Question

Challenge...

Prove 3 = 4,
without dividing or multiplying it by zero.... Also, without using that many variable addition thing....

You can just involve a single variable....

Let's see, who can do it..

Hint = ) You can skip a case, when there happen to be extra cases...

Answer 1

nice pic

Answer 2

thanks, but you need to answer the question for medal @Rainbow_dashie

Answer 3

ik im reading just wait

Answer 4

@hartnn ... wanna try ?

Answer 5

logic here

(1)guy weds (+) (1)gal =3 (one kid)

Answer 6

@gudden, You can't do this?

Answer 7

Guys, let me say something, don't tell the answers, just help. Please.

Answer 8

I can @Rainbow_dashie

And you can violate the rules of square root, ( that's one of the method.... and a great hint too )

Answer 9

Oh ok your testing us

Answer 10

\(3 =4\)


\(Hence Proved\)... \(Ans.!\) :P

Answer 11

Not testing you people... just gave you a great question, to spend your time at....

Trust me, it's easy...

Answer 12

@waterineyes ,,,, Kindly show the work!

Answer 13

Closely look, I showed.. :P

Answer 14

naah.! Show your work in at least 5 lines @ waterineyes...

and it must be not


3
=
4
hence
proved :p

Answer 15

Okay:

We have :
LHS as 3
RHS as 4
And you know:
LHS = RHS
=> 3 = 4 written 6 lines.. and not by your method too.. :P

Anything else you want? :P

Answer 16

Yeah....
proper.. work..

Answer 17

I want proper work...:P

Answer 18

How can you say that is not proper?? :P

Answer 19

cause you equated, LHS and RHS, without any reason....

Answer 20

Okay I got it..

Answer 21

So many people are staring at this foolish question :P

Answer 22

Should I pm you the answer or tell it here?

Answer 23

PM...

Answer 24

Starting with:

\(9-21 = 16-28\)
\[9 - 21 + \frac{49}{4} = 16 - 28 + \frac{49}{4}\]
\[3^2 - \frac{2.3.7}{2}= 4^2 - \frac{2.4.7}{2} + (\frac{7}{2})^2 +(\frac{7}{2})^2\]
\[(3 - \frac{7}{2})^2 = (4 - \frac{7}{2})^2\]

Answer 25

exactly....^_^

Answer 26

Take square root and cancel 7/2.

Answer 27

we skipped the fact of taking the modulus and still, the answer, looks correct to someone, who has no experience with this function...

Answer 28

This is clearly a cheating from my side:

http://engg.entrancecorner.com/old/discussion-forum/131-algebra/41606-how-to-prove-that-34.html

Answer 29

Now, I am looking at what I did there.. :)

The solution is not getting to the person who have posted it.. :P

Answer 30

It's impossible to prove it. Flawed proofs are not proofs.

Answer 31

haahhaha....