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# If that sounds like it makes no sense that's because... well, it doesn't

Correct, other than being a totally different problem with the exact opposite outcome.

posted by DU at 6:55 AM on July 2, 2013 [3 favorites]

Addition is associative so there's no reason to.

posted by edd at 7:05 AM on July 2, 2013

I already melted my own brain just to spite you.

posted by louche mustachio at 7:27 AM on July 2, 2013 [2 favorites]

Is "on" defined as "when photons are escaping the bulb"?

posted by 3FLryan at 7:49 AM on July 2, 2013

On the contrary: the two problems are difficult for the same reason, the reason that Hardy identified, which is that an infinite sum does not have a value until we give it one. We have choices about how to define infinite sums, and sometimes the best choice violates one or another of our intuitions. (Other times there is room for disagreement about the best choice, or there is a case to be made that there is no good choice.)

posted by escabeche at 8:01 AM on July 2, 2013 [8 favorites]

Add a slash to each equals sign, making them "not equals".

posted by Pararrayos at 9:51 AM on July 2, 2013 [2 favorites]

There's a better (but rather more demanding) answer than that.

posted by Paul Slade at 9:56 AM on July 2, 2013

I am getting really annoyed by articles telling me how I'm going to feel about them in the headline. It's like they've absorbed all the tropes of a certain type of spam in order to present their click-bait. "These New Man Tricks Will Melt Her Brain" "You Will Not Believe This Extraordinary Sex Enhancer" "The Most XXXXX Girls You Will See Tonight" "This Could Be The Most Amazing Pleasure Tip Ever".

posted by oneirodynia at 10:38 AM on July 2, 2013 [4 favorites]

If you wanted that to work wouldn't you have to write down some numbers as the limits of integration? Otherwise you'd end up with algebraic functions (which admittedly are equal to any number for for different values of the variable).

posted by Ned G at 12:34 PM on July 2, 2013

This is more what I meant, expressed more clearly. It seems like the differences he introduces between the three formulations of the 'same' problem are actually pretty significant, and it's not really the same equation three times at all. Therefore it's not brain melting when you get three different answers, especially when you stretch that out over infinity (creating an asymptotic graph, I guess?).

But, once again, I'm not super math literate.

posted by codacorolla at 12:41 PM on July 2, 2013

my brain keeps trying to visualize this make it stop

posted by Elementary Penguin at 2:35 PM on July 2, 2013 [1 favorite]

NOW MOVE THREE MATCHSTICKS TO MAKE IT EQUAL 12

posted by Joe in Australia at 7:15 PM on July 2, 2013

If one hasn't taken pre-calc or calculus, it's interesting because it makes you consider what an infinite series actually is, and what it means to sum one, and it's a simple example of a series that doesn't converge.

posted by empath at 12:17 AM on July 3, 2013

If you subtract infinity from infinity you still have infinity. (Subtract the odd numbers from the natural numbers, for example).

posted by empath at 12:18 AM on July 3, 2013

Except

posted by Eideteker at 7:17 AM on July 3, 2013

posted by dfan at 10:43 AM on July 3, 2013

Post

# If that sounds like it makes no sense that's because... well, it doesn't

July 2, 2013 6:38 AM Subscribe

This Simple Math Puzzle Will Melt Your Brain

"Adding and subtracting ones sounds simple, right? Not according to the old Italian mathematician Grandi—who showed that a simple addition of 1s and -1s can give three different answers."

"Adding and subtracting ones sounds simple, right? Not according to the old Italian mathematician Grandi—who showed that a simple addition of 1s and -1s can give three different answers."

Wikipedia, as always, has some nice relevant information.

posted by notpace at 6:48 AM on July 2, 2013

posted by notpace at 6:48 AM on July 2, 2013

Short version: Non-absolutely convergent series don't converge absolutely.

