Question Of The Day.

- anonymous

Question Of The Day.

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

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

- anonymous

algebraic*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

O + 4 = 5. solve for ooh.

- Callisto

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

- lgbasallote

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

- anonymous

U can answer in so many ways. Try all.

- anonymous

whoa...I'm keeping my bf

- anonymous

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

- 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.

- lgbasallote

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

- anonymous

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

- anonymous

Everything is possible in maths, lgba.

- anonymous

dividing by 0 is a fantasy?

- anonymous

Its fun time guys and girls. Lets enjoy

- anonymous

stuff just got real.

- anonymous

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

- Callisto

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

- apoorvk

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? ;)

- anonymous

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

- apoorvk

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)

- lgbasallote

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

- apoorvk

like this @ffm

- anonymous

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

- apoorvk

:D sorry for the troubles FFM

- lgbasallote

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

- anonymous

I am not notified with @ffm

- apoorvk

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

- anonymous

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

- 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

- apoorvk

So, Aadarsh where is the trouble. try guessing!

- anonymous

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

- lgbasallote

the fallacy works well

- Mani_Jha

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

- anonymous

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

- Mani_Jha

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

- anonymous

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

- KingGeorge

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

- anonymous

Hats off to @KingGeorge .

- Mani_Jha

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

- anonymous

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

- anonymous

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

- Mani_Jha

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

- 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, \(e=1\)

- anonymous

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

- anonymous

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

- 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.

- apoorvk

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

- anonymous

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

- apoorvk

Abel prize. lol

- apoorvk

you 'll get special academic razzies for that!

- anonymous

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

- 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.

- Mani_Jha

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

- anonymous

Oooooo! Really??

- anonymous

@shruti , @PRINCEOFPERSIA please try.

- anonymous

tumne apne info mei ye sab kya likh dala hai?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Smart answer, Prince Bhai.

- anonymous

Thanks @ Aadarsh..

- experimentX

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.