I might have just disproved math.

[gritch]

You lost a -1 in using the associative property. If you have:
0 = (1 - 1) + (1 - 1)...

You end up with the same number of 1's as -1's, but when you move the parentheses you need
0 = 1 + (-1 + 1) + (-1 + 1) ... - 1

The way you did it it have a unequal number of 1's and -1's, which is an incorrect associative property use.

That was my knee jerk reaction the first time I saw this too, but the problem is you really can't have an even or odd number of infinite terms. If one thinks of infinity in this case as a series of continuously repeating (-1+1), then yes your argument works but by using mathematics this is difficult to prove. Infinite and zero are odd concepts to humans.

 

[Blade1130]

That was my knee jerk reaction the first time I saw this too, but the problem is you really can't have an even or odd number of infinite terms. If one thinks of infinity in this case as a series of continuously repeating (-1+1), then yes your argument works but by using mathematics this is difficult to prove. Infinite and zero are odd concepts to humans.

No, you're misinterpreting what I'm saying. You're doing (1 - 1) to infinity, thus you MUST have the same number of +1's & -1's, an infinite number, but the same infinite of 1's as -1's. The way you did it, you have an extra +1 at the beginning without the -1 at the end to compliment it. You may have an infinite number of (1 - 1)'s, but when you do it that way, you have an extra +1 at the beginning. That's the mistake.

You're saying:
(1 - 1) * inf = 0
which is correct, but when you pull the associative property part you're getting an extra one and changing it to:
(-1 + 1) * inf + 1 != 0

Edit: Holy crap I just realized that when you substitute (1 - 1) with 0, you get 0 * inf = 0 but I just said 2 posts back:

0 * inf =~ 0.00000000001 * 999999999999 =~ 1 (Therefore 0 * inf = 1)

Either I just broke math too, or indeterminites really are indeterminite.

 

[Simple Bluff]

But I've got my own mathematical paradox for you guys. It has been disproven but I can't quite recall how.

0=0, this is correct.
0+0=0 is correct.
Therefore, I can add infinite amounts of zeroes together and the result will equal 0
0=0+0+0+0+... +0n
1-1=0 is correct.
Therefore I can add infinite amounts of (1-1) together and the result will equal 0
0=(1-1)+(1-1)+...+(1-1)n
Since this is addition, I can use the associate property and rearrange the parentheses as thus:
0=1+(-1+1)+(-1+1)+...+(-1+1)n
Since -1+1=0 and this series proceeds to infinity all (-1+1) terms can be reduced to 0, yielding the result:

0=1

That's very clever. Unfortunatly though, if you ever study Analysis, you'll find a find a simple (if, frankly, anticlimactic) way of disproving that logic - periodic sums like the one above are always "divergent" (ie, does not tend to a single, finite, value). Therefore it has no limit, which by extension means that it doesn't "behave" well. You can't simply factor out one of the ones, the same way that you cannot simply divide by zero, because operations like that are not defined in this context.
Bear in mind I don't have a degree in Maths, or anything else for that matter, so my explanation may not be entirely accurate. And even if it is, it's probobly not the best way of explaining it.

Anyway, here's another fun paradox, disproved by the same logic:

Let S = 1 + 2 + 4 + 8 + 16 +... (etc.)

Take 1 from both sides: (S - 1) = 2 + 4 + 8 + 16 +...

Factor out the 2 on the RHS: (S - 1) = 2(1 + 2 + 4 + 8 +...)

But recall S = 1 + 2 + 4 + 8 + 16..., so we have

(S - 1) = 2S

Rearrange to get 2S - S = -1

S = -1

But S = 1 + 2 + 4 + 8 + 16 +...

So 1 + 2 + 4 + 8 + 16 +... = -1 (!!!)

WHAT HAS SCIENCE DONE

 

[nuba km]

So, I have an interesting math based question. If you don't like/hate math or don't understand basic algebra(I understand if you don't) just hit the big THE ESCAPIST logo in the corner and that will bring you home.