Long version:

s = 1- 1 + 1 - 1 + 1 - 1 + 1 … is either:

Long version:

s = 1- 1 + 1 - 1 + 1 - 1 + 1 … is either:

- (1 - 1) + (1-1) + (1-1) + … = 0
- 1 + (-1 +1 ) + (-1 + 1) + … = 1
- 1 - (1 - 1 + 1 - 1 + 1 - 1 + …) = 1 - s, so s = 1/2

See also the suggestion 1+2+3+4... = -1/12. Obviously.

posted by edd at 6:53 AM on July 2, 2013 [2 favorites]

posted by edd at 6:53 AM on July 2, 2013 [2 favorites]

Take the series 1,0,1,0,1,...

If you look at every odd position in the series, it's a 1! But if you look at every even position, it's a 0!

Mind blown? Not so much.

posted by CaseyB at 6:54 AM on July 2, 2013 [9 favorites]

If you look at every odd position in the series, it's a 1! But if you look at every even position, it's a 0!

Mind blown? Not so much.

posted by CaseyB at 6:54 AM on July 2, 2013 [9 favorites]

*Reminds me of the old 0.9999999... = 1 arguments on sci.math.*

Correct, other than being a totally different problem with the exact opposite outcome.

posted by DU at 6:55 AM on July 2, 2013 [3 favorites]

It's just an infinite series of 1, 0. Any other definition is ridiculous.

posted by sonic meat machine at 7:01 AM on July 2, 2013

posted by sonic meat machine at 7:01 AM on July 2, 2013

I don't usually assign a definite value to this series, but when I do, I use the Cesàro summation (which in this case is 1/2). Stay thirsty, my friends.

posted by Elementary Penguin at 7:02 AM on July 2, 2013 [26 favorites]

posted by Elementary Penguin at 7:02 AM on July 2, 2013 [26 favorites]

As I as watching I thought 'wait, isn't there an order of operations for whether you subtract or add first?', and it turns out there isn't. Is there a reason that no distinction is made there?

Also, I don't get why, in the second formulation of the problem, he just added in plus signs seemingly out of nowhere.

posted by codacorolla at 7:04 AM on July 2, 2013 [1 favorite]

Also, I don't get why, in the second formulation of the problem, he just added in plus signs seemingly out of nowhere.

posted by codacorolla at 7:04 AM on July 2, 2013 [1 favorite]

codacorolla, think of subtraction as adding inverses.

So 1 - 1 + 1 - 1 + … is 1 + (-1) + 1 + (-1) + …. I don't know why he didn't do that in the video, since it would make that step a lot clearer.

posted by Elementary Penguin at 7:05 AM on July 2, 2013

So 1 - 1 + 1 - 1 + … is 1 + (-1) + 1 + (-1) + …. I don't know why he didn't do that in the video, since it would make that step a lot clearer.

posted by Elementary Penguin at 7:05 AM on July 2, 2013

*Is there a reason that no distinction is made there?*

Addition is associative so there's no reason to.

posted by edd at 7:05 AM on July 2, 2013

codacorolla the "extra" plus sign is just like this one:

a-b = a+ (-b)

posted by koolkat at 7:06 AM on July 2, 2013 [2 favorites]

a-b = a+ (-b)

posted by koolkat at 7:06 AM on July 2, 2013 [2 favorites]

Well I suppose I'm a bit wrong because subtraction isn't associative, but as Elementary Penguin makes clear you can think of it as addition of negative numbers, and then you just conventionally do it from left to right with equal precedence.

posted by edd at 7:08 AM on July 2, 2013

posted by edd at 7:08 AM on July 2, 2013

Oh, I see. I thought it was probably something like that, but I haven't taken a math course in 8 years, so I was a little rusty.

posted by codacorolla at 7:12 AM on July 2, 2013

posted by codacorolla at 7:12 AM on July 2, 2013

*This Simple Math Puzzle Will Melt Your Brain*

I already melted my own brain just to spite you.

posted by louche mustachio at 7:27 AM on July 2, 2013 [2 favorites]

