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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

To be able to regroup terms, you must first show what properity of the sequence?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Ricky wrote:

To be able to regroup terms, you must first show what properity of the sequence?

...convergence.

Bad speling makes me [sic]

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Exactly. Something I did not do. It is entirely possible to show convergence as 1/(i^2+I) < 1/(i^2), and 1/(i^2) converges. I just didn't do it.

Alright Dross, you're turn if you want it.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

I put it to you good people that a ladder, when pulled away from a wall by it's base at a constant speed, will fall infinitely fast.

Let a ladder of length L rest against a vertical wall. Let the horizontal distance from the base of the ladder to the base of the wall be given by x(t), and let the vertical distance between the base of the wall and the top of the ladder at time t be given by y(t).

Now, since the ladder, the distance x and the distance y form a right-angled triangle, we have by pythagoras that:

Now let the ladder be pulled away by it's base, such that the base moves away from the wall with a constant speed v. Differentiating the above with respect to t, we find that:

Now, as the ladder is pulled away at a constant speed, so x will approach L. Thus the numerator for the above expression for

will approach the non-zero term -Lv, whilst the denominator will approach zero. Thus will approach infinity, becoming arbitrarily large as x approaches L.*Last edited by Dross (2006-09-21 11:41:10)*

Bad speling makes me [sic]

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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

what does the dot above y and x mean?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

"Change in".

is distance, so is change in distance, or speed.As for proving you wrong, I don't suppose I'm allowed to just give a video of a ladder falling over as a counter-example?

Why did the vector cross the road?

It wanted to be normal.

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Kurre wrote:

what does the dot above y and x mean?

Sorry, I should have been more clear! Indeed Mathsyperson is right, denotes change in with respect to time, or - the speed of .

mathsyperson wrote:

As for proving you wrong, I don't suppose I'm allowed to just give a video of a ladder falling over as a counter-example?

Of course you aren't - that would just take all the fun out of it!

Bad speling makes me [sic]

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, **Dross**s problem is more to do with physics than mathematics, Id say. The mistake is in assuming that the ladder always makes a right-angled triangle with the wall and floor. This is true when the ladder is stationary, but when you pull the base away from the wall, the top of the ladder is going to be pulled away as well. Then the Pythagorean relation between

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Wow - I'd totally forgotten about this. I'm glad somebody got it!

I'd be interested in carrying on this thread, if you've got a false proof you can offer Jane?

Bad speling makes me [sic]

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well then, heres one I just made up. It should be quite easy, I hope.

*Last edited by JaneFairfax (2007-03-01 12:41:08)*

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**lightning****Real Member**- Registered: 2007-02-26
- Posts: 2,060

the hence part?????

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ill make it simpler. Ill prove instead that

Now everyone should get it. This was precisely what my first fakse proof was all about; the differentiation was redundant, intended no more than to throw you off and make the error less obvious.

*Last edited by JaneFairfax (2007-03-07 07:19:07)*

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**lightning****Real Member**- Registered: 2007-02-26
- Posts: 2,060

the -1=1 part ?????

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

JaneFairfax, lightning hasn't learned about that yet.

Cool thread by the way.

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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

I wouldn't know much about this, but I think it may have to do with the sin and cos functions being periodic and hence the x point at -1 is virtually the same as the x point at 1.

Nevermind.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Nope, thats not it. Look closely at the equations again.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It must have something to do with the square root having two possible values.

Why did the vector cross the road?

It wanted to be normal.

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**lightning****Real Member**- Registered: 2007-02-26
- Posts: 2,060

i know what a square root is just not a value or pie or x y stuff but in 3 months i'll be in first year of the high school

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Exactly **mathsyperson**!

The correct statement should be

.Now your turn to post a false proof.

*Last edited by JaneFairfax (2007-03-08 06:57:08)*

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**lightning****Real Member**- Registered: 2007-02-26
- Posts: 2,060

whats it gonna be math or math

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

card(X) means the cardinality of a set X.

Prove that if card(X) ≤ card(Y) and card(Y) ≤ card(X), then card(X) = card(Y).

Proof: By definition, since card(X) ≤ card(Y), there exists a 1-1 function f:X->Y. Similarly, there exists a 1-1 function g:Y->X. But then every element x in X maps to a unique element y in Y. Similarly every element y in Y maps to a unique element x in X. Consequently X and Y must have the same number of elements and the functions f and g are 1-1 and onto. Then by definition, card(X) = card(Y).

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oops, it looks like I missed Jane's confirmation. Which is a shame, because I have a really dandy false proof just waiting to be posted.

Unfortunately, it's going to have to wait because I can't find a flaw in Ricky's. The result of the proof is clearly correct, so it must just be that one of those steps is bending the rules of set theory.

As a random wild guess, I'll say that he needs to show that f = g[sup]-1[/sup], rather than just that two functions f and g exist.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

That is correct Jane. In general, if a mapping between two sets of the same cardinality is 1-1, then you are only guaranteed onto if their cardinality is finite.

Your turn Jane.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Woohoo, I got it right!

I dont have a new false proof now. Since it was Mathsys turn last time, he can go next.

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