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Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Prove that 1 = 2 , 0 = 1 and 1 = 2. I need clever answers.
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
not this one again D: algbebraic expressions is my final answer
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
algebraic*
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
All are requested to try. This is a puzzle question.
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
O + 4 = 5. solve for ooh.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
x = 2x Cross out x > 1=2 :S
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.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

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
U can answer in so many ways. Try all.
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
whoa...I'm keeping my bf
 2 years ago

Mathisnotfun Group TitleBest ResponseYou've already chosen the best response.0
x + x = 4x x(1+1) = 4x (1 + 1) = 4 2 = 4 1 = 2
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.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

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
it's actually a fallacy lol =)) dividing 0 by 0 is illegal :p
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.6
@Aadarsh: This is not a puzzle, this is wrong mathematics. Finding the error would be some fun though.
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Everything is possible in maths, lgba.
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
dividing by 0 is a fantasy?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Its fun time guys and girls. Lets enjoy
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
stuff just got real.
 2 years ago

gurvinder Group TitleBest ResponseYou've already chosen the best response.0
aad plzz aye koi question hai ..........
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
@ffm true, we cannot cancel the variables like that :P
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.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

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
@apoorvk smart reply. Newton would have given us gifts!!!!!
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.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

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@Callisto how do you tag @FoolForMath with ffm???
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
like this @ffm
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Really? Bansal people are copying from Brain Mapping Academy, Hyderabad.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
:D sorry for the troubles FFM
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i dont think @ffm is the same as @FoolForMath yes?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.6
I am not notified with @ffm
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
may be, aadarsh. or the other way around. someone always copies from somebody
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.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

Callisto Group TitleBest ResponseYou've already chosen the best response.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

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
So, Aadarsh where is the trouble. try guessing!
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
@apoorvk bhai, there is no mistake. Mistake der in assumption.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
the fallacy works well
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.0
@As a general rule, you shouldn't multiply or divide an equation by zero.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.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

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.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

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Mistake is in this: 12 = 21
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Hats off to @KingGeorge .
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.0
Haha. Good. I used a fallacy to prove another fallacy!
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
@King , @shruti , @neha2050 , @salini Please try.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.6
\(a^2 = b^2 \implies \pm a= \pm b \) So there are two solutions.
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.6
@KingGeorge: When a number is raised to a complex power, the result is not precisely defined.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.6
See here: http://en.wikipedia.org/wiki/Mathematical_fallacy.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
exactly. there is a mistake as @ffm just showed. how do you think are you dividing by '0'??
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.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

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
Abel prize. lol
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
you 'll get special academic razzies for that!
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Lets hope so!!!!!!!!!!!!!!!!!!!!!!!
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.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

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.0
@Aadarsh, You can't do that. That's why zero is called the 'hero' of Mathematics!
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Oooooo! Really??
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
@shruti , @PRINCEOFPERSIA please try.
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
tumne apne info mei ye sab kya likh dala hai?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Mera naam hai didi!!!!!! Yeh to maine dhamaal se copy kiya hai.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.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

PRINCEOFPERSIA Group TitleBest ResponseYou've already chosen the best response.0
Multiply zero on both sides. 1 x 0 = 2 x 0 0 = 0 1 = 2
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Smart answer, Prince Bhai.
 2 years ago

PRINCEOFPERSIA Group TitleBest ResponseYou've already chosen the best response.0
Thanks @ Aadarsh..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
1 = 1 1^0 = 2^0 = 3^0 = a^0 1 = 2 = 3 = a
 2 years ago
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