Case 2 just truncates the series early with an extra +1. If that's fair then there's an obvious Case 4 where you start with a -1: -1 + (-1 + 1) + (-1 +1 ) + ... = -1

posted by achrise at 7:27 AM on July 2, 2013

posted by achrise at 7:27 AM on July 2, 2013

To his "mind-bending" question at the end ... 1/2 isn't an option, is it? The whole rest of the video supposes all real numbers are in play, but a light bulb is binary. So, it's on or off, depending on where you "finished" your infinite sequence.

posted by 3FLryan at 7:31 AM on July 2, 2013

posted by 3FLryan at 7:31 AM on July 2, 2013

Theorem: Every horse has an infinite number of legs

Horses have an even number of legs. Behind they have two legs, and in front they have fore legs. This makes six legs, which is certainly an odd number of legs for a horse. The only number that is both odd and even is infinity. Therefore, horses have an infinite number of legs.

posted by 7segment at 7:32 AM on July 2, 2013 [32 favorites]

Horses have an even number of legs. Behind they have two legs, and in front they have fore legs. This makes six legs, which is certainly an odd number of legs for a horse. The only number that is both odd and even is infinity. Therefore, horses have an infinite number of legs.

posted by 7segment at 7:32 AM on July 2, 2013 [32 favorites]

I thought that the addition of little + signs was a bit of a slip-up in an otherwise careful demonstration. You do have to be careful with where you add parentheses since -a+b is not the same as -(a+b).

As to the light-bulb question, it is going on and off infinitely fast at 2 minutes. What does that mean? For a real light-bulb, it means its always off since it is turned back off before a single photon has time to escape. Otherwise, the light-bulb is in a superposition state - both on and off. |1>+|0>. So the best answer is 1/2 again.

posted by vacapinta at 7:36 AM on July 2, 2013 [2 favorites]

As to the light-bulb question, it is going on and off infinitely fast at 2 minutes. What does that mean? For a real light-bulb, it means its always off since it is turned back off before a single photon has time to escape. Otherwise, the light-bulb is in a superposition state - both on and off. |1>+|0>. So the best answer is 1/2 again.

posted by vacapinta at 7:36 AM on July 2, 2013 [2 favorites]

Yes, this is all well and good, but how do I put it into a pivot table?

posted by digitalprimate at 7:45 AM on July 2, 2013 [3 favorites]

posted by digitalprimate at 7:45 AM on July 2, 2013 [3 favorites]

*For a real light-bulb, it means its always off since it is turned back off before a single photon has time to escape.*

Is "on" defined as "when photons are escaping the bulb"?

posted by 3FLryan at 7:49 AM on July 2, 2013

“Whether finite or infinite, a number is even when it can be divided into pairs. For finite sets, this is the same as the ability to divide the set into two sets of equal size, since one may consider the first element of each pair and the second element of each pair... The smallest infinity ω is even, ω+1 is odd, ω+2 is even and so on. Every limit ordinal is even, and then it repeats even/odd up to the next limit ordinal.” — http://math.stackexchange.com/a/49046

Looking at the series, it's apparent that after any odd number of terms, the answer is 1 (and the light is on), and after any even number of terms, the answer is 0 (and the light is off).

In the “completed” set proposed in the paradox example, the infinite states of the light can be divided into two equal real world sets of states, the set where the light is on, and the set where the light is off, indicating that this particular real world infinity ω is even and therefore at the completion of states making up these two infinite sets, the number of terms is even and the light is off.

posted by Skeuomorph at 7:57 AM on July 2, 2013

Looking at the series, it's apparent that after any odd number of terms, the answer is 1 (and the light is on), and after any even number of terms, the answer is 0 (and the light is off).

