You good with math?
8/18/05, 01:02 PM
#1
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I am bored so I am going to mess with some of your heads, here's an old math brain teaser for all of you......
here is a proof to prove that 1=2 so 1+1=4 not 2
IF:
A = B
THEN:
A^2 = AB
A^2 - B^2 = AB - B^2
( A - B ) ( A + B ) = B ( A - B )
A + B = B
B + B = B
2B = B
2 = 1
Therefore 1 + 1 = 2 + 2, i.e. 1 + 1 = 4, not 2.
Now this is all fine and dandy for screwing with your algebra teachers head (since we all know they dont know jack) but now tell me why it is that this does not work. It's simple math you learn long before you learn proofs.
in algebra class I didnt like the teacher so I sat there one day and tried to debunk the whole stupid idea that if you assume some obviously wrong statements to be true that you can prove anything and thus algebra was a waste of time. Little did I know at the time that this proof already existed, but I came up with it anyway after sitting at my desk for 20 minutes and handed it to my teacher, stumping her and leaving her at a loss for words and just dismissing the whole idea that I had just proved that this proof thing was pointless because if under the right circumstances assumed you can prove anything and that cant be the case right? well after I handed it to her and she sat there pondering it and getting mad, I noticed that the proof was flawed because of some simple math, I never told her, I just let her go on looking dumb, oh well.
here is a proof to prove that 1=2 so 1+1=4 not 2
IF:
A = B
THEN:
A^2 = AB
A^2 - B^2 = AB - B^2
( A - B ) ( A + B ) = B ( A - B )
A + B = B
B + B = B
2B = B
2 = 1
Therefore 1 + 1 = 2 + 2, i.e. 1 + 1 = 4, not 2.
Now this is all fine and dandy for screwing with your algebra teachers head (since we all know they dont know jack) but now tell me why it is that this does not work. It's simple math you learn long before you learn proofs.
in algebra class I didnt like the teacher so I sat there one day and tried to debunk the whole stupid idea that if you assume some obviously wrong statements to be true that you can prove anything and thus algebra was a waste of time. Little did I know at the time that this proof already existed, but I came up with it anyway after sitting at my desk for 20 minutes and handed it to my teacher, stumping her and leaving her at a loss for words and just dismissing the whole idea that I had just proved that this proof thing was pointless because if under the right circumstances assumed you can prove anything and that cant be the case right? well after I handed it to her and she sat there pondering it and getting mad, I noticed that the proof was flawed because of some simple math, I never told her, I just let her go on looking dumb, oh well.
8/18/05, 01:47 PM
#4
How about the easy version?
Because you can't divide by zero.
Because you can't divide by zero.
8/18/05, 01:48 PM
#5
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Consider a ladder of length L leaning against a frictionless wall which is at right angles to the ground. You pull the bottom of the ladder horizontally away from the wall, at constant speed v. The claim is that this causes the top of the ladder to fall infinitely fast. Common sense tells us this can't possibly be true, but can you find the flaw in the following supposed "proof" of this claim?
Step 1: As shown, let x denote the horizontal distance from the bottom of the ladder to the wall, at time t.
Step 2: As shown, let y denote the height of the top of the ladder from the ground, at time t.
Step 3: Since the ladder, the ground, and the wall form a right triangle,
Step 4: Therefore,
Step 5: Differentiating, and letting x' and y' (respectively) denote the derivatives of x and y with respect to t, we get that
Step 6: Since the bottom of the ladder is being pulled with constant speed v, we have x' = v, and therefore
Step 7: As x approaches L, the numerator in this expression for y' approaches -Lv which is nonzero, while the denominator approaches zero.
Step 8: Therefore, y' approaches
as x approaches L. In other words, the top of the ladder is falling infinitely fast by the time the bottom has been pulled a distance L away from the wall.
Step 1: As shown, let x denote the horizontal distance from the bottom of the ladder to the wall, at time t.
Step 2: As shown, let y denote the height of the top of the ladder from the ground, at time t.
Step 3: Since the ladder, the ground, and the wall form a right triangle,
Step 4: Therefore,
Step 5: Differentiating, and letting x' and y' (respectively) denote the derivatives of x and y with respect to t, we get that
Step 6: Since the bottom of the ladder is being pulled with constant speed v, we have x' = v, and therefore
Step 7: As x approaches L, the numerator in this expression for y' approaches -Lv which is nonzero, while the denominator approaches zero.
Step 8: Therefore, y' approaches
as x approaches L. In other words, the top of the ladder is falling infinitely fast by the time the bottom has been pulled a distance L away from the wall.