I wanted to know zero divided by zero equals. I tried to do at algebraically. This is what I did:

-The answer I was trying to get will be represented by x
0/0=x
-I times both sides by zero
0=0x
-This equals out to be 0=0 because anything times 0 is 0.
-This proves that x can be any number. for example if 5=x than 0=5*0 still is 0=0
-I rearrange 0=0x to be:
0/x=0
-Now since x can be any number now lets say x=0
-That makes this:
0/0=0
-And since x=0/0 (Right in the beginning^) and 0=0/0 also then x=0
-If you fallowed so far and remember that x can be any number then that means zero can also be any and every number. So 0 can now equal 5 or any other number.

I realize something is most likely wrong here.
So tell me escapist, Did i Disprove math?
Edit: I see the error now. Its not that x equals 0 its that at one point x CAN = 0

you are wrong because 1 simple reason, you can't divide by zero, not even zero, even if you could that would give you the number 1 meaning x is equal to 1, no matter how you rearrange the equation the number 1 would be a solution an almost always the only solution.
also in your equation x can equal anything for one possible form of the equation. you rearranged x so it was only able to be equal to 0, x does not have a range of values (as there is no sign, and even if you used those signs the only possible values for x would be 0 respectively) it's ONLY possible value for your equation to work is 0.

also using algebra fro the square root of negative numbers is only there to represent an idea not a number like any 0.XXXX... repeating endlessly is more of a representation of an idea then a number, so it won't add up if used in a context out side of its meant usage and sometimes even within its usage as the reason we use the representation of an idea is because you can't actually use real numbers for it in the first place.

 

[Lukeje]

But I've got my own mathematical paradox for you guys. It has been disproven but I can't quite recall how.

0=0, this is correct.
0+0=0 is correct.
Therefore, I can add infinite amounts of zeroes together and the result will equal 0
0=0+0+0+0+... +0n
1-1=0 is correct.
Therefore I can add infinite amounts of (1-1) together and the result will equal 0
0=(1-1)+(1-1)+...+(1-1)n
Since this is addition, I can use the associate property and rearrange the parentheses as thus:
0=1+(-1+1)+(-1+1)+...+(-1+1)n
Since -1+1=0 and this series proceeds to infinity all (-1+1) terms can be reduced to 0, yielding the result:

0=1

Here, have a wikipedia article. It explains why the sum you're referring to changes its limit with different bracketing.

http://en.wikipedia.org/wiki/Grandi%27s_series

 

[Meshakhad_v1legacy]

Hmm, so if any number can equal zero, and if there's one of you, then that means you don't exist!

 

[gritch]

No, you're misinterpreting what I'm saying. You're doing (1 - 1) to infinity, thus you MUST have the same number of +1's & -1's, an infinite number, but the same infinite of 1's as -1's. The way you did it, you have an extra +1 at the beginning without the -1 at the end to compliment it. You may have an infinite number of (1 - 1)'s, but when you do it that way, you have an extra +1 at the beginning. That's the mistake.

You're saying:
(1 - 1) * inf = 0
which is correct, but when you pull the associative property part you're getting an extra one and changing it to:
(-1 + 1) * inf + 1 != 0

Ya. That's a pretty satisfactory explanation. I'd love to continue playing the devil's advocate here - we could argue about what exactly infinite is but I don't think it'll really get us anywhere useful.

How about another riddle?

Let's say we have a tortoise and a hare in a race. The hare can move twice the distance the tortoise can in half the time.

Let us suppose the tortoise has a head start of distance D. It takes the hare a time t to travel this distance. But in this time t the tortoise has traveled a distance of D/2+D. The hare has not caught up to the tortoise yet.

So then it takes the hare a time t/2 to travel the new distance D/2. In this time the tortoise has traveled a distance of, D/4+D/2+D now. The hare has not caught the tortoise.

This can continue on indefinitely.

The question is, will the hare ever catch up to the tortoise?

 

[gritch]

That's very clever. Unfortunatly though, if you ever study Analysis, you'll find a find a simple (if, frankly, anticlimactic) way of disproving that logic - periodic sums like the one above are always "divergent" (ie, does not tend to a single, finite, value). Therefore it has no limit, which by extension means that it doesn't "behave" well. You can't simply factor out one of the ones, the same way that you cannot simply divide by zero, because operations like that are not defined in this context.
Bear in mind I don't have a degree in Maths, or anything else for that matter, so my explanation may not be entirely accurate. And even if it is, it's probobly not the best way of explaining it.

