What intuitively obvious mathematical statements are false?

It's a great analogy. If you'd like to see more like this, check out The Code Book, by Simon Singh. In fact, he uses this very analogy in his public key chapter.

It's an absolutely fantastic read. I can't keep my hands on it- I keep giving my copy away to share it with people, then buying a new one.

Now give him a final, no cheat sheet, and fucking multitudes of encryption schemes.

This back and forth isn't really how any modern cryptographic system works, but it's neat anyhow.

but that requires you creating a code before she can listen to you... so she hasnt heard everything. you might as well recommend coming up with a new language and speaking in that language. its the same

description of cryptography

cryptscription

This same analogy was used in the children's educational uk series "how 2" on the 90s. Worth a watch.

He actually took it from a common example in books and whathaveyou. Except the locks are usually color coded as well to make it easier to keep track of when explaining the different encryptions.

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Diffie-Hellman exchange, roughly, right? There are different ways to do public key cryptography, another one is eliptic curve-based, and there is an relatively inefficient secure-hash-only based on too, not that i am that well-learned in this stuf...

Note that there are caveits, and "implementation concerns". And if Eve can change the messages, she do a man-in-the-middle attack.

These type of cryptographic constructions are known as three-pass protocols. You're right, integer multiplication three-pass protocols are completely insecure, because multiplication is about as computationally intensive as its inverse, and so the plaintext is trivially reconstructed from the three transmitted messages. I guess integer multiplication three-pass is pedagogically useful, though, because you get an intuition that your three-pass operation must be commutative, and, as you've deduced, asymmetric in some way, so that it's not so easy to calculate the inverse.

Real three-pass protocols use commutative operations that are computationally asymmetric, like exponentiation modulo a large prime, or exponentiation in the Galois field. Computing the inverse of these operations would effectively be equivalent to solving the discrete logarithm problem.

It doesn't work exactly like OP suggested. The message is actually scattered around a modulo group so it's not discernible what the actual product is.

The metaphor of the two locks is genius though, that's a good way to explain cryptography to non-math people.

Diffie helman exchange is an example of what he is describing: https://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange#Description As one of the other commenters mentioned, it ustilizes the fact that the discrete log problem is difficult to solve (i.e. what Eve has to do to decode the message).

Ignore the maths; it's just a bad example; also ignore the process, because that's wrong too. All that's any good is the analogy.

There are a number of encryption techniques known as public-key encryption. The most common involves very large prime numbers. This involves 3 numbers - 2 very large primes, and their product. There is a method of encrypting a message using the product of the primes in such a way that it can only be decrypted in a reasonable amount of time by someone who knows the original primes. Finding the primes from the product is possible, but not in a reasonable amount of time.

Alice has 2 very large primes, and knows their product. Bob wants to send her a message, and tells her so. Alice sends Bob her public key (the product) - these 2 crucial steps are missed out in the above simplistic example. Bob uses this to encrypt his message, and sends it to Alice. Alice can decrypt it using her private key (the 2 large primes). Eve knows everything that has passed between Alice and Bob but cannot decrypt the message because she doesn't have the private key. There is no need for Alice and Bob to meet, or communicate securely at any point, which is what makes public key encryption so immensely useful.

d'oh!

Ok, now read about Diffie-Hellman key exchange. That's how it's really done.

Yes, however typically encryption isn't quite this simple. A better example might be like this:

Alice has 2 sets of numbers aPub and aPriv, such that they are complements of each other. Bob has a similar set called bPub and bPriv. Both aPriv and bPriv are ONLY known by Alice and Bob respectively, while aPub and bPub are known by everyone.

Alice sends a message to Bob that looks like m * bPub. Bob then can read the message by doing m * bPub / bPriv. But it's important to remember that bPub =/= bPriv and is only really able to be decoded by the decryption algorithm that is being used that knows BOTH numbers. Or to go back to the locks analogy Bob hands locks out to everyone that only he has they key for.

***OR***

Alice would then send a message to Bob that looks like m * aPriv * bPub that could then be decrypted with m * bPriv * aPub. This however requires a bit of forethought on Alice and Bobs part as they need to have a previously established form of communications to build a decryption algorithm, but this method is MUCH more secure.

Good question! I was thinking exactly this.

You're not. This doesn't work. Don't trust anything you read in this sub.

Thank you.

You're thinking of Mallory. Eve is tetraplegic and mute.

