this post was submitted on 12 Dec 2023
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6÷2(1+2) (programming.dev)
submitted 11 months ago* (last edited 11 months ago) by wischi@programming.dev to c/memes@lemmy.ml
 

https://zeta.one/viral-math/

I wrote a (very long) blog post about those viral math problems and am looking for feedback, especially from people who are not convinced that the problem is ambiguous.

It's about a 30min read so thank you in advance if you really take the time to read it, but I think it's worth it if you joined such discussions in the past, but I'm probably biased because I wrote it :)

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[–] wischi@programming.dev 3 points 11 months ago* (last edited 11 months ago) (3 children)

"when in doubt" is a bit broad but left to right is a great default for operations with the same priority. There is actually a way to calculate in any order if divisions are converted to multiplications (by using the reciprocal value) and subtractions are converted to additions (by negating the value) that requires at least a little bit of math knowledge and experience so it's typically not taught until later to prevent even more confusion.

For example this: 6 / 2 * 3 can also be rewritten as 6 * 2⁻¹ * 3 and because multiplication is commutative you can now do it in any order for example like 3 * 6 * 2⁻¹

You can also "rearrange" the order without changing the meaning if you move the correct operation (left to the number) with it (should only be done with explicit multiplication)

6 / 2 * 3 into 6 * 3 / 2 (note that I moved the division with the 2)

You can even bring the two to the front. Just remember that left to the six is an "imaginary" (don't quote me ^^) multiplication. And because we can't just move "/2" to the beginning we have to insert a one (empty product - check Wikipedia) like so:

1 / 2 * 6 * 3

This also works for addition and subtraction

7 + 8 - 5

You can move them around if you take the operation left to the number with it. With addition the "imaginary" operation at the beginning is a plus sign and the implicit number you use is zero (empty sum - check Wikipedia)

8 - 5 + 7

or like this

0 - 5 + 8 + 7

because with negative numbers you can use the minus sign to indicate negative numbers you can even drop the leading zero like this

-5 + 8 + 7

That's not really possible with multiplication because "/2" is not a valid notation for "1/2"

[–] MBM@lemmings.world 4 points 11 months ago (1 children)

6 / 2 * 3

Semi-related: something in me wants to read that as 6 / (2*3), because 6 * 3 / 2 feels like a much more 'natural' way to write it

[–] SmartmanApps@programming.dev 1 points 8 months ago

Semi-related: something in me wants to read that as 6 / (2*3)

100% related actually, since that's the actual next line of working out. i.e. you cannot remove brackets unless there is only 1 term left inside, a mistake which those who have prematurely removed brackets have made and ended up with the wrong answer (because it flips the 3 from being in the denominator to being in the numerator).

[–] Brak@hexbear.net 2 points 11 months ago

this is beautiful but my brain glazed over when i saw so many numbers, back to eating glue for me!

[–] SmartmanApps@programming.dev 0 points 8 months ago

6 / 2 * 3 into 6 * 3 / 2 (note that I moved the division with the 2)

And note that it doesn't work if the multiply was an addition. e.g. 6/2+3=6 but 6+3/2=7.5. Multiplication and division are both binary operators, and you can't move them around unless you also move the term to the left with it. i.e. 6/2+3=6. 3+6/2=6.

Just remember that left to the six is an “imaginary” (don’t quote me ^^) multiplication

No, to the left of the 6 is an actual plus sign, but we don't write plus signs if it's at the start of an expression. +6 and x6 aren't the same thing at all (and, since x is a binary operator, you couldn't write just x6 anyway - there would have to be a term to it's left). No expression ever starts with x6.

That’s not really possible with multiplication because “/2” is not a valid notation for “1/2”

It's not a valid notation for multiplication either - both multiplication and division are binary operators and must be written with 2 terms.