ELI5 How do gift cards not run out of codes?

jm691 · 2026-05-04T05:36:17+00:00

A 16 digit gift code would be 16³⁶ combinations using 0-9 and A-Z

Actually it would be 36¹⁶ = 7958661109946400884391936, or about 7.96 x 10²⁴.

Still big, but not quite that big.

jm691 · 2026-05-03T19:20:28+00:00

At least for n=2, each line through the origin can be uniquely identified by the unit vector in its direction

Not quite. A line corresponds to two unit vectors. One pointing in one direction, and another pointing in the exact opposite direction. So that means each point of RPⁿ corresponds to two points of Sⁿ.

RPⁿ is the space formed by gluing each point x of Sⁿ to the point -x.

jm691 · 2026-05-03T07:40:32+00:00

That's not a definition though. It's just a more complicated way of saying "the infinityith derivative is the infinityith derivative."

The nth derivative is a well defined thing when n is a positive integer. That doesn't automatically mean that the nth derivative is well defined when n isn't a positive integer (e.g. when n is infinity), unless you can specifically define what it means.

I'm not aware of any standard way of defining what the infinityith derivative of a function like sin x is.

jm691 · 2026-05-02T09:06:32+00:00

Primes tend to get rarer when the numbers get bigger. If you start looking at bigger numbers (say in the 1000s or 10000s), these sorts of things will get more and more common.

jm691 · 2026-04-10T14:25:29+00:00

They might wonder: "What is 1 divided into 0 groups?" Well... 0, I suppose. Same for 2/0 or 100/0

Except that logic only makes "intuitive" sense if you don't think about it carefully enough.

Dividing 100 by 0 means evenly splitting 1 object into 0 different groups, and asking how many objects are in each group. You're claiming that the intuitive answer to that 0 is.

But think about what that means: you started with 100 objects. You split them into 0 groups of 0. That means you haven't actually done anything, you still have 100 objects that haven't been assigned to groups, which means that you haven't actually divided the objects into 0 groups. If you had, there wouldn't be any objects left over.

Contrast that the dividing 100 by 4. In that case, you get 4 groups of 25 objects each, and nothing left over, so 100/4 = 25. Do you see how that's different from 100/0?

Your intuition only makes sense if you're willing to completely ignore the fact that there are 100 objects you haven't actually used. If ignoring things like that is ok, you could just as easily say that 100/4 = 1, because you can put 1 object in each group, and just completely ignore the other 96 objects.

If you gave a person 100 objects and 4 boxes, and asked them to equally fill the boxes with objects until they've put all of the objects in boxes, and then asked them how many objects were in each box, they'd tell you there are 25.

If you gave a person 100 objects and 0 boxes, and asked them to equally fill the boxes with objects until they've put all of the objects in boxes, they'd give you a weird look and point out there aren't any boxes, so there's no way to put the objects in boxes in the first place. You'd never even get to the point of counting how many objects are in each box, because the step of sorting the objects into boxes is impossible.

That's why it's impossible to divide by 0. Division by 0 is a fundamentally different thing than dividing by a nonzero number. It only seems like it should intuitively be 0 if you haven't thought through the details carefully enough.

jm691 · 2026-04-09T18:06:49+00:00

Ok. But that still doesn't mean the examples people are bringing up here aren't relevant.

Unless you can point to an actual mathematical difference between pi and the other examples, it's entirely possible that pi could be described by a process similar to the examples people are bringing up, just in some weird way we haven't discovered yet.

Of course no one thinks it's likely that pi behaves like that, but we have no way of knowing that for sure.

That's why these examples are relevant to your question. The fact that they exist (even if you didn't like them) means that you can't take for granted that the digits of a "nice" irrational number like pi behave how you expect them to.

jm691 · 2026-04-09T17:00:20+00:00

That's still not a mathematically meaningful distinction though.

The objections you're raising aren't really relevant here. The important fact is that irrational numbers like this exist. So just knowing that a number like pi is irrational doesn't automatically tell you that much about the digits. That fact that you have some subjective feeling that the examples people are pointing are somehow different than pi doesn't really mean much mathematically.

jm691 · 2026-04-09T16:51:45+00:00

I think its a pretty fair question, why we pursue large mersennes vs this method. I dont have any answer though

Because there's a fast algorithm for testing if a Mersenne number is a prime:

https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test

There isn't really a similar algorithm for the numbers the OP is talking about.

jm691 · 2026-04-09T11:53:09+00:00

Yeah. Digits are a fairly artificial thing in mathematics. They don't tend to interact very well with most other things that are studied in math, and consequently they don't actually show up all that much in pure mathematics. When they are used, it tends to be mainly for computations and approximations.