Public key crypto assumes that Alice and Bob know how each other's locks look like before they start communicating.

In the analogy, the locks are the public keys and, as you correctly figured out, you need to exchange the public keys through a trusted (but not necessarily secret) medium before you start encrypting. You might meet up face to face beforehand or delegate the trust to a third party who knows both the public keys.

The assumption was:

Eve listens to absolutely everything

The assumption was not:

Eve listens to absolutely everything and modify it.

Man in the Middle attacks (MITM) require assumption 2.

In reality math problems like RSA are used. They're strongly resistant to that analysis

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I'm going to use the analogy example explain this, but here are the variables

M	a	b
2015	217645199	492876847
	32452867	275604547
	236887691	179424673
	982451653	694847539

M : This is your message

a : These are your locks

b : These are the recipient's locks

You lock the box {M} by multiplying it with your locks. This makes locked box {Ma} with a value of {3312309379967778134280375206895560885}. You send this to the recipient.

Then the recipient adds their locks making the box {Mab} with a value of {56095416572385525154713578876611339168291668429150410898641475603328355}. This is returned to you.

Now you undo your locks creating locked box {Mb} with a value of {34124911482289254484502370986393738345}.

Finally the recipient unlocks their locks leaving an open box {M} with the value {2015}.

The weakness lies in that a Man-in-the-Middle (MitM) would have seen {Ma}, {Mab} and {Mb}. So now they have all the tools to reverse your locks and the recipient's locks.

Mathematically, {Mab}/{Ma} creates {b-composite} with a value of {16935439941582756567991251109872823}.

MitM does not know the values of {b}, but does not need them to unencrypt. Now take {Mb} and divide by {b-composite} to create {M} with the value, as ever, {2015}.

Strong encryption knows this weakness and therefor does not use straight multiplication, but by this analogy you are indeed correct. If the MitM misses any transmission though, the contents are secured

yes. The described system is completely useless. If you'd like I can describe a system that actually works, but it's a little more complicated.

She doesn't know the primes used though. There is no 'key' attached to the box for her to use.

no

Doesn't have to be. RSA uses them, but ECC and NTRU and McEliece are also options

For that given experiment it doesn't have to be primes, but because every number can be defined by primes and they also play a key role in cryptography, people tend to use them in such examples.

They're way harder to guess is basically what it boils down to

doesn't in this case. In fact, the example given is totally bogus and insecure. Decrypting the message without given only ma, mb, and mab is exactly as hard as "encrypting" them in the first place.

Unfortunately OP doesn't know much, has half remembered a very well known analogy, and made a statement of how it can be used ("by multiplying by a bunch of primes") that isn't remotely secure. Cryptography doesn't work like this at all.

But I guess that not many people have heard about the two locks technique for exchanging something securely and think it is really cool.

The result of two prime numbers multiplied together can only be factored back (factoring = finding the original multiplied numbers) to those two primes. Multiply two non-prime numbers together, and there are multiple solutions to the factoring problem.

Exactly my question... Eve should easily crack this....

Because prime number divide roundly with only one combination. It keeps people from guessing the encryption key.

Go look how RSA works. In this instance it makes absolutely no difference, the encryption does not work whatever numbers you use.

Exponentation is technically multiplication.

Primes are used because the algorithm depends on factoring - figuring out which two whole numbers were multiplied together to arrive at another number. If you have the result and one of the factors, determining the other is simple - you just divide the result by the factor you have - but if you don't, it's impossible to determine what the original factors were without brute-force guessing (which for large numbers, takes a LONG time).

Since primes have only two factors (one and itself), a number that is result of multiplying two primes can only be the result of multiplying those two primes - no two other numbers could result in the same product. If one or both of the numbers is not prime, then there are multiple solutions to the factoring problem, making it far easier to "crack".

For example: 7 times 11 equals 77, but because 7 and 11 are both prime numbers, there are no other numbers (other than 1 and 77) that can be multiplied together to get that result. On the other hand, 7 times 10 equals 70, but because 10 is not prime - it's divisible by 2 and 5 - it's also true that 5 times 14 equals 70, and 35 times 2 equals 70 as well.

Yeah. Of all the examples I have ever heard in all my years of schooling, this one struck me as the most intuitively eye opening.

What's in the fucking box?!

Fucking Mallory sets the box on fire.

Multiply it by whatever you want. Any number is the product of primes.

