DEAD ENDS : Part III : Justified lies

Here I am !
Bored of 2=1s aren't ya'll ? Here are some more of classic lies, told in the most decieving manner !


(1) *** classic *** ( Some PHYSICS for a change )
There in the figure, we have a ladder leaning against the brown wall. Its length is l & the two ends are labelled as A & B. End B is pulled in the positive direction of the x-axis with a constant speed V. The velocity of end A in the downward direction is U ( variable ). x,y are the respective coordinates of the two ends.

x² + y² = l²

differentiating w.r.t. time we get :

2.x.∂ x/ ∂t + 2.y. ∂ y/ ∂t = 0

2.x.∂ x/ ∂t + 2.y. ∂ y/ ∂t = 0
2.x.V = - 2.y.U(x)
U(x) = - x .V/ y

Now, when ladder falls completely, and becomes flat, then y = 0 & x = l.

To calculate the veocity of end A when ladder just becomes flat, we put the values in equation for U(x) getting :

U ( l ) = - l.V/0 = - ∞

So the velocity of end A becomes infinite when the ladder becomes flat ? !

(2)
We are all aware of the theorem of Mathematical Induction. However, none would have seen its misuse. Here is one such misuse which is horribly difficult to disproove !!!

Statement to be verified is :

S(n) : In any set of n people, all would have equal ages.

Now we proceed :

S(1) is true ( Since in a group(set) of 1 person, all have equal age. )

Let S(k) be true.

Now ;

In a group(set) of (k+1) people, we make all possible sets of k people. In all these possible set of n people, S(k) would be true (as per assumption) and hence all people in each of the sets would have equal members. Now since all such possible sets would have members of the same age, this is only possible if all the members in the (k+1) member be equally aged. Therefore All people in the set of (k+1) people have equal ages.

Hence S(1) is TRUE & S(k+1) is TRUE whenever S(k) is TRUE.

Hence prooved by mathematical induction , the TRUTH of statement S(n) for all n belonging to N.

Application of the above theorem for the set of BLOG MEMBERS "B" We conclude that All members of this blog are of equal age.
QED

(3) *** classic ***

ABCD is a square. BE = BC.
PQ bisects CD,AB.
OR is perp.bisect. of DE.
PQ, RO intersect at O.
∠ ABE is an OBTUSE ANGLE. ( clearly )

Δ ORD ≡ Δ ORE (SAS)
:. OD = OE

Δ OQA ≡ Δ OQB (SAS)
:. OA = OB & ∠ OAB = ∠ OBA

Δ OAD ≡ Δ OBE (SSS)
:. ∠ OAD = ∠ OBE
∠ ABE = ∠ OBE - ∠ OBA
= ∠ OAD -∠ OAB

But ∠ OAB - ∠ OAB
= 90 °

:. ∠ ABE = 90 °
However, we've assumed the angle to be obtuse. :. Every obtuse angle is a right angle. Similarly, It can be shown that every acute angle is a right angle.
QED

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This is part III and I'll let ya'll know that 3's my lucky number. So hereby, I quit with this series. Not that the resources are over ! Infact they're endless. But today eve, I'm leaving for Delhi & I won't be able to be online so regularly. Though I'll surely stay in touch atleast twice a week !

A last message. I believe that us KVamPYs are special in some way. I also believe, that we can be a great strength when united ! So here's a request. Mayb you can't come online very often but do stay in touch by atleast going through the blog once a week !!!
This blog is for us KVamPYs, by us KVamPYs & of us KVamPYs.

So letz ROCK ... letz RULE !!!

& Ill borrow Rash's line...
SwItCh On Ur BuLbS !!!

~ Twishmay Shankar ~
( Twish ;-) )

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I HaVe ThE POWER!!

Well, does the title remind you of the classic “He-Man” Cartoon series?? Well I’m not posting about cartoons but this has to do with “Power.” But that’ll come later.

First of all, I wanted to post my intro too but was waiting for Twish to mail me the pics from the camp. So that’s me saying: ”I have the Power!!”. You want a formal Intro? Here you go:

Name: Rasagy Sharma (a.k.a. ‘Rash’ The Gr81)

Residing in: Delhi

Interests: Dramatics (Done loads of TV and Radio Shows), PlayingSoccer (Some of you who played with me know that – It’s a second religion for me!), of course anything to do with Computers (I just luv Computers!) and last but not the least, making Friends (and continuing the friendship too!)

Enough about me. So after reading Bravura’s post, I was sitting in front of my computer wondering about what went wrong there. Got some ideas, but couldn’t join them all up. And then a short chat with Twish only raised my ideas to higher powers… Yes. It was powers about which I have been thinking all night. Well, most of such weird looking problems arise from the fact that what we have been taught about powers, multiplication etc. is too basic (coz we were too young to understand complex nos. at that time). And even now, we haven’t updated our information about such elementary things, though we do know complex nos. etc. And thus arise such ques.

