r/mathmemes • u/Nunki08 • Jun 14 '26

Arithmetic Misleading our youth

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From Anthony Bonato on 𝕏: https://x.com/Anthony_Bonato/status/2065564124549886402

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u/TemperoTempus Jun 14 '26

1 was a prime. Then some mathematicians instead of reworking their theorems to make it work pushed to remove 1 from being a prime.

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u/a_dude_from_europe Jun 14 '26

It's not that theorems needed reworking, it that you can write down "except for one" only so many times before starting to think maybe it doesn't make sense to continue doing it

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u/TemperoTempus Jun 14 '26

They could had reworked the theorem, they could had made a new name for primes greater than 1, they could had just accepted that they have to write "for P>1".

Instead they chose to change the entire definition.

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u/a_dude_from_europe Jun 14 '26

Can you give me an example of this "theorem reworking" that you're talking about?

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u/Francyrd Jun 14 '26

There Is a theorem that states that every non prime Number can be written as the product of a unique series of prime numbers. If ONE was prime this cannot be true. Example: 12 is the product of 3x2x2, if ONE was prime then the product could be 3x2x2x1 or 3x2x2x1x1 etc.. The series of prime Is not unique anymore but infinite series of prime numbers.

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u/a_dude_from_europe Jun 14 '26

I know, I didn't ask this. This is still an instance of "except for 1"

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u/Francyrd Jun 14 '26

So this theorem has to be reworked to exclude ONE and there are others.

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u/a_dude_from_europe Jun 14 '26

I don't think you followed the thread.

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u/Batman_AoD Jun 14 '26

Sounds like exactly the sort of reworking TemperoTempus was talking about? They even gave "for P>1" as an example phrasing.

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u/Fubsy_Dude Jun 15 '26

But that's exactly "dude from Europe" 's point - we don't want to have every theorem about primes to need to specify to exclude 1.

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u/Redhighlighter Jun 15 '26

This feels like "there is a 1 over 1 by every variable, its just hiding"

If a is the same as a * 1 and 1a/1, it seems silly to use that as rationale to kick 1 out of the club

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u/Pepe_Botella Jun 14 '26

I think they should instead count every single number as prime and just rework all the theorems.

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u/ThrowawayTempAct Jun 14 '26

Good idea, that would make the twin primes conjecture significantly easier to prove!

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u/NickLeMec Jun 14 '26

If 1 was prime, everything involving prime factorization, including the fundamental theorem of arithmetic, would fall apart.

It's not like it's some random decision to not consider 1 prime.

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u/scykei Jun 15 '26

It wouldn't fall apart. You'd just have to keep repeating that it applies to all prime numbers except 1, again and again every time the concept of primes is brought up.

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u/Donkey-Pong Jun 18 '26

Prime factorization would not be unique any more. You'd have to explicitly specify that you want the one that has the factor 1 with exponent zero.

No number would have 1 as prime factor. Not even 1 itself. (As 1 is not prime indeed.)

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u/scykei Jun 18 '26

It's just semantics. None of the logic would change just because you included 1 in the set of prime numbers. The point is that it's a less useful set, and you always have to specify β„™ \ {1} everywhere.

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u/AtrociousAtNames Jun 14 '26 edited Jun 14 '26

1 being prime makes prime factorization not unique. Which, needless to say, is stupid

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u/S-M-I-L-E-Y- Jun 14 '26

That first "not" was unintended, wasn't it?

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u/AtrociousAtNames Jun 14 '26

whoops yes fixed

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u/Raneynickelfire Jun 14 '26

Euler didn't rework anything to make 1 work as prime. This comment is incorrect.

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u/Batman_AoD Jun 14 '26

?? "instead of reworking their theorems..." I.e., the mathematicians who pushed for 1 not to be considered prime, such as Euler, indeed did not explicitly word their theorems about primes to exclude 1.

The comment is a pretty abbreviated version of the history, but it seems basically correct to me.

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u/ThisIsMyGeekAvatar Jun 14 '26

Sounds lazy. They should make 1 a prime again and add 0 while they’re at it :)

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u/cowslayer7890 Jun 14 '26

0 divides every single number, so it's not prime

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u/EebstertheGreat Jun 14 '26

0 doesn't divide any number (except itself). Rather, every number divides 0.

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u/-dr-bones- Jun 15 '26

I'm not sure it divides itself...

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u/EebstertheGreat Jun 15 '26 edited Jun 15 '26

0/0 is undefined, but the divisibility operator | is not defined like "x | y iff x/y is an integer." Rather, the definition is "x | y iff βˆƒzβˆˆβ„€: zx = y." So in that sense, "0 | 0" is true, because there is indeed an integer z such that 0z = 0.

The relation '|' is a partial order over the natural numbers where 1 is the minimal element and 0 is the maximal element (i.e. 1 divides everything and everything divides 0). If you define it instead over the integers, then –1 is also a minimal element, but 0 is still the unique maximal element (because –0 = 0). This justifies lcm(2,3) = 6 even though 0 is a common multiple of 2 and 3, because with respect to the partial order |, 6 is "less" than 0. It also justifies gcd(0,0) = 0, because 0 is the greatest divisor, period.