I believe the idea is that primes are harder to "accidentally" divide by. For example, if you multiply by 60, there are many ways to divide that. 6x10, 2x3x10, 60x1, 30x2, 15x4 and such. If it's a prime then there's only one way to divide it and so the chance of Eve accidentally finding a way to unlock the lock is lesser. (I have no idea about cryptography though, this may be totally wrong)

My knowledge of cryptography is limited, but so far nobody has found an efficient algorithm for factoring integers. (Well, without using quantum computers.)

Factoring a number is hard because basically the algorithm is to try to divide for every prime number, and the ones used are really large; thus the operation is expensive.

That said, the example does not show RSA, but some very silly and crackable in 3seconds scheme.

After seeing the messages ax and abx you know the value of "b", then when you see the message "bx" you know the value of x. Prime numbers or not, nothing makes that scheme working.

To make the math work and make the problem hard to solve. Here's a much better explanation that doesn't get the problem wrong like OP: http://youtu.be/YEBfamv-_do. (Important part starts around 3 minutes with a much better analogy)

Because otherwise, you could just brute force divide out each prime factor of a or b

Given enough time, everything is solvable.

That's the struggle man, the encyptors v the decryptors.
A never ending battle

A lot? That thing is not RSA, it would take perhaps 3 seconds to crack it.

KKKKAAAAAHHHHNNNNN!!!!!!

(Also, it's khan academy)

That isn't even close to the tip of the iceberg. Good encryption systems are complex and multi-layered. They require trust, secrecy, and privacy, and what is described above barely falls into secrecy and doesn't cover all the minutia required to get real secrecy.

The only way for a computer to find the prime factors of a number is to try them (up to the square root of the number). So you'd have to check 2, then 3, then 4, etc until you find a number that is a factor of the number you're trying to crack. If your number is prime, you won't have any success until you actually get to the number. So if your number is prime and very large, it's going to take a very long time to find the factors of it.

If it wasn't prime, for example an even number, you find a factor when you get to 2. Now you can divide the number you're trying to get to by 2 and you've cut the time to find the factors of that number in half.

A trivial example - if you use the number 7. You have to try 2 through 7 before you find the only factor (7). So, 6 steps. If you use the number 8, you find a match on your first try with number 2. Now you have 2 and 4. You hit again with your first try at factoring 4 and you've got all of your factors. 2, 2, and 2. And it only took 2 steps. So using a prime made it take twice as many steps to factor. And in this case, factoring the number is the equivalent to unlocking the lock.

All locks are pickable, it just depends on how long it takes, and it takes less time to pick a bunch of small simple locks than a more complex one.

Imagine a large prime number being a large complex key to a large lock. The larger the prime, the more complex the key, and lock. All non-prime numbers can be factored into smaller primes (6 = 2 x 3, 16 = 2 x 2 x 2 x 2). If you did not use primes, you're effectively using a bunch of small less complex keys to a bunch of small locks.

In encryption books you normally use Alice and Bob as examples of people trying to send secure messages to each other. Eve is eavesdropping on the conversation, trying to break the encryption. Mallory is someone who is generally just trying to mess things up for all parties. It makes it easy to use those names when describing things since everyone knows the roles.

No. But at the start of an encrypted session, you need to have an exchange like this to exchange private encryption keys.

Yes, but only if the key is at least as long as the message. Otherwise, the encryption is going to be quite weak.

You never exchange the primes.

Only I know the primes I used, and the receiver knows the primes they used.

The whole point is, only I lock and unlock my own lock, and you do the same for yours. But because we do it in the way described by /u/UlyssesSKrunk the package is always locked by at least one of us.

You mean she can make copies of that locked box and keep one for herself ^hint hint-NSA does this^, but she cannot open it. She needs a special locksmith who can break the lock and that lock smith has to be the lockmaker/keymaker themselves. ^hint hint-microsoft and facebook^

We don't actually use multiply in encryption because it's a two way function, you can guess the input if you know the output. y = f(x), if you know y and function f() you can guess what the input x is.

What we use is an one way function. In real encryption: function f(), the encryption protocol is publicly available. Message y is also publicly transfered through the internet. But Eve cannot guess the content x inside without knowing the primes you and I used (called private key)

---

Edit: Depends on the importance of message x we can use a longer private key or a more complex function f(). Eve can try to guess the private keys with brute computational force of a super computer, but it is ineffective to do so (i.e. she spent 20 years to crack an encrypted email between you and I, which could be worthless by that time).

