing sequence calculation, but he was a nutjob and shouldn't be taken too seriously. that might have been a joke, or it might have been true that someone put in data wrong leading to a severe error Im partial to the proof that there are more numbers between 0 and 1 then there are numbers in the sequence 1,2,3,4,5... Ill post it when I have more time. Orthodoxy is unconciousnessthe only ones who should kill are those who are prepared to be killed. Link to comment Share on other sites More sharing options...

Lenticular_J Posted May 15, 2009 Share Posted May 15, 2009 You guys are gonna make the world implode. WHY CAN'T MATH JUST BE 1, 2, 3, 4. catch it now so you can like it before it went so mainstream Link to comment Share on other sites More sharing options...

thrash-boy Posted May 15, 2009 Share Posted May 15, 2009 You guys are gonna make the world implode. WHY CAN'T MATH JUST BE 1, 2, 3, 4. if it was you wouldn't be sitting in front of a computer now. you wouldn't have a roof over your head. chances are you wouldn't even have clothes on your back. I love maths. Everything makes sense and is logical unlike languages, especially English. Why do most words have to have a u after the q if q is the first letter? I really hate that kind of thing. With maths i can see things that actually have value, and manipulate data to calculate all kinds of wonderful and useful things, without worrying about laws like the one i just mentioned in English that doesn't seem to have any logical reasoning. I think im about the only one i know that did the hard maths in school this year only because i thought it was fun. The kinds of things i want to do in life wont require maths anywhere near as complex as what im learning. I still see a point to learning it, even if the point is just to do something that gives me a fuzzing rewarding feeling when i do something right. Maths rocks. Link to comment Share on other sites More sharing options...

tintin113 Posted May 15, 2009 Share Posted May 15, 2009 In the end, does it really matter? I hope you all don't get ulcers over things like this. Haha that epitimises (sp?) these forums. =D> And thank you to -insert thread-maker's name here- for making this thread 'cos I'm getting pretty annoyed of people trying to proof things due to their misunderstanding of algebra. Algebra is not to be underestimated but many people fool themselves into thinking that they understand what they're talking about with it. They don't. Thanks to Quarra for the awesome sig!Xbox360 Gamertag = Tintin113 Link to comment Share on other sites More sharing options...

EdgedThesis Posted May 15, 2009 Share Posted May 15, 2009 I don't have a mind for maths, unfortunately. Whenever I think of a mathematical problem I do best to keep it simple and, if possible, think of each unit as a physical object. For example, I rationalise dividing by zero to be nonsensical because you can't divide a number of objects into zero groups. They'll always be in groups, even if it's just groups of one. Then you might have a mind for physics, where every number has set significance and you are not left in a void of variables and symbols with no meaning. But I don't want to go among mad people!Oh, you can't help that. We're all mad here..." Link to comment Share on other sites More sharing options...

20Rice04 Posted May 15, 2009 Share Posted May 15, 2009 If you can find the error here I'll be bloody impressed :P -1 = -1 (1/-1) = (-1/1) (1^(1/2)/-1^(1/2)) = (-1^(1/2)/1^(1/2)) multpily by 1^(1/2).-1^(1/2) and you get 1= -1 Q.E.D PLAWKS? Look guys... I absolutely must be a mass baby-seal murderer! Link to comment Share on other sites More sharing options...

fastortoise Posted May 15, 2009 Share Posted May 15, 2009 ^ can't take the root of a negative number because there is no number which, multiplied by itself, gives a negative. Link to comment Share on other sites More sharing options...

brunokiller Posted May 15, 2009 Share Posted May 15, 2009 Actually before you state that I use wrong calculations, read the topic as I clearly said 'Now to make people stop believing 1/0=infinity. I suggest you read this. ^ can't take the root of a negative number because there is no number which, multiplied by itself, gives a negative. i If you can find the error here I'll be bloody impressed :P -1 = -1 (1/-1) = (-1/1) (1^(1/2)/-1^(1/2)) = (-1^(1/2)/1^(1/2)) multiply by 1^(1/2).-1^(1/2) and you get 1= -1 Q.E.D PLAWKS? (1^(1/2)/-1^(1/2)) = (-1^(1/2)/1^(1/2)) 1/i =- i/1 1/i = -i As 1^0.5 x -1^0.5 = i,both sides times i. 1 = 1. 3) Parentheses with no reason to them. 2 + 7 x 4 = 30. Order of operations. (2 + 7) x 4 = 36. Brackets need a REASON to be there. Yeah, if you use + and x through each other, THEN you can't use brackets at random. I only used x 's so I can use as much brackets as I want. Last time I checked , 2 x 7 x 4 = 56 2 x (7 x 4)= 56 (2 x 7) x 4 = 56 1. a = b 2. a^2 = ab 3. a^2 - b^2 = ab - b^2 4. (a-B)(a+B) = b(a-B) 5. a+b = b 6. b+b = b 7. 2b = b 8. 2 = 1 If I apply 1. a = b and 4. (a-B)(a+B) = b(a-B) Then a-b = 0 Thus 0 x (a+B) = B x 0 0 = 0 What you've done is basically a divide by zero error. You gotta try a tiny bit harder when you try to disprove me. My blog Link to comment Share on other sites More sharing options...

Kiriyama Posted May 15, 2009 Share Posted May 15, 2009 ^ can't take the root of a negative number because there is no number which, multiplied by itself, gives a negative. He never took the square root though. Denizen of Darkness| PSN= sworddude198 Link to comment Share on other sites More sharing options...

brunokiller Posted May 16, 2009 Share Posted May 16, 2009 ^ can't take the root of a negative number because there is no number which, multiplied by itself, gives a negative. He never took the square root though. In conventional mathematics without complex numbers, you can't even write down -1^0.5 My blog Link to comment Share on other sites More sharing options...

Kiriyama Posted May 16, 2009 Share Posted May 16, 2009 ^ can't take the root of a negative number because there is no number which, multiplied by itself, gives a negative. He never took the square root though. In conventional mathematics without complex numbers, you can't even write down -1^0.5 Doesn't it just work out to be -1 though? Denizen of Darkness| PSN= sworddude198 Link to comment Share on other sites More sharing options...

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