Anyway, here's another fun paradox, disproved by the same logic:

Let S = 1 + 2 + 4 + 8 + 16 +... (etc.)

Take 1 from both sides: (S - 1) = 2 + 4 + 8 + 16 +...

Factor out the 2 on the RHS: (S - 1) = 2(1 + 2 + 4 + 8 +...)

But recall S = 1 + 2 + 4 + 8 + 16..., so we have

(S - 1) = 2S

Rearrange to get 2S - S = -1

S = -1

But S = 1 + 2 + 4 + 8 + 16 +...

So 1 + 2 + 4 + 8 + 16 +... = -1 (!!!)

WHAT HAS SCIENCE DONE

Indeed I have studied Analysis. That's where I first came across this problem. I do remember disproving it rather simply but series where never really my forte so to speak. Just an interesting example of using vague mathematical concepts to incorrectly "prove" outlandish concepts.

 

[Il_Exile_lI]

I think you lost all mathematical credibility as soon as you used "times" as a verb.

I think he's doing something perfectly normal, at least in America. Many people say "You times/timesd these numbers". It gets on my nerves, but from what I've seen, they don't realize it's "multiplied".

OT: Good points. I think you did make some errors, though. But I'm not a math teacher.

Since when has "well other dumb people also say this thing wrong" ever been a valid excuse for stupidity and poor English. When I was in high school, I too heard many, let's call them "lesser students", use the word "times" as a verb, but common usage doesn't make it any less wrong or idiotic.

 

[Blade1130]

How about another riddle?

Let's say we have a tortoise and a hare in a race. The hare can move twice the distance the tortoise can in half the time.

Let us suppose the tortoise has a head start of distance D. It takes the hare a time t to travel this distance. But in this time t the tortoise has traveled a distance of D/2+D. The hare has not caught up to the tortoise yet.

So then it takes the hare a time t/2 to travel the new distance D/2. In this time the tortoise has traveled a distance of, D/4+D/2+D now. The hare has not caught the tortoise.

This can continue on indefinitely.

The question is, will the hare ever catch up to the tortoise?

I accept this challenge, largely because I have nothing better to do at the moment.

Here I don't think you made any math errors, but it appears that the hare will never catch up to the tortoise on the surface, when he really will. I would take more of a Physics problem approach. (Assuming constant velocity for both animals)

If the tortoise moves a distance D / 2 in time t, with an initial position of D then it has a position equation of
s[sub]T[/sub](t) = (D / 2)t + D

If the hare moves a distance D in time t then it has a position equation of
s[sub]H[/sub](t) = Dt

Setting these two equations equal to each other gets you
s[sub]T[/sub](t) = s[sub]H[/sub](t)
(D / 2)t + D = Dt
D = t(D - (D / 2))
D = t(D / 2)
D / (D / 2) = t
t = 2?

That would be my guess.

As for a retaliatory riddle hmm... I recall an old word problem I got a long time ago.
3 people split a room in a hotel. It is $30 for a room, so they each pay $10. The manager feels that he has ripped them off, so he has a bell boy give them back $5. The men then don?t know how to split it so they tip the bell boy $2, then split the $3 among the 3 of them. So each man paid $9 which makes $27 + the $2 they gave to the bell boy makes $29... What happened to the extra dollar?

 

[TheNaut131]

-I times both sides by zero

Please don't ever say that again. Please. Ever. Please.

Spoiler: Ahem

THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS THIS

I may not be the best mathematician, but Mother of God, I hate that phrase.

Times? No. You don't say times, because it's not really a verb and it sounds extremely stupid. I mean really, just say multiply! It's not that hard and it sounds a lot better than times. Seriously, how can you say "I times-" without hearing how awkward it sounds?

 

[Donnie Restad]

So, I have an interesting math based question. If you don't like/hate math or don't understand basic algebra(I understand if you don't) just hit the big THE ESCAPIST logo in the corner and that will bring you home.