No, it assumes that Eve can't perform two integer divisions.

Really, she should be like Janet in Accounting who doesn't give a fuck.

:-)

Sure, if you're speaking a secret language that Eve doesn't understand, then obviously, the statement is false. Don't need four paragraphs to explain that one!

However, since you asked to be corrected if you make an error...

Let's say you're sending an encrypted email, and watching some encrypted video files, and generally sending encrypted information back and forth between your computer and remote servers each day. Your computer is not going to bother with things like "Oh, this looks like an email! I'll just translate it to Schminglish real quick, and send it to this other server that understands Schminglish when no one else does!" That would be crazy. Schminglish translates emails, but might not work with video files or audio. More importantly: Schminglish is required to be understood by one other server, and one other server only, which means that your computer would be required to "know" an entirely different language for each server it talks to AND each server would be required to know an entirely different language for each computer it talks to! As you can imagine, the number of languages would grow astronomically huge. This is actually a well-studied situation in cryptography, and is most often quickly dismissed as impractical :)

So how can you communicate with another remote server, secretly, even when it understands the language but is unable to understand the message? That's what this whole "prime numbers thing" solves.

Imo this wouldn't be false because there would be identifiable pattern in the communication, in your example "238" means "I miss my chicken and cheese breadsticks", or "524,710" means "I'm going to be late for school today. So wherever those numbers appear in the structure of the communication they will mean the same thing. This is because Both A and B have learned a language and and understand the structure.

Encryption attempts to remove all structure and render the message indecipherable, using your example "238" means "I miss my chicken and cheese breadsticks" but later In the paragraph "238" could mean "I'm going to be late for school today" making it exponentially harder to decipher. The keys determine the meaning based on position, when the keys reach a certain size you effectively making the task impossible with today's computing powers. The keys define what the numbers mean relative to there position.

The point is that is not the language that makes the communication secure but the process. The locking the box analogy, both A an B both understand that they need to apply there own lock and send it back in order for the other partner to unlock the box. In your example the box had a combination lock and both A and B have been told the code. It possible for the person in the middle to guess the code. But with the locks the need to have both keys. The only thing A and B need to know is the process, and even with the process being publicly available, the person in the middle still can't unlock the box.

Commutative operations is needed when you do three passes like here, just as others said. Except they must be hard in one direction, which multiplication isn't. ECC operations for example are hard in the reverse direction.

You never exchange keys in this scheme

Well, as Eve, if I could guess what numbers to divide the encrypted message by to get the orginal, then I can decrypt the message. If the numbers are prime, then there are only two possible divisors that would work. If the numbers are not prime then there are more possible answers that will also give the correct result.

basically nothing. The protocol he described isn't a good one, as others have pointed out.

One of the simpler algorithms that's actualy used is called the diffie-hellman key exchange. Let's say you and I want to agree on a key or password that we can use to encrypt messages we send to each other.

We start by agreeing on a prime number, that we'll call p, and a special number g, that we call a "generator". next, you and I both pick a random number (a and b respectively) between 1 and p-1. We keep those secret

Next, I take the number g^a, divide it by p, and take the remainder (this is called taking g^a mod p). You do the same, with g^b, and then we exchange g^a and g^b.

So now I take the g^b you sent me, and do (g^a )^b mod p, and at the end I'm left with g^ab mod p, and you have g^ba mod p, and it turns out these two things are equal. So we now have a number between 1 and p-1 that we can use as a password.

Here's why this works. if you have numbers (g^a mod p) and (g^b mod p), it turns out to be really hard (we think) to compute (g^ab mod P). This is called the diffie-hellman problem. But, if you have b, and g^a mod p, then it's trivially easy, so Bob and Alice can both find the secret key, but Eve cannot.

The reason p has to be prime in this case, is because if p is composite, and Eve can factor p, then it is actually easy to solve this problem.

RSA needs it. Other algorithms like ECC don't.

Oh God, you actually teach encryption and you can't tell that this scheme is both completely broken and nothing to do with RSA?

The original statement was discussing asymmetric crypto, which does not rely on any secret exchange. OTP doesn't have this nice property.

I think he meant "for any instance of Eve with fixed computing power, there are instances of this scheme she cannot break".

Ah, yeah, to expand on that other explanation...

