1+1=1
1+1=1
I ran across this in some old old high-school papers, it actually can be disproved quite easily, and i'll explain it after the "proof"
1 + 1 = 1
a = 1
b = 1
a = b
a2 = b2
a2 - b2 = 0
(a-b)(a+b) = 0
(a-b)(a+b)/(a-b) = 0/(a-b)
1(a+b) = 0
(a+b) = 0
1 + 1 = 0
2 = 0
1 = 0
1 + 1 = 1
Why this proof is not "real":
you cannot divide by zero; if you could, you could pretty much prove anything. But this was an extra credit problem on a test we took, we basically had to explain if this was a valid proof or not and why.
At any rate, figured some of you might could get a kick out of it.
1 + 1 = 1
a = 1
b = 1
a = b
a2 = b2
a2 - b2 = 0
(a-b)(a+b) = 0
(a-b)(a+b)/(a-b) = 0/(a-b)
1(a+b) = 0
(a+b) = 0
1 + 1 = 0
2 = 0
1 = 0
1 + 1 = 1
Why this proof is not "real":
you cannot divide by zero; if you could, you could pretty much prove anything. But this was an extra credit problem on a test we took, we basically had to explain if this was a valid proof or not and why.
At any rate, figured some of you might could get a kick out of it.
Re: 1+1=1
Indeed, the equation is flawed but you say multiplying two negatives doesn't work? How so?
If you do use complex numbers along with the standard notation that i=√-1 where i^2=-1 , then:
√-1 x √-1 = i x i = i^2 = -1
*pokes tongue out at Omni*
The only true 'flaw' as such is the seperation of one constant into two but that remains ambiguous.
Here's another one for you Omni, just cause I likes to see you gnash your teeth
------
Hmm, how's about I show that a ladder will fall infintely fast if pulled from a wall.
Let's see, consider a ladder (with length L) which leans against a frictionless wall which is at a right angle to the ground. I pull the ladder horizontally away from the wall at a constant speed (let's call it v). The claim i make about this scenario is that the top of the ladder will fall infintely fast. Obviously this can't be true but can you prove me wrong?
Let x represent the horizontal distance from the bottom of the ladder to the wall, at time t.
Let y represent the height from the top of the ladder to the ground, at time t.
The ladder, the wall and the ground form a right angle triangle. Pythagoras tells us: x^2 + y^2 = L^2
Therefore, y = √(L^2 - x^2)
Differentiating this and letting x' and y' represent the derivitives of x and y respectivly with respect to t gives us:
y' = - xx'/√(L^2 - x^2)
As the bottom of the ladder is being pulled at the constant speed v, we have that x' = v and therefore:
y' = - xv/√(L^2 - x^2)
As x approaches L. the numerator in the fractional expression for y' approaches -Lv which is of course non-zero, whilst the denominator approaches zero.
Therefore, y' approaches negative infinity as x approaches L. To simplify, the top of the ladder is falling infinitly fast by the time the bottom has been pulled L distance away from the wall.
-------
Your turn Omni, find the fallacy and i'll give you a chocolate chip cookie!
EDIT: Corrected spelling
If you do use complex numbers along with the standard notation that i=√-1 where i^2=-1 , then:
√-1 x √-1 = i x i = i^2 = -1
*pokes tongue out at Omni*
The only true 'flaw' as such is the seperation of one constant into two but that remains ambiguous.
Here's another one for you Omni, just cause I likes to see you gnash your teeth
------
Hmm, how's about I show that a ladder will fall infintely fast if pulled from a wall.
Let's see, consider a ladder (with length L) which leans against a frictionless wall which is at a right angle to the ground. I pull the ladder horizontally away from the wall at a constant speed (let's call it v). The claim i make about this scenario is that the top of the ladder will fall infintely fast. Obviously this can't be true but can you prove me wrong?
Let x represent the horizontal distance from the bottom of the ladder to the wall, at time t.
Let y represent the height from the top of the ladder to the ground, at time t.
The ladder, the wall and the ground form a right angle triangle. Pythagoras tells us: x^2 + y^2 = L^2
Therefore, y = √(L^2 - x^2)
Differentiating this and letting x' and y' represent the derivitives of x and y respectivly with respect to t gives us:
y' = - xx'/√(L^2 - x^2)
As the bottom of the ladder is being pulled at the constant speed v, we have that x' = v and therefore:
y' = - xv/√(L^2 - x^2)
As x approaches L. the numerator in the fractional expression for y' approaches -Lv which is of course non-zero, whilst the denominator approaches zero.
