anonymous
  • anonymous
Question Of The Day.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Prove that 1 = 2 , 0 = 1 and 1 = 2. I need clever answers.
anonymous
  • anonymous
not this one again D: algbebraic expressions is my final answer
anonymous
  • anonymous
algebraic*

Looking for something else?

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

More answers

anonymous
  • anonymous
All are requested to try. This is a puzzle question.
anonymous
  • anonymous
O + 4 = 5. solve for ooh.
Callisto
  • Callisto
x = 2x Cross out x -> 1=2 :S
lgbasallote
  • 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
  • anonymous
U can answer in so many ways. Try all.
anonymous
  • anonymous
whoa...I'm keeping my bf
anonymous
  • anonymous
x + x = 4x x(1+1) = 4x (1 + 1) = 4 2 = 4 1 = 2
anonymous
  • 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
  • lgbasallote
it's actually a fallacy lol =)) dividing 0 by 0 is illegal :p
anonymous
  • anonymous
@Aadarsh: This is not a puzzle, this is wrong mathematics. Finding the error would be some fun though.
anonymous
  • anonymous
Everything is possible in maths, lgba.
anonymous
  • anonymous
dividing by 0 is a fantasy?
anonymous
  • anonymous
Its fun time guys and girls. Lets enjoy
anonymous
  • anonymous
stuff just got real.
anonymous
  • anonymous
aad plzz aye koi question hai ..........
Callisto
  • Callisto
@ffm true, we cannot cancel the variables like that :P
apoorvk
  • 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
  • anonymous
@apoorvk smart reply. Newton would have given us gifts!!!!!
apoorvk
  • 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
  • lgbasallote
@Callisto how do you tag @FoolForMath with ffm???
apoorvk
  • apoorvk
like this @ffm
anonymous
  • anonymous
Really? Bansal people are copying from Brain Mapping Academy, Hyderabad.
apoorvk
  • apoorvk
:D sorry for the troubles FFM
lgbasallote
  • lgbasallote
i dont think @ffm is the same as @FoolForMath yes?
anonymous
  • anonymous
I am not notified with @ffm
apoorvk
  • apoorvk
may be, aadarsh. or the other way around. someone always copies from somebody
anonymous
  • anonymous
@Mani_Jha , @AravindG , @payalvsangle , @heena , @sheena101 , @gurvinder , @Ishaan94 , please try. I have to present this before my seniors and get cash prize.
Callisto
  • 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
  • apoorvk
So, Aadarsh where is the trouble. try guessing!
anonymous
  • anonymous
@apoorvk bhai, there is no mistake. Mistake der in assumption.
lgbasallote
  • lgbasallote
the fallacy works well
Mani_Jha
  • Mani_Jha
@As a general rule, you shouldn't multiply or divide an equation by zero.
anonymous
  • 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
  • Mani_Jha
(1−2)^2=(2−1)^2=1 1−2=2−1 2=4 1=2 Find out the mistake in this.
anonymous
  • anonymous
Mistake is in this: 1-2 = 2-1
KingGeorge
  • 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
  • anonymous
Hats off to @KingGeorge .
Mani_Jha
  • Mani_Jha
Haha. Good. I used a fallacy to prove another fallacy!
anonymous
  • anonymous
@King , @shruti , @neha2050 , @salini Please try.
anonymous
  • anonymous
\(a^2 = b^2 \implies \pm a= \pm b \) So there are two solutions.
Mani_Jha
  • 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
  • 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
  • anonymous
@KingGeorge: When a number is raised to a complex power, the result is not precisely defined.
anonymous
  • anonymous
See here: http://en.wikipedia.org/wiki/Mathematical_fallacy.
KingGeorge
  • 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
  • apoorvk
exactly. there is a mistake as @ffm just showed. how do you think are you dividing by '0'??
anonymous
  • 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
  • apoorvk
Abel prize. lol
apoorvk
  • apoorvk
you 'll get special academic razzies for that!
anonymous
  • anonymous
Lets hope so!!!!!!!!!!!!!!!!!!!!!!!
KingGeorge
  • 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
  • Mani_Jha
@Aadarsh, You can't do that. That's why zero is called the 'hero' of Mathematics!
anonymous
  • anonymous
Oooooo! Really??
anonymous
  • anonymous
@shruti , @PRINCEOFPERSIA please try.
anonymous
  • anonymous
tumne apne info mei ye sab kya likh dala hai?
anonymous
  • anonymous
Mera naam hai didi!!!!!! Yeh to maine dhamaal se copy kiya hai.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
anonymous
  • anonymous
cool...yhi kaam kro aur ye kya qestn post kar rakha hai. i felt sumth imp. is there but is'nt
anonymous
  • anonymous
Multiply zero on both sides. 1 x 0 = 2 x 0 0 = 0 1 = 2
anonymous
  • anonymous
Smart answer, Prince Bhai.
anonymous
  • anonymous
Thanks @ Aadarsh..
experimentX
  • 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.