8/18/05, 02:03 PM
#7
I remember reading the ladder one before. Something about integrating the equation would give you a funky answer. I don't really feel like trying to remember Diff Eq right now :P ... I mean the obvious answer is that y decreases the same rate x increases. The differential version just has me :scratch: at the moment. Its been a while
8/18/05, 02:07 PM
#9
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Originally posted by SurfnSoCal@August 18, 2005, 1:51 PM
Consider a ladder of length L leaning against a frictionless wall which is at right angles to the ground. You pull the bottom of the ladder horizontally away from the wall, at constant speed v. The claim is that this causes the top of the ladder to fall infinitely fast. Common sense tells us this can't possibly be true, but can you find the flaw in the following supposed "proof" of this claim?
Step 1: As shown, let x denote the horizontal distance from the bottom of the ladder to the wall, at time t.
Step 2: As shown, let y denote the height of the top of the ladder from the ground, at time t.
Step 3: Since the ladder, the ground, and the wall form a right triangle,
Step 4: Therefore,
Step 5: Differentiating, and letting x' and y' (respectively) denote the derivatives of x and y with respect to t, we get that
Step 6: Since the bottom of the ladder is being pulled with constant speed v, we have x' = v, and therefore
Step 7: As x approaches L, the numerator in this expression for y' approaches -Lv which is nonzero, while the denominator approaches zero.
Step 8: Therefore, y' approaches as x approaches L. In other words, the top of the ladder is falling infinitely fast by the time the bottom has been pulled a distance L away from the wall.
Consider a ladder of length L leaning against a frictionless wall which is at right angles to the ground. You pull the bottom of the ladder horizontally away from the wall, at constant speed v. The claim is that this causes the top of the ladder to fall infinitely fast. Common sense tells us this can't possibly be true, but can you find the flaw in the following supposed "proof" of this claim?
Step 1: As shown, let x denote the horizontal distance from the bottom of the ladder to the wall, at time t.
Step 2: As shown, let y denote the height of the top of the ladder from the ground, at time t.
Step 3: Since the ladder, the ground, and the wall form a right triangle,
Step 4: Therefore,
Step 5: Differentiating, and letting x' and y' (respectively) denote the derivatives of x and y with respect to t, we get that
Step 6: Since the bottom of the ladder is being pulled with constant speed v, we have x' = v, and therefore
Step 7: As x approaches L, the numerator in this expression for y' approaches -Lv which is nonzero, while the denominator approaches zero.
Step 8: Therefore, y' approaches
as x approaches L. In other words, the top of the ladder is falling infinitely fast by the time the bottom has been pulled a distance L away from the wall.the basis of the whole thing going wrong is that you are using just calc and forgetting that it is a physics problem, first to prove this in any way we need to know the mass of the ladder and if it is evenly distributed along the ladder, also is there friction between the ladder and the ground, what about air resistance, since when can you simply substitute distance with velocity (v=d/t), we dont know the elapsed time or any of the directional vectors of anything, so in phsyics the problem probably doesnt have enough to be solved anyway, but in calc there are also some fallacies here that are just bad math, I will respond in greater detail after I am done cleaning my car for the show in a couple of hours, if I have time,
8/18/05, 02:13 PM
#10
I've got a couple math-riddles too.
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
8/18/05, 02:16 PM
#11
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Originally posted by freebass55@August 18, 2005, 1:23 PM
It's been a while. Explain this one to me. Please.
It's been a while. Explain this one to me. Please.
A = B
THEN:
A^2 = AB
A^2 - B^2 = AB - B^2
{so A^2 - B^2 = ( A - B ) ( A + B ) and AB - B^2 = B ( A - B )}
( A - B ) ( A + B ) = B ( A - B )
{Divide both sides by common ( A - B ) and ( A - B )/( A - B ) = 1 leaving}
A + B = B
B + B = B
2B = B
2 = 1
Therefore 1 + 1 = 2 + 2, i.e. 1 + 1 = 4, not 2.
8/18/05, 03:13 PM
#12
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Originally posted by rrobello@August 18, 2005, 1:10 PM
OMG there are so many things wrong with this one......
the basis of the whole thing going wrong is that you are using just calc and forgetting that it is a physics problem, first to prove this in any way we need to know the mass of the ladder and if it is evenly distributed along the ladder, also is there friction between the ladder and the ground, what about air resistance, since when can you simply substitute distance with velocity (v=d/t), we dont know the elapsed time or any of the directional vectors of anything, so in phsyics the problem probably doesnt have enough to be solved anyway, but in calc there are also some fallacies here that are just bad math, I will respond in greater detail after I am done cleaning my car for the show in a couple of hours, if I have time,
OMG there are so many things wrong with this one......
the basis of the whole thing going wrong is that you are using just calc and forgetting that it is a physics problem, first to prove this in any way we need to know the mass of the ladder and if it is evenly distributed along the ladder, also is there friction between the ladder and the ground, what about air resistance, since when can you simply substitute distance with velocity (v=d/t), we dont know the elapsed time or any of the directional vectors of anything, so in phsyics the problem probably doesnt have enough to be solved anyway, but in calc there are also some fallacies here that are just bad math, I will respond in greater detail after I am done cleaning my car for the show in a couple of hours, if I have time,
OMG....reading way too much in to it. the calc is correct, but not even needed. think about the premis, and the flaw will become obvious. ladder resting against a frictionless wall. bottom is pulled out.
think about that.