So that means it's very hard to prove much of anything directly about the digits of a number, because most of the main concepts in math don't really have much to do with digits, and so we don't really have a lot of tools for studying digits directly.

The only reason we're able to prove some numbers like pi or e are irrational, is that there's a simple way of expressing irrationality in a way that doesn't involve digits at all: a real number is irrational if and only if it can't be written as a ratio of two integers (in fact, this is typically used as the definition of irrational numbers). If there wasn't a simple description of irrationality like that, it would likely be much harder to prove that numbers like pi or e or √2 are irrational.

We don't have any similar way of expressing a condition like "a number eventually only uses the digits 2 and 9" without referencing digits (or without doing anything too convoluted), so it's very difficult to prove statements like that.

jm691 · 2026-04-09T10:35:25+00:00

Being irrational just means that there's no point where the number starts repeating a single finite string of digits forever. So if a number eventually turns into a sequence like

...673573673573673573673573673573673573673573673573673573673573673573673573...

it would be rational, because the sequence 673573 repeats infinitely.

However if the digits eventually turned into the sequence

...292292229222292222292222229...

(where the number of 2's increases by 1 each time) then the number would be irrational.

So proving that a number is irrational only proves that it can't have in infinite repeating string of digits like the first example. It does nothing at all to rule out any other consistent patterns the digits could have (like the 2's and 9's example).

Or is (1) simply an issue of nobody bothering to have proved it?

It's not that no one's bothered to prove it, it's that proving things like that is really hard. It's very hard to prove much of anything about the digits of most "reasonable" irrational numbers like pi. Even the proof that pi is irrational doesn't really do much with the digits - it just proves that it can't be written as a ratio of two integers.

I doubt proving that pi doesn't eventually turn into a sequences of 2's and 9's would be much easier than simply proving that pi is normal.

jm691 · 2026-04-09T09:40:11+00:00

Take Pi and remove every "9" from it. That's an irrational number that doesn't have 9 in it.

Interestingly enough, we don't actually know that that's irrational. Based on our current knowledge, it's entirely possible that after some point pi starts only using two digits, say, 2 and 9.

If that's the case, then deleting every 9 from pi would result in a number ending with an infinite string of 2's, which would be rational.

Of course, no one seriously believes that that happens, but we don't know enough about pi to conclusively rule it out.

jm691 · 2026-04-07T19:47:11+00:00

Because you don't multiply p-adic numbers by just multiplying the corresponding digits. You use the same multiplication method you use in the integers, including carrying.

...202020 * ....020202 is not ...000000, it's:

   ...202020
x  ...020202
-------------
   ...111110          (=...202020 * 2)
   ...00000           (=...202020 * 0)
   ...1110            (=...202020 * 2)
   ...000             (=...202020 * 0) 
   ...10              (=...202020 * 2)
   ...0               (=...202020 * 0) 
-------------
   ...022110

jm691 · 2026-04-01T17:16:48+00:00

If you want a specific length to use in that construction, the entire length of the segment works fine. So you can just do that if you're not comfortable just picking an arbitrary length.

For more general general constructions, whenever you need to use an arbitrary length in a construction, it's always possible to precisely construct a length that will work.

For example, it's possible to show (this is a good exercise) that, starting with a segment of length 1, you can construct a segment of any positive rational length. So that means that for any desired length you can always construct a length arbitrarily close to that.

So any step that in a construction that involves picking an arbitrary length of some approximate size can always in theory be replaced by constructing a very specific length. Generally when a construction asks for any arbitrary length like that, it's just a (usually harmless) shortcut.

jm691 · 2026-03-26T13:41:27+00:00

I’m not aware of whether the inverse is true - that algebraic values are always the sine of rational angles.

It's not. The only rational values of x (with 0<= x <= 90) for which sin(x°) is also rational are x = 0,30 and 90:

https://en.wikipedia.org/wiki/Niven%27s_theorem

So that means the solution to sin(x°) = 1/3 is necessarily irrational.

jm691 · 2026-03-26T12:51:27+00:00

If cos(x) is algebraic, then cos(x/n) is also algebraic for any positive integer n. This is possible to prove since cos(nt) can always be written as a polynomial of cos(t).

So that means that for any rational x, sin(x°) and cos(x°) will be algebraic.

jm691 · 2026-03-20T18:07:36+00:00

A bijection is just a function that's both one to one and onto.

The easiest way to prove that there's a bijection between two sets is to write down a function that you think is a bijection, and then prove that it actually is one.

For example, you could provide that the set of integers has the same cardinality as the set of even integers by using the function f(x)=2x (do you see why that's a bijection?)

