anonymous 4 years ago Question Of The Day.

1. anonymous

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

2. anonymous

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

3. anonymous

algebraic*

4. anonymous

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

5. anonymous

O + 4 = 5. solve for ooh.

6. Callisto

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

7. anonymous

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

8. anonymous

U can answer in so many ways. Try all.

9. anonymous

whoa...I'm keeping my bf

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

Everything is possible in maths, lgba.

15. anonymous

dividing by 0 is a fantasy?

16. anonymous

Its fun time guys and girls. Lets enjoy

17. anonymous

stuff just got real.

18. anonymous

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

19. Callisto

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

20. anonymous

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? ;)

21. anonymous

22. anonymous

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)

23. anonymous

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

24. anonymous

like this @ffm

25. anonymous

26. anonymous

:D sorry for the troubles FFM

27. anonymous

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

28. anonymous

I am not notified with @ffm

29. anonymous

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

30. anonymous

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

31. Callisto

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

32. anonymous

So, Aadarsh where is the trouble. try guessing!

33. anonymous

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

34. anonymous

the fallacy works well

35. Mani_Jha

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

36. anonymous

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

37. Mani_Jha

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

38. anonymous

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

39. KingGeorge

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

40. anonymous

Hats off to @KingGeorge .

41. Mani_Jha

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

42. anonymous

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

43. anonymous

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

44. Mani_Jha

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

45. KingGeorge

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, $$e=1$$

46. anonymous

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

47. anonymous
48. KingGeorge

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.

49. anonymous

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

50. anonymous

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

51. anonymous

Abel prize. lol

52. anonymous

you 'll get special academic razzies for that!

53. anonymous

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

54. KingGeorge

@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.

55. Mani_Jha

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

56. anonymous

Oooooo! Really??

57. anonymous

58. anonymous

tumne apne info mei ye sab kya likh dala hai?

59. anonymous

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

60. anonymous

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

61. anonymous

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

62. anonymous

63. anonymous