An easy question

bilal azhar · 03-03-2005, 11:46 PM

sam i think u have misunderstood y2.it is y square.i have subtracted y square from both sides and not y two.

so if u factorize x2-y2 then answer is (x+y)(x-y)

bilal

bilal azhar · 03-04-2005, 12:27 AM

if somebody proves that 2=1 then he should be given a noble prize.
The general misconception among the people are that accuntants are very good in maths but the fact is something else(i am not saying this simply because of my above post).

Take the example of above prove,nobody is being able to detect a simple basic mathematical mistake in the above prove.

bilal

Desert Sleet · 03-05-2005, 02:05 AM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by bilal azhar
 if somebody proves that 2=1 then he should be given a noble prize.
The general misconception among the people are that ac****ants are very good in maths but the fact is something else(i am not saying this simply because of my above post).

Take the example of above prove,nobody is being able to detect a simple basic mathematical mistake in the above prove.

bilal
<hr height="1" noshade id="quote"></blockquote id="quote">
Well, not sure what the official rebuttle for this is, but from what I remember from high school algebra there is only one number that can make this statement true
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x = x <hr height="1" noshade id="quote"></blockquote id="quote">
and based on that number you cannot perform this operation
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x/x = x/x <hr height="1" noshade id="quote"></blockquote id="quote">

There is only one number which can proove this statement true and its 0.
<blockquote id="quote">quote<hr height="1" noshade id="quote">(x+y)(x-y) = y(x-y)<hr height="1" noshade id="quote"></blockquote id="quote">
At this point, if you apply the rule, x = y, both sides of the = are 0, and of course every thing after that is correct, if you assume x = 0.
<blockquote id="quote">quote<hr height="1" noshade id="quote">x+y = y
x+x = x
2x = x<hr height="1" noshade id="quote"></blockquote id="quote">
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x/x = x/x

2 = 1<hr height="1" noshade id="quote"></blockquote id="quote">
At this point you have just divided 0 by 0, on both sides of the equal sign. Now, I have no idea if that mean they're both 0, undefined, or 1.

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If I could... Then I would... Turn back time!!

bilal azhar · 03-05-2005, 08:43 AM

no,desert sleet this in not the right answer.

TO give u another hint,everything in the above prove is right except one step which is neither at the start nor at the end.

bilal

Desert Sleet · 03-05-2005, 04:23 PM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by bilal azhar
 no,desert sleet this in not the right answer.

TO give u another hint,everything in the above prove is right except one step which is neither at the start nor at the end.

bilal
<hr height="1" noshade id="quote"></blockquote id="quote">
No, you only proved that 0=0.
Go to the step where you have (x+y)(x-y) = y(x-y)
Since x=y that is the same as saying (x+y)(0)=y(0) which of course is 0=0.

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If I could... Then I would... Turn back time!!

bilal azhar · 03-06-2005, 12:26 AM

here is the answer
as x = y so this means that x-y =0

now look at this step (x+y)(x-Y)= y(x-y)
(x+Y)(x-y)/(x-y) = y(x-y)/(x-y)
which means x+y = y

now as i have divided x-y with x-y but x-y is equal to zero and there is mathematical rule that anything divided by zero is equal to infinity,so this prove terminates from the middle.

bilal

Desert Sleet · 03-11-2005, 08:01 PM

So lets continue

Okay, you're face to face with a ferocious tiger. You can't run, you can't hide from him. What do you do? You have no gun either.

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When The Going Gets Tough ... The Tough Gets Going ...

Huma · 03-12-2005, 02:49 AM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by sam
 the tiger is in the cage...
<hr height="1" noshade id="quote"></blockquote id="quote">
i will lie down and hold my breath, IF i do not faint.. [8)][8)][8)]

Shahid_fss · 03-12-2005, 06:42 AM

maybe I am in the cage....

