Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Aadarsh Group Title

Question Of The Day.

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Prove that 1 = 2 , 0 = 1 and 1 = 2. I need clever answers.

    • 2 years ago
  2. rebeccaskell94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    not this one again D: algbebraic expressions is my final answer

    • 2 years ago
  3. rebeccaskell94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    algebraic*

    • 2 years ago
  4. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    All are requested to try. This is a puzzle question.

    • 2 years ago
  5. dpaInc Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    O + 4 = 5. solve for ooh.

    • 2 years ago
  6. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    x = 2x Cross out x -> 1=2 :S

    • 2 years ago
  7. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Let a= b Then a^2 = ab a^2 + a^2 = a^2 + ab 2a^2 = a^2 + ab 2a^2 - 2ab = a^2 + ab - 2ab 2a^2 - 2ab = a^2 - ab 2(a^2 - ab) = a^2 - ab --------- ------- a^2 - ab a^2 - ab 2 = 1

    • 2 years ago
  8. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    U can answer in so many ways. Try all.

    • 2 years ago
  9. rebeccaskell94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    whoa...I'm keeping my bf

    • 2 years ago
  10. Mathisnotfun Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    x + x = 4x x(1+1) = 4x (1 + 1) = 4 2 = 4 1 = 2

    • 2 years ago
  11. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @lgbasallote answer is thrilling. I was searching for ur answer, LGBA!!!!! Now proceed and u can solve that 1=0 and also 0 = 2.

    • 2 years ago
  12. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    it's actually a fallacy lol =)) dividing 0 by 0 is illegal :p

    • 2 years ago
  13. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    @Aadarsh: This is not a puzzle, this is wrong mathematics. Finding the error would be some fun though.

    • 2 years ago
  14. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Everything is possible in maths, lgba.

    • 2 years ago
  15. dpaInc Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    dividing by 0 is a fantasy?

    • 2 years ago
  16. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Its fun time guys and girls. Lets enjoy

    • 2 years ago
  17. rebeccaskell94 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    stuff just got real.

    • 2 years ago
  18. gurvinder Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    aad plzz aye koi question hai ..........

    • 2 years ago
  19. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @ffm true, we cannot cancel the variables like that :P

    • 2 years ago
  20. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    let x=0 then 20x = 100x or x = 5x or 1 = 5 :P :p Okayy... proving is one thing. I ll ask you guys this: wherein here is the trouble. Which step did we make the mistake at, so we get such erratic answers that would cause Newton to suffer a heart attack if he ever saw this? ;)

    • 2 years ago
  21. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @apoorvk smart reply. Newton would have given us gifts!!!!!

    • 2 years ago
  22. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh yeah @foolformath already asked this. so where is the error, aadarsh can you think? (this question was presented in Bansal classes entrance test multiple times)

    • 2 years ago
  23. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Callisto how do you tag @FoolForMath with ffm???

    • 2 years ago
  24. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    like this @ffm

    • 2 years ago
  25. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Really? Bansal people are copying from Brain Mapping Academy, Hyderabad.

    • 2 years ago
  26. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    :D sorry for the troubles FFM

    • 2 years ago
  27. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont think @ffm is the same as @FoolForMath yes?

    • 2 years ago
  28. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    I am not notified with @ffm

    • 2 years ago
  29. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    may be, aadarsh. or the other way around. someone always copies from somebody

    • 2 years ago
  30. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @Mani_Jha , @AravindG , @payalvsangle , @heena , @sheena101 , @gurvinder , @Ishaan94 , please try. I have to present this before my seniors and get cash prize.

    • 2 years ago
  31. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmmm... I was trying to show that I was responding his words, so .... it doesn't matter if I have really tagged him :P

    • 2 years ago
  32. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So, Aadarsh where is the trouble. try guessing!

    • 2 years ago
  33. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @apoorvk bhai, there is no mistake. Mistake der in assumption.

    • 2 years ago
  34. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    the fallacy works well

    • 2 years ago
  35. Mani_Jha Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @As a general rule, you shouldn't multiply or divide an equation by zero.