I wanted to know zero divided by zero equals. I tried to do at algebraically. This is what I did:

-The answer I was trying to get will be represented by x
0/0=x
-I times both sides by zero
0=0x
-This equals out to be 0=0 because anything times 0 is 0.
-This proves that x can be any number. for example if 5=x than 0=5*0 still is 0=0
-I rearrange 0=0x to be:
0/x=0
-Now since x can be any number now lets say x=0
-That makes this:
0/0=0
-And since x=0/0 (Right in the beginning^) and 0=0/0 also then x=0
-If you fallowed so far and remember that x can be any number then that means zero can also be any and every number. So 0 can now equal 5 or any other number.

I realize something is most likely wrong here.
So tell me escapist, Did i Disprove math?
Edit: I see the error now. Its not that x equals 0 its that at one point x CAN = 0

MAN, some people can be rude about math. Let me try to explain.

Your edit is correct. By your initial statement, X can and does equal absolutely everything.

Think of it like this: Zero divided by anything is zero, correct?
And anything divided by zero (which can and does happen) is either infinity or negative infinity.
And anything divided by itself is 1. ie, 7/7=1, 5/5=1, you get the picture.

So here's the problem. 0/0 is something divided by zero, it's zero divided by something, AND it's something divided by itself. So, essentially, it's infinity, negative infinity, zero, and one, all at the same time.

Calculus does this all the time. In its simplest form, the limit definition of a derivative is nothing more than zero divided by zero.

 

[Cid Silverwing]

You can't divide by zero.

Why not?

If you can't understand this, I'll be shocked if you're not failing your math classes.

Dividing shit by zero ends the equation in a zero. You're essentially hitting Ctrl+A+Backspace. You can't divide something by nothing, same way you can't add, subtract or multiply by nothing, although adding and subtracting won't end in a zero, multiplying will.

Otherwise, go back to basic math.

 

[SmegInThePants]

my take, coming from little mathematics in my background:

100 divided by 25, i.e. how many multiples of 25 can be made to fit in to 100 = 4

100 divided by zero, i.e. how many multiples of nothing_at_all can be made to fit in to 100? even infinity wouldn't be enough.

numbers normally represent iterations of a thing, there are 3 apples, 5 oranges, 100 corrupt senators. Zero represents a hole, a lack of iterations, a lack of a number.

and thus to divide by zero is fruitless.

 

[Maze1125]

which is correct, but when you pull the associative property part you're getting an extra one and changing it to:
(-1 + 1) * inf + 1 != 0

There's no such thing as an "extra 1" when it comes to infinity. There are an infinite number of 1s and an infinite number of -1s.

infinity + 1 = infinity
There's absolutely nothing wrong with him producing that extra 1, infinity is such that it allows that.

If you haven't already, see Lukeje's link for more details.

 

[targren]

Zero is a callous *****. She doesn't adhere to the same rules as normal numbers do.

I read this and I laughed at the idea of personifying numbers.

"1 is a little slut. She'll multiply with anyone, even zero or her own negative square root, but still never ends up with anything of her own identity afterwards."

"2 is an irritating little hipster. He was prime before it was cool and he's STILL the only even number to bear that distinction, and all the other even wannabes answer to him."

Sorry, it's a whimsical evening...

 

[rabidmidget]

Needs work.

See me after class.

 

[Blade1130]

infinity + 1 = infinity
There's absolutely nothing wrong with him producing that extra 1, infinity is such that it allows that.

Not true, infinity + 1 does in fact equal infinity, however the second infinity is greater than the infinity used first.

To use overly complicated mathematics to prove something conceptually quite simple.