In the “completed” set proposed in the paradox example, the infinite states of the light can be divided into two equal real world sets of states, the set where the light is on, and the set where the light is off, indicating that this particular real world infinity ω is even and therefore at the completion of states making up these two infinite sets, the number of terms is even and the light is off.

posted by Skeuomorph at 7:57 AM on July 2, 2013

The great British number theorist G.H. Hardy covered this in his book

"…it does not occur to a modern mathematician that a collection of mathematical symbols should have a ‘meaning’ until one has been assigned to it by definition. It was not a triviality even to the greatest mathematicians of the eighteenth century. They had not the habit of definition: it was not natural to them to say, in so many words, `by X we mean Y.’ … it is broadly true to say that mathematicians before Cauchy asked not ‘How shall we define

1 – 1 + 1 – 1 + …

but

‘What is 1 – 1 + 1 – 1 + …

and that this habit of mind led them into unnecessary perplexities and controversies which were often really verbal.”

posted by escabeche at 7:57 AM on July 2, 2013 [16 favorites]

*Divergent Series*:"…it does not occur to a modern mathematician that a collection of mathematical symbols should have a ‘meaning’ until one has been assigned to it by definition. It was not a triviality even to the greatest mathematicians of the eighteenth century. They had not the habit of definition: it was not natural to them to say, in so many words, `by X we mean Y.’ … it is broadly true to say that mathematicians before Cauchy asked not ‘How shall we define

1 – 1 + 1 – 1 + …

but

‘What is 1 – 1 + 1 – 1 + …

and that this habit of mind led them into unnecessary perplexities and controversies which were often really verbal.”

posted by escabeche at 7:57 AM on July 2, 2013 [16 favorites]

*Reminds me of the old 0.9999999... = 1 arguments on sci.math.*

Correct, other than being a totally different problem with the exact opposite outcome.

Correct, other than being a totally different problem with the exact opposite outcome.

On the contrary: the two problems are difficult for the same reason, the reason that Hardy identified, which is that an infinite sum does not have a value until we give it one. We have choices about how to define infinite sums, and sometimes the best choice violates one or another of our intuitions. (Other times there is room for disagreement about the best choice, or there is a case to be made that there is no good choice.)

posted by escabeche at 8:01 AM on July 2, 2013 [8 favorites]

I share the lightbulb conundrum with my Calculus classes at the end of the semester, as a lead up to Calc 2. It's a good way to see who is paying attention (the ones who are bothered by it all) and the ones who aren't (those who just keep taking notes).

posted by monkeymadness at 8:05 AM on July 2, 2013

posted by monkeymadness at 8:05 AM on July 2, 2013

It seems fair to me to suppose that if a binary switch has somehow been flicked an infinite number of times in a finite period, the state it's left in will be indeterminable or random.

posted by wobh at 8:07 AM on July 2, 2013 [1 favorite]

posted by wobh at 8:07 AM on July 2, 2013 [1 favorite]

Even better, 1+0-1+1+0-1+...=2/3.

Put that in your pipe and smoke it.

(This is why absolutely convergent series are your friend.)

posted by Hactar at 8:36 AM on July 2, 2013

Put that in your pipe and smoke it.

(This is why absolutely convergent series are your friend.)

posted by Hactar at 8:36 AM on July 2, 2013

Well, you can't take 3 from 2; 2 is less than 3, so you look at the 4 in the tens' place. Now, that's really 4 tens, so you make it 3 tens; regroup; and you change a ten to ten 1s, and you add it to the 2 and get 12, and you take away 3, that's 9.

Is that clear?

posted by grubi at 8:42 AM on July 2, 2013 [4 favorites]

Is that clear?

posted by grubi at 8:42 AM on July 2, 2013 [4 favorites]

My favorite is

1-1/2+1/3-1/4+1/5…

which can equal any real number if you rearrange the terms (sadly this one is actually convergent to The natural log of two, so not as ridiculous).

posted by Elementary Penguin at 8:45 AM on July 2, 2013

1-1/2+1/3-1/4+1/5…

which can equal any real number if you rearrange the terms (sadly this one is actually convergent to The natural log of two, so not as ridiculous).

posted by Elementary Penguin at 8:45 AM on July 2, 2013

If a light bulb has been cycled on and off an infinite number of times in two minutes, it will be off at the end of the two minutes, as it will have burned out.

