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Shocked to learn that

1+2+3 = 1×2×3.

What's the point of having different operations if the result is always the same?

@codingquark Oh you're probably right! "New is always better."

@esclear @erou ah! Perhaps different printing process or slightly incompatible doc formats or different versions of LaTeX!

@codingquark @erou
🤔 Could it be version-dependent?

Maybe TeX version x rotates all plus symbols by x radians?
So the left sum could have been produced by a very recent (or early or pretty-much-exactly-in-between) version of TeX and the right sum would have been produced around version 0.7853982 or version 2.356194 ?

@erou Right? Such as for the Knuth operators 2 [n] 2 = 4 for all n in the natural numbers. That is like basically always. Lol

@erou I'd like to know how you're defining "always."

1+2+4 is not the same as 1x2x4.

@erou almost never, in this particular case. There are only four "perfect numbers" known under ten thousand, and I would expect them to become vanishingly rare as you go higher.

britannica.com/science/perfect

@WizardOfDocs Well if we want to be serious about this, even the other perfect numbers do not satisfy this kind of equality, because the product of all the proper factors of a number is not equal to the number itself, except if there are only two distinct prime numbers (like 6).

For example 12 ≠ 1×2×3×4×6.

So you have to be perfect and semi-prime!

@WizardOfDocs @erou those are edge cases and can be ignored. For the sake of simplicity and ease of use we should settle for either + or ×.

@Little_Rascal @erou the chance of encountering a perfect number among all the naturals is zero

therefore, addition and multiplication are different

@WizardOfDocs @Little_Rascal But you do not use all natural numbers, just the small ones!

And we saw that it happens with small numbers, so I agree with @Little_Rascal, we should just pick one!

@erou @WizardOfDocs cant even remember the last time i used any other digits then 0-9, and like ⅓ of those adhere to the rule we just established. Imo thats good enough with tolerances in mind.

@don Well that also proves my point! Thank you for your investigation.

@erou finding another set of consequtive integers that work with other operators is left as a challenge for the reader

@erou this is actually the subject of a recent Mathologer video, which clarifies that there is actually an infinite number of (x, y, z) triples that satisfy x + y + z = xyz!

youtube.com/watch?v=IguNXoCjBE

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