## Aadarsh Group Title Question Of The Day. 2 years ago 2 years ago

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

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

algebraic*

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

5. dpaInc Group Title

O + 4 = 5. solve for ooh.

6. Callisto Group Title

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

7. lgbasallote Group Title

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

U can answer in so many ways. Try all.

whoa...I'm keeping my bf

10. Mathisnotfun Group Title

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

@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. lgbasallote Group Title

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

13. FoolForMath Group Title

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

Everything is possible in maths, lgba.

15. dpaInc Group Title

dividing by 0 is a fantasy?

Its fun time guys and girls. Lets enjoy

stuff just got real.

18. gurvinder Group Title

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

19. Callisto Group Title

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

20. apoorvk Group Title

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? ;)

22. apoorvk Group Title

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. lgbasallote Group Title

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

24. apoorvk Group Title

like this @ffm

26. apoorvk Group Title

:D sorry for the troubles FFM

27. lgbasallote Group Title

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

28. FoolForMath Group Title

I am not notified with @ffm

29. apoorvk Group Title

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

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

31. Callisto Group Title

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. apoorvk Group Title

So, Aadarsh where is the trouble. try guessing!

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

34. lgbasallote Group Title

the fallacy works well

35. Mani_Jha Group Title

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

36. FoolForMath Group Title

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 Group Title

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

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

39. KingGeorge Group Title

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

Hats off to @KingGeorge .

41. Mani_Jha Group Title

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

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

43. FoolForMath Group Title

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

44. Mani_Jha Group Title

@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 Group Title

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. FoolForMath Group Title

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

47. FoolForMath Group Title
48. KingGeorge Group Title

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. apoorvk Group Title

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

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

51. apoorvk Group Title

Abel prize. lol

52. apoorvk Group Title

you 'll get special academic razzies for that!

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

54. KingGeorge Group Title

@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 Group Title

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

Oooooo! Really??

58. shruti Group Title

tumne apne info mei ye sab kya likh dala hai?

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

60. shruti Group Title

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

61. PRINCEOFPERSIA Group Title

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