Therefore, y' approaches negative infinity as x approaches L. To simplify, the top of the ladder is falling infinitly fast by the time the bottom has been pulled L distance away from the wall.
-------
Your turn Omni, find the fallacy and i'll give you a chocolate chip cookie!
EDIT: Corrected spelling
- Quaternion
- Posts: 113
- Joined: Sat Dec 20, 2003 2:57 pm
Re: 1+1=1
I wanted to elabourate a bit more on why the proof of 1= -1 is incorrect. You say that its becuase of the ambiguity of √-1, I would say this isn't true if you were to actually define √ correctly . Representing a complex number geometrically in terms of a vector on a complex plane one can write any complex number as r x exp(iO) = r x (cos(O)+isin(O) where r is the length of the vector and O is the angle taken in an anticlockwise direction from the possitive x axis. The √1 then has infinitely many solutions as one can rotate any solution by a phase that is a multiple of 2 * pi infinitely many amout of times. The fact that the function isn't single valued is what causes the problem in your proof. However in reality when working with complex functions the standard practice is to restrict the domain of the angle O to an angle -pi<O<=pi. The line O = -pi is called a branch cut, and moving across it one moves onto a seperate but identical (isomorphic) complex plane. So in short the problem lies in the fact that -1 although a solution to √1 it lies on a different complex plane and thus cant be used.
On a side note, I defined the angle O oriented from the positive axis, it doesn't really matter where you take this angle from as long as you keep track of where the branch cut moves to and in doing so restrict your domain to a single complex plane.
Quat.
On a side note, I defined the angle O oriented from the positive axis, it doesn't really matter where you take this angle from as long as you keep track of where the branch cut moves to and in doing so restrict your domain to a single complex plane.
Quat.
Re: 1+1=1
*yawn*
There is the implicit assumption here that the top of the ladder remains resting against the wall. However, that is not always true. Once the ladder has reached a sufficiently small angle to the horizontal, your pulling of the bottom away from the wall will actually cause the top to pull away from the wall too. When this happens, there is no longer the relationship , because x, y, and L no longer form the sides of a closed right triangle.
There is the implicit assumption here that the top of the ladder remains resting against the wall. However, that is not always true. Once the ladder has reached a sufficiently small angle to the horizontal, your pulling of the bottom away from the wall will actually cause the top to pull away from the wall too. When this happens, there is no longer the relationship , because x, y, and L no longer form the sides of a closed right triangle.
Re: 1+1=1
I decided that it's very fun to revive old topics!
EDIT: oh, and Xaphy darling...that ladder...it can't fall infinitely fast if you pull the bottom out, because the speed that the top of the ladder falls at, is directly related to the speed that you are pulling at the bottom (even if the wall that the ladder is resting against is very very slick). If you can prove to me that you can pull the ladder bottom at an infinite speed (say...faster than the speed of light even) then, perhaps this might work...but, I don't think it's possible for any mortal (without help from a "super machine") to drag the bottom of a ladder out from underneath it at lightspeed (or faster.....)
*imagines xaphy's arms moving too fast to see, and one second the ladder is standing, and the next, the top is already on the ground...note that the ground is blasted apart due to the impact of the infinitely fast top of the ladder speeding downward and hitting the pavement. Watches as Xaphy is thrown backward at an infinite speed....*
EDIT: oh, and Xaphy darling...that ladder...it can't fall infinitely fast if you pull the bottom out, because the speed that the top of the ladder falls at, is directly related to the speed that you are pulling at the bottom (even if the wall that the ladder is resting against is very very slick). If you can prove to me that you can pull the ladder bottom at an infinite speed (say...faster than the speed of light even) then, perhaps this might work...but, I don't think it's possible for any mortal (without help from a "super machine") to drag the bottom of a ladder out from underneath it at lightspeed (or faster.....)
*imagines xaphy's arms moving too fast to see, and one second the ladder is standing, and the next, the top is already on the ground...note that the ground is blasted apart due to the impact of the infinitely fast top of the ladder speeding downward and hitting the pavement. Watches as Xaphy is thrown backward at an infinite speed....*
Re: 1+1=1
to mate? what are you saying Tee? :pTee wrote:how do you know omni is a "she" ? had omni ever really come out and stated if "it" was female to mate ? not to my knowledge. The gender of omni will be unknown till "it" states it.
Are you related to Migam by any chance? <EG>
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Re: 1+1=1
Going to fix these typo's as well there Tee?Tee wrote:Omni is a "it" and *sticks tongue out at omni* fixxed my typos
Yet more proof of a Migam connection! :p
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Send me Ideas, Suggestions, Bugs or Complaints!
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