8/18/05, 03:16 PM
#13
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Originally posted by Enfynet@August 18, 2005, 1:16 PM
I've got a couple math-riddles too.
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
I've got a couple math-riddles too.
Here's one that is appropriate with my username...
++Hotel Infinity:
One of the qualifications for a hotel clerk at Hotel Infinity is a working knowledge of infinity. Paul applied, was interviewed, and started work the following evening. He wondered why the hotel required that all its clerks know about infinity, infinite sets, and transinfinite numbers. He figured since it was an infinitely roomed hotel it would be no problem finding rooms for its guests. After his first night on the job, he was glad he had that knowledge.
When he relieved the day clerk, she informed him that there were an infinite number of rooms presently occupied. As she left, a new guest walked in with a reservation. He needed to decide which room to give the guest. He thought for a moment, then moved each occupant to a room with the next highest number, and was therefore able to vacate room #1. He felt good about his solution, but just then an infinite bus load of new guests arrived. How would he give then their rooms?
easy.
A: Refer them to the other infinite hotel down the street.
8/18/05, 04:28 PM
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ok real quick before I get redy for the show are you saying its a riddle or a math question Surfn???? if its a riddle then the ladder wouldnt be able to stay up against the wall to begin with if there is no friction, so you cant have the question, but if its a math question I will have to look again later, since we are both running a little behind for the show. see you there.
8/18/05, 04:39 PM
#15
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Ok, its a math problem....but its how they solve it and why they can't solve it that way. you are close with-
"its a riddle then the ladder wouldnt be able to stay up against the wall to begin with if there is no friction.."
"its a riddle then the ladder wouldnt be able to stay up against the wall to begin with if there is no friction.."
8/19/05, 04:13 AM
#17
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ok Surfn Ive looked at it more now that I am finally home and assuming that none of the way obvious things wrong with this are the answer you are looking for, I went step by step through it to figure out where it would go wrong mathematically, and I am not sure if this is what you are looking for since it is because of an assumption that is made at the begining that isnt true that makes the math no longer work, you see if you forget the physics involved then the whole thing would work if the ladder were to fall straight down, but since you are pulling on it at the bottom more than likely it will fall down and be pulled away from the wall, so the problem falls apart in step 3, because as it is pulled away from the wall it will no longer be a the same triangle (the hypotenuse will be extended) or its not a triangle if you count that it is no longer touching the wall therefore all the sides do not touch, but anyway back to the math, the equation for right triangles of x^2+y^2=L^2 will no longer apply thus not being able to continue on with the problem and not being able to solve it, but using physics it could actually be figured out how fast it would fall, and the answer would never be infinately, the answer would be 9.8m/s^2 here on earth unless the force being used to pull the ladder was stronger than the force of gravity on the vertical vector of the end falling (the force being split up into the horizontal force and verticle force being applied as you pulled) thus making you have to pull that much harder on the ladder. But this would also have a limit of not being able to exceed the terminal velocity. I can write a formula for you to solve the problem under all circumstances if you would like, but it wont be tonight since it is very late and I am tired.
8/19/05, 05:17 AM
#18
After all these years,
My C/T still sucks!
My C/T still sucks!
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Gawd How I despise math.
In 9th grade I decided to play with Algebra......Rigggggggggght.
So I saved for summer school, bought a calculator to prepare for my regular math class and spent the rest of the semester reading books about Jim Morrison.
In 9th grade I decided to play with Algebra......Rigggggggggght.
So I saved for summer school, bought a calculator to prepare for my regular math class and spent the rest of the semester reading books about Jim Morrison.
8/19/05, 09:58 AM
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Answer: 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.
good job bro, you got it.
good job bro, you got it.
8/19/05, 11:34 AM
#20
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Um, I can correct the spelling in the posts, and THATS as far as I can get with all this greek.
This is getting scary, talking about quadratic angles, and hypotneuses, and and that other stuff! I think I am going blind!
This is getting scary, talking about quadratic angles, and hypotneuses, and and that other stuff! I think I am going blind!