If your adversary (Eve) has arbitrarily great computing capability, you may simply use sufficiently large primes in your classical crypto scheme. Sufficiently large meaning "large enough to retain any arbitrary level of security you desire."

How exactly you get those sufficiently large primes is another question entirely, and certainly relies on your processing power not being fixed. Actually, for practical public-key crypto, we use Fermat pseudoprimes, which can be generated relatively quickly. However, the amount of computation you have to do to use this scheme grows much more slowly than the amount of computation your adversary needs to break it. That's the fundamental idea of practical cryptography -- though exactly how much is "much more slowly" depends on the scheme.

Now, if your adversary has infinite capability, traditional cryptosystems do not hold -- even if you, too, have infinite capability.

Thank you.

... that has no bearing on this example at all.

Good explanation. Thanks.

However big your primes are, you only have to perform two divisions.

It's not the size of the numbers that matters; OC is using them wrong.

Yeah I think your right after some research I was thinking of stuff like encryption systems that use a 128-bit random number for the key attached to an account. It also really depends upon what you are using encryption with.

No worries! StevenXC's construction is the typical counterexample, and it is a pretty strange set.

Let U be any set of real numbers. Any point x in U is called an interior point if U contains some open interval centered at x. For example, the point 0.9 is an interior point of the set [0, 1) since the interval (0.85, 0.95) centered at 0.9 is entirely contained in [0, 1). However, 0 is not an interior point of [0, 1) since no open interval centered at 0 lies entirely within [0, 1) (any such interval contains points less than 0).

A set is called open if all of its elements are interior points. Thus, if U is an open set in R, you can take any point in U and "wiggle around" a bit (i.e. come up with a small open interval about that point) while remaining in U. For example, open intervals are open, but half-open and closed intervals are not, because their endpoints have no "wiggle room".

The incorrect claim is: if a set U is open in R and contains every rational number, it must be (more or less) all of R. (This seems intuitive since the rationals are dense in R.)

iirc there are models of ZF where R as a vector space over Q doesn't have a base

Not the above poster, but I would guess it's similar to the proof that the rationals are countable but it's like 2 AM and I'm too tired to math now.

AC => An infinite set has the same cardinality as any Cartesian product of a finite number of copies of itself.

In fact, Tarski's theorem shows that the converse also holds, so this is just another form of the Axiom of Choice.

Assuming B has the same Cardinality as R (I'm pretty sure it does; someone correct me if I'm wrong), there is a rather straightforward bijection between RxR and R. Represent each R as an infinite string of digits, and construct a new infinite string of digits by interleaving the digits from your original two numbers. There is a little bit more involved to account for ambiguous reorientations (The real number 1 can be represented by either "10000000..." or "999999999...", for instance), but that is the gist of it.

They're not isomorphic as vector spaces over R, but they are isomorphic as vector spaces over Q. The reason we forget scalar multiplication is simply because a vector space is just an abelian group where we have the extra structure provided by scalar multiplication, and so to get the result we want, we just realize we're already done by saying "oh wait, we don't need this scalar multiplication business" at the end.

When working more abstractly with vector spaces (usually not done in a first course in linear algebra), you have to be very aware of what field you're working over.

Reread the post above yours

Certainly not, 2^aleph_0 = aleph_1 is the continuum hypothesis and is known to be independent of ZFC. Aleph_1 is the next cardinal after aleph_0 but 2^aleph_0 might be equal to aleph_37 or pretty much any other cardinal.

Yeah, it's not 0 if you look at it on a molecular level - I meant an idealized dartboard, which I should've made more clear.

P(X=x) = 0 ∀ x ∈ Ω is kind of unnecessary, but we get your point.

Ω

What set does this represent?

There are no infinitesimal real numbers except 0. Probability is a real number. (And yeah, I'm ignoring the fact that the tip is blunt, the fact that the dartboard is made out of molecules...)

Even if you include the infinitesimals it doesn't allow regularity.

Zero.

You're talking about the real world, he's talking about a random variable.

How are continuous random variables artificial? And no, there isn't.

There are no infinitely small real numbers. Probabilities are real numbers. And it's not infinitely small, it's 0.

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Sounds about right..

Well in both it means that the probability would be approaching 1. It would be a limit question wouldn't it?

But as long as there is one example where it won't happen you could never actually get P=1.