The definition of L'Hopital's Rule is:
(where f(x) & g(x) are continuous, differentiable functions and f'(x) is the derivative of f(x) and g'(x) is the derivative of g(x))

If Lim[sub]x->a[/sub](f(x) / g(x)) = 0 / 0 || inf / inf then
Lim[sub]x->a[/sub](f(x) / g(x)) = Lim[sub]x->a[/sub](f'(x) / g'(x))

We can use this rule to prove how one function grows greater than another and thereby how one infinity is greater than another. Consider the example:

If f(x) = e[sup]x[/sup] and g(x) = x

Lim[sub]x->inf[/sub](e[sup]x[/sup] / x) = e[sup]inf[/sup] / inf = inf / inf (therefore L'Hopital's Rule applies, thus we derive the top and bottom and find the limit)

Lim[sub]x->inf[/sub](e[sup]x[/sup] / 1) = e[sup]inf[/sup] / 1 = e[sup]inf[/sup] = inf

The answer of infinity tells us that f(x) = e[sup]x[/sup] grows faster than g(x) = x, which can be seen quite obviously by graphing both equations. But if we simply plug in infinity to both equations we get:
f(inf) = e[sup]inf[/sup] = inf AND g(inf) = inf

By this logic f(inf) = g(inf), but as L'Hopital's Rule proves, f(x) grows faster than g(x) and (after checking initial values) there is no point x on the interval (-inf, inf) in which g(x) >= f(x). Therefore f(x) != g(x) on any x. As such infinity does not equal infinity.

Infinity can equal infinity, but it must be the same infinity. The infinity that is the result of f(x) is greater than the infinity created by g(x). Although f(inf) does equal infinity, it is a different infinity than the one in g(inf).

By the way there is a small error in there, I'm curious if anyone can find it and call me out on it. I'm at a loss of how to explain it without that error, but lets see if anyone finds it to begin with.

EDIT: To go back to the original example he is saying (1 - 1) * inf = inf - inf = 1 which is not true.
(1 - 1) * inf = inf - inf = 0 as it should. Think about it in a way that does not use infinities.

(1 - 1) + (1 - 1) + (1 - 1) = 0 = true

Shifting all parentheses to the right

1 + (-1 + 1) + (-1 + 1) + (-1) = 0 = still true

He is forgetting that last -1, the fact that it is (1 - 1) * inf, means that EVEN TO INFINITY, there are the same number of 1's as -1's and therefore equals 0.

 

[Cerdog]

infinity + 1 = infinity
There's absolutely nothing wrong with him producing that extra 1, infinity is such that it allows that.

Not true, infinity + 1 does in fact equal infinity, however the second infinity is greater than the infinity used first.

That's not quite correct. Infinity isn't a number, so it's quite hard to use it in a situation like this, but the size of an infinity depends on the cardinality. For example, the natural numbers (0, 1, 2, ...), the integers (... -2, -1, 0, 1, 2, ...) and the rational numbers (fractions) are all the same "level" of infinity; as weird as it sounds, they are all the same size. On the other hand, the real numbers are larger.

 

[gritch]

I accept this challenge, largely because I have nothing better to do at the moment.

Here I don't think you made any math errors, but it appears that the hare will never catch up to the tortoise on the surface, when he really will. I would take more of a Physics problem approach. (Assuming constant velocity for both animals)

If the tortoise moves a distance D / 2 in time t, with an initial position of D then it has a position equation of
s[sub]T[/sub](t) = (D / 2)t + D

If the hare moves a distance D in time t then it has a position equation of
s[sub]H[/sub](t) = Dt

Setting these two equations equal to each other gets you
s[sub]T[/sub](t) = s[sub]H[/sub](t)
(D / 2)t + D = Dt
D = t(D - (D / 2))
D = t(D / 2)
D / (D / 2) = t
t = 2?

That would be my guess.

As for a retaliatory riddle hmm... I recall an old word problem I got a long time ago.
3 people split a room in a hotel. It is $30 for a room, so they each pay $10. The manager feels that he has ripped them off, so he has a bell boy give them back $5. The men then don?t know how to split it so they tip the bell boy $2, then split the $3 among the 3 of them. So each man paid $9 which makes $27 + the $2 they gave to the bell boy makes $29... What happened to the extra dollar?

... I'm really not good at riddles apparently. I can't figure your's (or even my own out). I will assume your calculations were correct for my riddle but we still have a issue. Logically, the way the problem is stated one one theorize that the hare and tortoise never meet up but physics as well as common sense tells us otherwise.

One can use the concept of the Harmonic Series to prove this but I can't for the life of me find it in my damn Calculus notes.

I'm sorry, I can't find a way to solve your riddle.

Note: Pardon the delay, had to feed myself.