Regardless of whether Schroedinger's cat was observing the light bulb or not.

posted by jefflowrey at 8:52 AM on July 2, 2013

Regardless of whether Schroedinger's cat was observing the light bulb or not.

posted by jefflowrey at 8:52 AM on July 2, 2013

... which is that an infinite sum does not have a value until we give it one.

Wait, Formalism? Today's Tuesday, isn't it? Didn't you get the schedule?

M, T, W, F - Platonist

Th - Intuitionist

Formalism is reserved for Sunday, when the world is watching closely.

(Saturday is Ultimate and Beer.)

posted by benito.strauss at 8:57 AM on July 2, 2013 [5 favorites]

Wait, Formalism? Today's Tuesday, isn't it? Didn't you get the schedule?

M, T, W, F - Platonist

Th - Intuitionist

Formalism is reserved for Sunday, when the world is watching closely.

(Saturday is Ultimate and Beer.)

posted by benito.strauss at 8:57 AM on July 2, 2013 [5 favorites]

Is this mind-blowing in the sense that there's no real rigor in the definition of the question, just a bunch of manipulations and appeals to intuition? Because that's the kind of thing that would get my blood boiling as an undergraduate. It's a wonder I managed to not bludgeon either the half-competent instructors who would spring such "puzzles" on the audience - at the limits of their mathematical depth already - or the trained puppies in the front row yelping for attention and begging to give their already memorized answer.

Yes, I was very popular as an undergraduate, why do you ask?

posted by Dr Dracator at 8:58 AM on July 2, 2013 [2 favorites]

Yes, I was very popular as an undergraduate, why do you ask?

posted by Dr Dracator at 8:58 AM on July 2, 2013 [2 favorites]

Well, 1-1 is really just shorthand for 1+ -1 but that kind of makes the whole thing fall apart doesn't it.

I suppose I could also argue that by adding brackets, he has created a different series. In other words this:

(1 - 1) + (1-1) + (1-1) + … = 0

1 + (-1 +1 ) + (-1 + 1) + … = 1

1 - (1 - 1 + 1 - 1 + 1 - 1 + …) = 1 - s, so s = 1/2

Is really three different equations with three difference answers.

posted by VTX at 9:21 AM on July 2, 2013

I suppose I could also argue that by adding brackets, he has created a different series. In other words this:

(1 - 1) + (1-1) + (1-1) + … = 0

1 + (-1 +1 ) + (-1 + 1) + … = 1

1 - (1 - 1 + 1 - 1 + 1 - 1 + …) = 1 - s, so s = 1/2

Is really three different equations with three difference answers.

posted by VTX at 9:21 AM on July 2, 2013

Okay. Horses then. They have a diagonal gait, which means they move an odd leg and an even leg at the same time, then vice versa, which cancels out, so, logically they don't move at all.

Unless they are pacers or camels, where they move left side legs together, then right side legs, which is odd.

I can tell you from personal experience that camels are odd. I'm pretty sure that's an axiom.

posted by mule98J at 9:23 AM on July 2, 2013

Unless they are pacers or camels, where they move left side legs together, then right side legs, which is odd.

I can tell you from personal experience that camels are odd. I'm pretty sure that's an axiom.

posted by mule98J at 9:23 AM on July 2, 2013

It makes perfect sense that a non converging series doesn't have a single value. It's almost the definition of it.

The 0.5 value also makes sense. It's just the closest thing the series has to a limit.

As far as the light on/off question goes, I'd answer it as an EE this way. The voltage applied to the light is a square wave with a duty cycle of 50%. So the light is "half on". Practically, the illumination of the light would be somewhere between 0 to around 50% depending on type of light. For a perfectly linear light source (perfect LED, maybe), the illumition would be 1/2 that of the 100% on light.

