Question

Question Of The Day.

Answer 1

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

Answer 2

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

Answer 3

algebraic*

Answer 4

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

Answer 5

O + 4 = 5. solve for ooh.

Answer 6

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

Answer 7

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

Answer 8

U can answer in so many ways. Try all.

Answer 9

whoa...I'm keeping my bf

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Everything is possible in maths, lgba.

Answer 15

dividing by 0 is a fantasy?

Answer 16

Its fun time guys and girls. Lets enjoy

Answer 17

stuff just got real.

Answer 18

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

Answer 19

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

Answer 20

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? ;)

Answer 21

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

Answer 22

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)

Answer 23

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

Answer 24

like this @ffm

Answer 25

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

Answer 26

:D sorry for the troubles FFM

Answer 27

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

Answer 28

I am not notified with @ffm

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

So, Aadarsh where is the trouble. try guessing!

Answer 33

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

Answer 34

the fallacy works well

Answer 35

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

Answer 36

let x=0

then 20x = 100x
or x = 5x

or 1 = 5 (How are you cancelling x? This is division by zero)

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

Hats off to @KingGeorge .

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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, \(e=1\)

Answer 46

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

Answer 47

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

Answer 48

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.

Answer 49

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

Answer 50

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

Answer 51

Abel prize. lol

Answer 52

you 'll get special academic razzies for that!

Answer 53

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

Answer 54

@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.

Answer 55

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

Answer 56

Oooooo! Really??

Answer 57

@shruti , @PRINCEOFPERSIA please try.

Answer 58

tumne apne info mei ye sab kya likh dala hai?

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

Smart answer, Prince Bhai.

Answer 63

Thanks @ Aadarsh..

Answer 64

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