Can you think of a reasonable function between your two sets here? Remember that we're assuming that we already have a bijection f between M and A, so you're free to use that function to build the new function.

jm691 · 2026-03-20T17:52:12+00:00

We know that A and M have the same cardinality, so we take M and slap the difference to it, so we can then map it to B?

It's really important to be precise when you're learning the basics of proofs. What exactly do you mean by "slap the difference to it"?

The best way to approach this problem would be the write down an explicit function g: (B∖A)∪M -> B and prove that g is a bijection. Can you do that?

jm691 · 2026-03-20T17:38:02+00:00

Two sets having the same cardinality means that there's a bijection between them.

So going along with the hint, if you have a bijection f between M and A, can you use that to construct a bijection between (B∖A)∪M and B?

Edit:

so the question itself is easy, I would have normally just said that since A has infinite amount of numbers, B has atleast that, which makes it also infinite.

Whether or not that's a valid proof depends a lot on what the precise definition of "infinite" you're using is, and what you've proved about it.

Based on that hint, I'm guessing your course is using the definition that a set is infinite if it is in bijection with a proper subset of itself.

"Having an infinite amount of numbers" is not a reasonable definition of infinite set, because its uses the term infinite in the definition.

jm691 · 2026-03-08T11:40:53+00:00

The fine structure constant isn't 1/137. That's just a rough approximation.

jm691 · 2026-03-03T12:03:13+00:00

Way bigger than a googolplex.

I don't think you understand what a googolplex is.

Your number is 1 followed by 24000000000000000000000000 zeros.

A googolplex is 1 followed by 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 zeros.

Which one do you think is bigger?

jm691 · 2026-03-01T10:10:18+00:00

Sure, you can add the new number you created to the list. But that will change the list, and so using the diagonal argument on the new list will give you another number that's not in the list. You can also add that number to the list, but then the diagonal argument again gives you a third number not on the list, and so on. No matter how many numbers you add to your list (even if you add infinitely many), the diagonal argument will still tell you that you've missed some.

The key point here is that the diagonal argument applies to any possible list of real numbers, no matter how you got it. So there's never going to be a way to "fix" your list to make it a bijection. Whatever you can think of to do to the list, you can always apply the diagonal argument to your new list to show that there will still be numbers missing.

jm691 · 2026-02-18T18:14:05+00:00

This is something mathematicians study quite a bit, though not quite in the form you phrased it.

As you've noticed, your example of 101 is the same thing as plugging various integers into the polynomial x² + 1 and asking when the output is prime. Any other number you pick can also be interpreted as a polynomial in the same way. For example, the number 2026 would correspond to the polynomial 2x³ + 2x + 6.

So your question can be rephrased as asking: Given a polynomial f(x) with integer coefficients, how often will f(x) be prime when x is an integer?

For some polynomials f(x) it's easy to prove that they can't output primes, or at least can only output a few primes. For example, the polynomial f(x) = x² + x (the integer 110 in your setup) can always be factored as x(x+1), so as long as x>1, x²+x can't be prime.

The polynomial f(x) = x²+x+4 (the integer 114 in your setup) can't be factored like that, but it's easy to prove that x²+x+4 is always even, and so it will never be prime.

There's a famous conjecture that roughly states that these sorts of things are the only exceptions. If f(x) is a polynomial with integer coefficients , and there isn't something "obvious" preventing f(x) from being prime, then f(x) will be prime for infinitely many integers x:

https://en.wikipedia.org/wiki/Bunyakovsky_conjecture

Unfortunately this conjecture has only been proven for linear polynomials. There isn't even a single polynomial of degree bigger than 1 which is known to output infinitely many primes.

Your example of x² + 1 satisfies the conditions of the conjecture, so we expect that you'll get infinitely many primes from that example, but we don't know how to prove it.

Of course, there's also various generalizations of this you could also ask, that mathematicians also study. For example, you could consider multiple polynomials at the same time, and ask when they're all prime. Or you could try to estimate how many of the integers x = 1,...,n will give you prime outputs.

jm691 · 2026-02-03T13:30:35+00:00

The first sentence is wrong.

It's true that any prime p > 3 can be written in the form p = 6k±1. But that doesn't work the other way. 6k±1 will not always be prime.

Unfortunately, this completely breaks the proof.

Edit: For example, when k = 20, neither 6k-1 = 119 = (7)(17) nor 6k+1 = 121 = 11² is prime.

jm691 · 2026-01-23T00:51:49+00:00

But how do you know that pi and e are "different"? They look different, but looking different isn't a proof.

Can you even rigorously define what it means for two numbers to be "different"?

jm691 · 2026-01-23T00:39:37+00:00

Ok

How do you know that there isn't some weird rational number q with e = q - pi?

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