Real lights are actually controlled this way. This is how dimmers work. It is called Pulse Width Modulation. Controlling voltage is somewhat difficult and inneficient. But turning a voltage on and off is easy and efficient. DC motors can also be controlled this way.

posted by jclarkin at 9:44 AM on July 2, 2013 [1 favorite]

The 0.5 value also makes sense. It's just the closest thing the series has to a limit.

As far as the light on/off question goes, I'd answer it as an EE this way. The voltage applied to the light is a square wave with a duty cycle of 50%. So the light is "half on". Practically, the illumination of the light would be somewhere between 0 to around 50% depending on type of light. For a perfectly linear light source (perfect LED, maybe), the illumition would be 1/2 that of the 100% on light.

Real lights are actually controlled this way. This is how dimmers work. It is called Pulse Width Modulation. Controlling voltage is somewhat difficult and inneficient. But turning a voltage on and off is easy and efficient. DC motors can also be controlled this way.

posted by jclarkin at 9:44 AM on July 2, 2013 [1 favorite]

*German maths problem*

Add a slash to each equals sign, making them "not equals".

posted by Pararrayos at 9:51 AM on July 2, 2013 [2 favorites]

The idea that one plausible answer is ½ makes this otherwise stupid arithmetic problem sort of interesting. At least it's better than those dumb 6÷2(1+2)= ? memetic viruses that keep showing up on Facebook. (See also How Many Squares). I have a distant family member who loves being #29,731 to comment on these things when they make the rounds, so I see far too many. Although I haven't seen any lately, maybe Facebook finally gave out some condoms?

posted by Nelson at 9:53 AM on July 2, 2013

posted by Nelson at 9:53 AM on July 2, 2013

*German maths problem*

Add a slash to each equals sign, making them "not equals".

posted by Pararrayos at 9:51 AM on July 2 [+] [!]

Add a slash to each equals sign, making them "not equals".

posted by Pararrayos at 9:51 AM on July 2 [+] [!]

There's a better (but rather more demanding) answer than that.

posted by Paul Slade at 9:56 AM on July 2, 2013

I don't know much about math, but I know that a light that spends half its time on and the other half off is at a 50% duty cycle, and appears roughly half as bright to the human observer.

posted by hellphish at 10:19 AM on July 2, 2013

posted by hellphish at 10:19 AM on July 2, 2013

*This Simple Math Puzzle Will Melt Your Brain*

I already melted my own brain just to spite you.

I already melted my own brain just to spite you.

I am getting really annoyed by articles telling me how I'm going to feel about them in the headline. It's like they've absorbed all the tropes of a certain type of spam in order to present their click-bait. "These New Man Tricks Will Melt Her Brain" "You Will Not Believe This Extraordinary Sex Enhancer" "The Most XXXXX Girls You Will See Tonight" "This Could Be The Most Amazing Pleasure Tip Ever".

posted by oneirodynia at 10:38 AM on July 2, 2013 [4 favorites]

Paul Slade: If you integrate all the numbers and add the integrals together it works, for certain values of c. Although I'm not sure the resulting mess is solvable.

posted by Grimgrin at 11:19 AM on July 2, 2013

posted by Grimgrin at 11:19 AM on July 2, 2013

It's 1 half the time, and 0 half the time, so it's 1/2.

Can I vote in Mississippi now?

posted by nicwolff at 11:20 AM on July 2, 2013 [1 favorite]

Can I vote in Mississippi now?

posted by nicwolff at 11:20 AM on July 2, 2013 [1 favorite]

So mathematicians don't have an expression evaluation convention? To a developer, an expression without parens, consisting only of + and -, is interpreted left to right. So the series is

(((((1) - 1) + 1) - 1) + 1) ...

which of course never converges.

A C programmer would consider the expression wasteful and just write

(((((1++)--)++)--)++)...

posted by zompist at 12:10 PM on July 2, 2013

(((((1) - 1) + 1) - 1) + 1) ...

which of course never converges.