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```*``` I love stars;
*`*`*`* Shining;
`*`*`*` and Smiling;
*`````* Always.
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[email protected]
http//www.shahid-fss.tk
========================

Desert Sleet · 03-13-2005, 03:07 AM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by Shahid_fss
 maybe I am in the cage....

========================
```*``` I love stars;
*`*`*`* Shining;
`*`*`*` and Smiling;
*`````* Always.
------------------------
[email protected]
http//www.shahid-fss.tk
========================
<hr height="1" noshade id="quote"></blockquote id="quote">
Spot right. You are in a Zoo and lion is in cage. [D]

Lets Continue... lets have a difficult one

Suppose we have two balls of the same radius. I'll call them the "rolling ball" and the "fixed ball".

The fixed ball is not allowed to move.

The rolling ball must touch the fixed ball - and it's allowed to roll without slipping or twisting on the surface of the fixed ball.

Now

1) Start with the rolling ball touching the north pole of the fixed ball.

2) Roll it down to the equator along a line of longitude.

3) Then roll it along the equator for an arbitrary distance.

4) Then roll it back up to the north pole along a line of longitude.

The question is when we carry out this process, can the rolling ball come back *rotated* relative to its original orientation ... or not?

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When The Going Gets Tough ... The Tough Gets Going ...

farazthegreat · 03-15-2005, 05:01 PM

The ball would have been rotated horizontally, but not vertically.

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A fine is a tax for doing wrong. A tax is a fine for doing well.

bilal azhar · 03-15-2005, 11:55 PM

i dont know the answer of the above one but here is another question.would anybody like to answer it?

Question Standing on the moon a person wants to see earth.where
will he look.up,down,right,left?

bilal

Desert Sleet · 03-16-2005, 12:07 AM

your answer is correct (the ball does not change orientation), but I think that your explanation is incomplete. Drawing a mark on the north pole only demonstrates that the poles do not move, but it still leaves some doubt as to whether the moving ball might have rotated around the north-south axis relative to the fixed ball.

Just in case anyone's interested, here's an abbreviated solution, using quaternions. Quaternions are of the form a+bi+cj+dk where a,b,c,d are real; ij = k; jk = i; ki = j; ii = jj = kk = ijk = -1. The coolest thing about them is that rotations can be handled with them. A rotation of w radians around the unit vector (x,y,z) is expressed by q1 = cos(w/2) + sin(w/2)(xi+yj+zk). Multiple rotations are obtained by multiplying the quaternions together, in reverse order.

I will set my coordinates so that the first rotation is from the z axis towards the x axis. It's a rotation of pi radians, around the y axis. So q1 = cos(pi/2) + sin(pi/2)j = j.

The second rotation is from the x axis towards the y axis, around the z axis, some angle 2t. The quaternion q2 = cos(t) + sin(t)k.

The last one is a rotation of pi radians back to the top, around the vector (sin(t) -cos(t) 0). So q3 = cos(pi/2) + sin(pi/2)(sin(t)i-cos(t)j) = sin(t)i - cos(t)j.

Combining them
Q = q3q2q1
= (sin(t)i - cos(t)j) (cos(t) + sin(t)k) (j)
= (sin(t)i - cos(t)j) (cos(t)j - sin(t)i)
= sin(t)cos(t)k +sin(t)sin(t) + cos(t)cos(t) - cos(t)sin(t)k
= 1, which corresponds to a rotation of 0 radians.

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When The Going Gets Tough ... The Tough Gets Going ...

Desert Sleet · 03-16-2005, 12:48 AM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by bilal azhar
 i dont know the answer of the above one but here is another question.would anybody like to answer it?

Question Standing on the moon a person wants to see earth.where
will he look.up,down,right,left?

bilal
<hr height="1" noshade id="quote"></blockquote id="quote">Pretend you are on earth and want to see the moon. Look out. The earth will be where the moon normally is.

Magic isn't it. Where'd you find the nice smilie?

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When The Going Gets Tough ... The Tough Gets Going ...