It's like Σ[¹/₂]ⁿ , n={0,1,2...}

It isn't just = 2, it's limit as n->∞ = 2. That is the supremum for the set. But it wouldn't be in the set. It "converges" to its limit.

At least, this is what I have got out of all of my undergraduate. Maybe the profs are lying to us to keep us content until we take a more complete course. :p

Exactly, and the dart tip has non-zero area, and so on - I'm talking about an idealised dartboard and an idealised dart. Essentially what I'm saying is "pick a random point in [0,1] x [0,1]" - but darts and a dartboard feels more intuitive.

I think he's imagining a situation where you consider a set of points and the in a discrete topological space with a metric and say the open ball is the set of points of distance less than 1 from the origin and the closed ball is the set of points of distance <= 1 from the origin. If the points of the space were set up so that there are at least a few points exactly at distance 1 from the origin then his statements follow.

My analysis book defines an open ball only with radius greater than 0 and so does Wikipedia. If we allow r=0, lots of things about open balls don't work.

Yeah I have a hard time with jokes too

They were just joking.

I also didn't get /u/chrzan 's joke until I'd looked at it for a while. It was an excellent joke, though!

redditors downvote those who are not in on the joke; it appears it's more important to be funny than helpful to most redditors who upvote or downvote.

Yes but that is rather ambiguous considering "!" as notation in western mathematics strictly refers to permutations, i.e. 5! = 5 x 4 x 3 x 2 x 1

Edit: And in the obvious case of 1! = .000000001 x .000000000000001 etc.

Friendly discussion? On reddit? Nay, sir, you must be mistaken...

the problem here is you give a damn about fake points

I like this one. LaTeX notation also works.

\neq

=/= looks hideous. != works in many contexts.

It can be expressed as 1/1, so it's in the rationals.

It's the sum of 9/10ⁿ from 1 to infinity. It's a sum of rationals so it would have to be a rational.

When people question if 1 = 0.999... they're generally interested in what is essentially a variation on Zeno's paradoxes, the standard conventions and definition of the reals be damned.

I doubt that's what's happening. Most people don't think that hard about math, and hold plenty of contradictory beliefs about math at the same time. "Infinity is a number" and "if x is a number then x+1 is a bigger number", "there is a smallest positive ("infinitesimal") number" and "there's always a number between two different numbers", etc. If asked, the same people would be perfectly fine with real numbers.

0.999... isn't convergent in the p-adics.

.999... is true in any base-10 system, is it not? It's a rational number, so extending Q differently doesn't change that fact.

Where it wouldn't be true is in a base other than base 10.

!= means "not equals" in some programming languages.

Yeah. I know its pretty common in programming. not sure about anywhere else.

X = 0.33333333... = 1/3

3x = 0.99999999... = 3/3 =1

There's a few ways to do it, always helpful to be able to explain things different ways to help others understand :)

There are some subtle issues with this method. Namely, it relies on the fact that 0.999...=1.

How do we know 10*(0.999...) = 9.99...? You're using a pattern that comes from the multiplication algorithm, but you can't use that algorithm with infinite digits, so we have to go deeper. 0.99... is by definition an infinite series (think back to your place value stuff) and you can only multiply through by the 10 like that if you know the series converges, so let's address that first.

We see this as the sum of terms 9*.1ⁿ (if n starts at 1), so now we have to backtrack a little further and examine the series .1ⁿ (which is the definition of .111....). We recognize this is a geometric sequence which converges to 1/9 (see calculus). So now we have that .111... = .1 + .01 + .001 + ... = 1/9. Since we now have convergence, 9.1 + 9.01 + 9.001+... = 9(.1+.01+.001+...) = 9*1/9 = 1.

Now we could go back to multiplying 10*(0.999...) but we've already proven that 0.999... = 1.

Edit: Fixed the random italicizing caused by my *s

algebra is where the party is at tho :(

however, on point with applied.

Hahaha, this is what I realized, every week I ask myself why... It's lovely but I can now see why it would drive someone insane xD

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Cantor's function is the canonical counter example. It turns out that Cantor's function is continuous everywhere.

Also, you can't just rearrange a few terms and and get a different sum. In order to produce a different sum from a series, you have to rearrange infinitely many terms.

If you only rearrange the first hundred, or million, or billion terms, then commutativity kicks in and the sums converge to the same thing.

Oh my god, pi is the square root of 2

Well there's a serious difference between sucking at high school math and feigning an educational history. I can tolerate the first, but definitely not the latter.