A C programmer would consider the expression wasteful and just write

(((((1++)--)++)--)++)...

posted by zompist at 12:10 PM on July 2, 2013

*If you integrate all the numbers and add the integrals together it works*

If you wanted that to work wouldn't you have to write down some numbers as the limits of integration? Otherwise you'd end up with algebraic functions (which admittedly are equal to any number for for different values of the variable).

posted by Ned G at 12:34 PM on July 2, 2013

*Is really three different equations with three difference answers.*

This is more what I meant, expressed more clearly. It seems like the differences he introduces between the three formulations of the 'same' problem are actually pretty significant, and it's not really the same equation three times at all. Therefore it's not brain melting when you get three different answers, especially when you stretch that out over infinity (creating an asymptotic graph, I guess?).

But, once again, I'm not super math literate.

posted by codacorolla at 12:41 PM on July 2, 2013

It isn't that "mathematicians don't have an expression evaluation convention", it's that evaluation order isn't supposed to affect the result when you're dealing entirely with addition. (And subtraction, which is just a notational convenience for addition of the additive inverse.) This is what's meant by the "addition is associative". (A+B)+C == A+(B+C). That's just as true in programming as it is in number theory.

And yeah, the reason it fails here is that the sequence in question doesn't actually have a sum. Even your proposal of only paying attention to left-to-right evaluation fails to produce an unambiguous value. Mathematicians are comfortable with this, in the same way that they're comfortable with 0/0 not having a value.

posted by baf at 12:46 PM on July 2, 2013 [2 favorites]

And yeah, the reason it fails here is that the sequence in question doesn't actually have a sum. Even your proposal of only paying attention to left-to-right evaluation fails to produce an unambiguous value. Mathematicians are comfortable with this, in the same way that they're comfortable with 0/0 not having a value.

posted by baf at 12:46 PM on July 2, 2013 [2 favorites]

This is cool and all but I was promised a puzzle! Not a video of a guy explaining math stuff. One of my pet peeves is that math is often "explained" to us, instead of allowing us to manipulate it ourselves. Of course it's tough to grok and/or boring when you can't get your brain all up in there and dirty with the concepts themselves.

We demand puzzles!

posted by hapax_legomenon at 1:27 PM on July 2, 2013 [1 favorite]

We demand puzzles!

posted by hapax_legomenon at 1:27 PM on July 2, 2013 [1 favorite]

*get your brain all up in there and dirty*

my brain keeps trying to visualize this make it stop

posted by Elementary Penguin at 2:35 PM on July 2, 2013 [1 favorite]

Well I think it's negative infinity.

Operate as follows: there are an infinite number of (1) terms at positions 0, 2, …, 2k, … Pair them up one-to-one with just half of the (-1) terms at positions 1, 5, …, 4k+1, … and get the sum of zero. But wait, you're left with the other half of the (-1) terms at positions 3, 7, …, 4k+3, …, which sum to -∞. QED.

If you disagree, well, that's just your opinion.

posted by jepler at 3:51 PM on July 2, 2013

Operate as follows: there are an infinite number of (1) terms at positions 0, 2, …, 2k, … Pair them up one-to-one with just half of the (-1) terms at positions 1, 5, …, 4k+1, … and get the sum of zero. But wait, you're left with the other half of the (-1) terms at positions 3, 7, …, 4k+3, …, which sum to -∞. QED.

If you disagree, well, that's just your opinion.

posted by jepler at 3:51 PM on July 2, 2013

(you can also prove it's any integer you care to. e.g., to prove it's 7, pair the (1) terms at positions 14, 16, …, 2k, … with the (-1) terms at positions 1, 3, …, 2k+1, …, which sums to zero. But you're left with the (1) terms at positions 0, 2, …, 12, which sum to 7)

posted by jepler at 4:06 PM on July 2, 2013

posted by jepler at 4:06 PM on July 2, 2013

There are already three answers to 1+1, so I'm not surprised.

