Let us take two variables *a *and *b*

Let *a=b*

Therefore, *a ^{2}=ab*

Therefore, *a ^{2}-b^{2}=ab-b^{2 }*(Subtracting

*b*from both sides)

^{2}Therefore,

*(a+b)(a-b)=b(a-b)*(using identiy

*a*

^{2}-b^{2}=*(a+b)(a-b)*

Therefore, *a+b=b *(Dividing* (a-b) *from both sides)

Since *a=b*, we can substitute *a* in place of *b*

Therefore, *a+a=a*

Therefore, *2a=a*

Therefore, *2=1*

In out world, two is equal to one.

Hence, proved.

Advertisements

See, this is absolutely wrong

you assumed let a=b

so this implies a-b=0

When you are dividing by a-b on both sides: you are dividing by zero on both sides.

Which is in fact not allowed. Division by zero is not allowed!!

Well, in fact, you caught it. In class, we found out the mistake. The fun is, that everybody believes it at first sight. Everyone is stumped.

Are you in AVTE? I remember Vipin sir talking about this in class once. Question is, can you figure out the error? 🙂 Its pretty simple, read it again

Er, what is AVTE?

Haha. nice. 🙂 ur blogs are fun dude!

Nice trick. But division with zero is not defined. If dividing or multiplying with zero was accepted, you would need not study how to prove anything in mathematics!!

To prove: LHS =RHS

Proof: LHS×0=0 and also RHS×0=0

Hence, proved that LHS=RHS!

Thanks for ur help… My algebra II teacher is always making us do this hard stuff in class. Ur blog really helped! =)

Heh. But this stuff is pretty ridiculous. Don’t take it as serious math.

another one of those, more trickier than dividing by zero:

x^2 = x + x + x … x times

differentiating both sides wrt x

2x = 1 + 1 + 1 … x times

=> 2x = x

=> 2 = 1 or x = 0 (which is not true)

=> 2 = 1

Er, first of all, thank you for commenting! Second, I’ve have no clue what differentiation means.

Thirdly, won’t x2=x*x*x* rather than x+x+x?

x^2 = x * x

a * b = a + a + a +… b times

=> x * x = x + x + x…x times

you’ll get to know derivatives in class 11th, and the soln of this prob in 12th. try it then

Its Not MGM I Hope You Get Who I Am.

Here’s Another Trick.

Let x=y

So 2x-x=2y-y

So 2x-2y=x-y

So 2(x-y)=x-y

Say Goodbye To x-y

Hence 2=1

But As Shekhar Stated x-y=0

And Division With 0 Is Not Defined

Hence These Are All Useless

Good discussion. That was a pretty differential calculus demonstration for some interested bunch of Montfortians. The real fun is in partial differentiation and implicit functions. Who is this impersonator, by the way?

It was to be a “pretty ‘lame’ diff…”. Some errors there.

That imposter is my classmate. Gee. I don’t even know what calculus is, so I can’t comment.