    • 2 years ago
  36. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    let x=0 then 20x = 100x or x = 5x or 1 = 5 (How are you cancelling x? This is division by zero)

    • 2 years ago
  37. Mani_Jha Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    (1−2)^2=(2−1)^2=1 1−2=2−1 2=4 1=2 Find out the mistake in this.

    • 2 years ago
  38. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Mistake is in this: 1-2 = 2-1

    • 2 years ago
  39. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    How about a proof that shows \(e=1\)? \[\Large e=e^{2\pi i \over 2\pi i}=(e^{2\pi i})^{1 \over 2 \pi i}=1^{1 \over 2 \pi i}=1\]

    • 2 years ago
  40. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hats off to @KingGeorge .

    • 2 years ago
  41. Mani_Jha Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha. Good. I used a fallacy to prove another fallacy!

    • 2 years ago
  42. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @King , @shruti , @neha2050 , @salini Please try.

    • 2 years ago
  43. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    \(a^2 = b^2 \implies \pm a= \pm b \) So there are two solutions.

    • 2 years ago
  44. Mani_Jha Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @KingGeorge, finding the value of anything to a complex power doesn't make sense. You can't be sure that: \[1^{1/2\pi i}=1\]

    • 2 years ago
  45. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    An alternative proof I found online: Let \(2e=f\). Then \(2^{2 \pi i}e^{2 \pi i} =f^{2\pi i}\). Since \(e^{2\pi i}=1\), we know that \(2^{2\pi i}=f^{2\pi i}\). This means that \(2=f\). From that we substitute \(f\) back in, and get \[2e=2\]Thus, \(e=1\)

    • 2 years ago
  46. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    @KingGeorge: When a number is raised to a complex power, the result is not precisely defined.

    • 2 years ago
  47. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    See here: http://en.wikipedia.org/wiki/Mathematical_fallacy.

    • 2 years ago
  48. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It's the basically the same deal for the second one. \(f^{2\pi i}\) can get really funky. \(2^{2\pi i}\) is already funky.

    • 2 years ago
  49. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    exactly. there is a mistake as @ffm just showed. how do you think are you dividing by '0'??

    • 2 years ago
  50. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Can we not try to prove that division by zero is indeed possible? At least by this, we can get the Abel Prize.

    • 2 years ago
  51. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Abel prize. lol

    • 2 years ago
  52. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you 'll get special academic razzies for that!

    • 2 years ago
  53. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Lets hope so!!!!!!!!!!!!!!!!!!!!!!!

    • 2 years ago
  54. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @Mani_Jha The actual problem with my original proof is that \[\Large e^{{2 \pi i} \over {2 \pi i}} \neq (e^{2 \pi i})^{1 \over {2 \pi i}}\]Complex exponents don't do that.

    • 2 years ago
  55. Mani_Jha Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Aadarsh, You can't do that. That's why zero is called the 'hero' of Mathematics!

    • 2 years ago
  56. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oooooo! Really??

    • 2 years ago
  57. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @shruti , @PRINCEOFPERSIA please try.

    • 2 years ago
  58. shruti Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    tumne apne info mei ye sab kya likh dala hai?

    • 2 years ago
  59. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Mera naam hai didi!!!!!! Yeh to maine dhamaal se copy kiya hai.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    • 2 years ago
  60. shruti Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    cool...yhi kaam kro aur ye kya qestn post kar rakha hai. i felt sumth imp. is there but is'nt

    • 2 years ago
  61. PRINCEOFPERSIA Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Multiply zero on both sides. 1 x 0 = 2 x 0 0 = 0 1 = 2

    • 2 years ago
  62. Aadarsh Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Smart answer, Prince Bhai.

    • 2 years ago
  63. PRINCEOFPERSIA Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks @ Aadarsh..

    • 2 years ago
  64. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    1 = 1 1^0 = 2^0 = 3^0 = a^0 1 = 2 = 3 = a

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.