The plus operator sometimes denotes OR. Of course there is the traditional arithmetic sum. You can think of the two ones as unique sets, and the union as a third unique thing.

posted by Chuckles at 5:08 PM on July 2, 2013

The plus operator sometimes denotes OR. Of course there is the traditional arithmetic sum. You can think of the two ones as unique sets, and the union as a third unique thing.

posted by Chuckles at 5:08 PM on July 2, 2013

*(1 - 1) + (1-1) + (1-1) + … = 0*

NOW MOVE THREE MATCHSTICKS TO MAKE IT EQUAL 12

posted by Joe in Australia at 7:15 PM on July 2, 2013

*Is this mind-blowing in the sense that there's no real rigor in the definition of the question, just a bunch of manipulations and appeals to intuition?*

If one hasn't taken pre-calc or calculus, it's interesting because it makes you consider what an infinite series actually is, and what it means to sum one, and it's a simple example of a series that doesn't converge.

posted by empath at 12:17 AM on July 3, 2013

*Operate as follows: there are an infinite number of (1) terms at positions 0, 2, …, 2k, … Pair them up one-to-one with just half of the (-1) terms at positions 1, 5, …, 4k+1, … and get the sum of zero. But wait, you're left with the other half of the (-1) terms at positions 3, 7, …, 4k+3, …, which sum to -∞. QED*

If you subtract infinity from infinity you still have infinity. (Subtract the odd numbers from the natural numbers, for example).

posted by empath at 12:18 AM on July 3, 2013

Yeah, the whole point of the video is that this

posted by Elementary Penguin at 2:50 AM on July 3, 2013

*was*mind blowing when mathematicians were first dealing with it. The solution mathematicians came up with is best summarized by escabeche's Hardy quote above.posted by Elementary Penguin at 2:50 AM on July 3, 2013

*"If you look at every odd position in the series, it's a 1! But if you look at every even position, it's a 0!"*

Except

`0! = 1! = 1`.

posted by Eideteker at 7:17 AM on July 3, 2013

(HAMBURGER factorial if that wasn't obvious)

If you're pairing them (an infinity) up with half of another infinity, that's not one-to-one (IIRC, the term is bijection). This is where the cardinality of infinite sets comes in. So no, not opinion. (And that's on the

posted by Eideteker at 7:31 AM on July 3, 2013

*"Operate as follows: there are an infinite number of (1) terms at positions 0, 2, …, 2k, … Pair them up one-to-one with just half of the (-1) terms at positions 1, 5, …, 4k+1, … and get the sum of zero. But wait, you're left with the other half of the (-1) terms at positions 3, 7, …, 4k+3, …, which sum to -∞. QED"*If you're pairing them (an infinity) up with half of another infinity, that's not one-to-one (IIRC, the term is bijection). This is where the cardinality of infinite sets comes in. So no, not opinion. (And that's on the

**R**.)posted by Eideteker at 7:31 AM on July 3, 2013

This is what's meant by the "addition is associative". (A+B)+C == A+(B+C). That's just as true in programming as it is in number theory.It's a little less true in programming than it is in number theory, for numerical precision reasons.

posted by dfan at 10:43 AM on July 3, 2013

That's an understatement:

posted by Rhomboid at 5:18 AM on July 4, 2013 [2 favorites]

If you're implementing a computer language, you generally must treat floating point operations as non-associative. You can't write an optimizer that freely rearranges expressions based on mathematical axioms, for instance. You must obey the order of operations exactly as written in the source code, because not doing so would break code that has been explicitly arranged to avoid catastrophic cancellation like in this example and other forms of loss of precision.Python 3.3.2 (v3.3.2:d047928ae3f6, May 16 2013, 00:03:43) [MSC v.1600 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> A = 0.5 >>> B = -0.5000000000000001 >>> C = 0.00000000000000005 >>> (A + B) + C -6.102230246251566e-17 >>> A + (B + C) -1.1102230246251565e-16

posted by Rhomboid at 5:18 AM on July 4, 2013 [2 favorites]

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