Well here are some examples:

1. 1⁰ = 2⁰

Then won’t 1 = 2 ?? (Say we have 2^x=2², then we say that x = 2, don’t we??!!)

2. To prove: 3 = 4
Proof:
Let a+b = c
4a-3a + 4b-3b = 4c-3c
:.4a+4b-4c=3a+3b-3c
:. 4*(a+b-c)=3*(a+b-c)
:. 3 = 4

Maybe most of you can find what’s wrong up there (I think you all Should find that out easily!). But ever wondered why even your computer’s scientific calculator can’t compute (-8)^1/3 ? Even though the answer is an obvious -2 ? (It says invalid input for the function: Got some clue??). Though its fine that it can't compute (-8)^1/2.

Another nice problem (which Twish told me during the camp) is the following:

We know that

4*4 = 4 + 4 + 4 + 4

:. 4*4= 4 + 4 +… 4 times.

:. x*x = x + x + … x times.

Differentiating,

2*x = 1 + 1 + 1 … x times.

:. 2*x = x

Hence 2 = 1!!!

All such problems arise because we forget our assumptions – the definitions of the functions – the domains of the functions and we tend to generalize a formula we learnt in junior classes here.

Now what I was thinking was this:

x = 1^1/3

Then x³ = 1. And we get the solutions for x. So as this n increases, the no. of solutions for x also increase (which is n itself).

But if n-> infinity , we get only one value for x = 1. (Actually here we can’t replace n by infinity as n-> infinity but n is not equal to infinity.)

Also,

2 ⁰ = 1
1 ⁰ = 1
0 ⁰ = ?

Well infact 0 ⁰ an indeterminate form. So we can treat it as an exception.

But what if n is a complex no.?? According to Twish, it’ll lead to infinite solutions for x. I reckon that’s what messes everything in Bravura’s problem. So can some body explain to me what’ll happen with complex powers? (Do we have any physical interpretation other than just saying Z = 2 ⁱ finding Z??)

Sorry for writing such a Llllllooooonnnngggg post, but I think it'll compensate for my absence in the coming few days when I'll go out of station. So I hope to find many posts by then (posts by members other than Twish's Dead Ends - come on others plz post!)

And before ending, a bit of cheerleading… (yeah I’m at least good at that!) U ppl are doing a gr8 job posting stuff, but do visit the blog regularly (Too busy studying huh?) And do give more ideas as to what more to add in the blog..

So SwItCh On YoUr FuSeD BuLbS!!

~RashTheGr81~

DEAD ENDS : Part II : Burning Reality

Hi Again
As promised, here is PART II ...
Bewarethough !!! I'd certainly say this one is more wierd, more interesting but still a LIE which has been imposed wrongly upon reality :

(1)
Every1 here knows the pretty little Binomial thorem going as :

(a+b)ⁿ = aⁿ + n*b*a^n-1 + ... + n*b^n-1*a + bⁿ

Observing we find that leaving the first and last term, all other terms are a multiple of n.

For some wierd results, let us put n = 0 ;

LHS = (a+b)⁰ = 1

RHS = a⁰ + 0 ... + 0 + b⁰

= 1 + 0 ... 0 + 1

= 2

:. since LHS = RHS

2 = 1

QED

(2)

Consider the integral :

I = ∫ 1/x ∂x

Intigrating by parts ;

I = ∫ 1*1/x ∂x

= x*(1/x) - ∫ x*(-1/x²) ∂x

= 1 + ∫ 1/x ∂x

= 1 + I

:. 0 = 1

or, 2 = 1

QED

(3) *** CLASSIC ***

Have a look at the figure. O is the origin. C is the centre of the circle. P a variable point (x,y).

t = OP ; R = CP ; a = OC

The equation of the circle would be:

x² + y² -2*a*x + a² - R²

Now, t² = x² + y²

Also putting P(x,y) in circles equation we get :

t² - 2*a*x + R² - a² = 0

:. t² = 2*a*x + R² - a²

2*∂t/ ∂ x = 2*a

Now for the point with minimum distance , The above expression equals 0.

For that to be possible, a = 0, that is O and C are the same points !

Hence Whenever O is not the center of the circle, there is no point on the circle from which the distant of O is maximum or minimum.

QED

There is still more to come. Before thinking about the faults, appreciate the beauty of these propositions, how decievingly they proove the unproovable !

This in not the end.

~ Twish ~

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NOTE : Those having difficulty typing math visit this link for keyboard codes :

http://www.planetburrito.com/tb/alt_codes.html

This is Somani

Hi ,
This is Somani,
Call me by any other nickname you please(just inform me about it )
For identification 'The girl whose hair stood up due to shock' should be enough.

I AM

Hi Every1
I'm Sunita Panda from IIT Kharagpur campus
u might remember me as
-the girl who used to ask stupid questions during lectures.
-the self appointed guide at the science city (till my job was outsourced to 2 strangers)
I think that should be enough