this post was submitted on 03 Dec 2023
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[–] And009@reddthat.com 17 points 11 months ago (3 children)

BDMAS bracket - divide - multiply - add - subtract

[–] Tlaloc_Temporal@lemmy.ca 15 points 11 months ago (2 children)

BEDMAS: Bracket - Exponent - Divide - Multiply - Add - Subtract

PEMDAS: Parenthesis - Exponent - Multiply - Divide - Add - Subtract

Firstly, don't forget exponents come before multiply/divide. More importantly, neither defines wether implied multiplication is a multiply/divide operation or a bracketed operation.

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

neither defines wether implied multiplication is a multiply/divide operation or a bracketed operation

The Distributive Law says it's a bracketed operation. To be precise "expand and simplify". i.e. a(b+c)=(ab+ac).

[–] And009@reddthat.com 0 points 11 months ago (2 children)

Exponents should be the first thing right? Or are we talking the brackets in exponents..

[–] Tlaloc_Temporal@lemmy.ca 4 points 11 months ago (1 children)

Exponents are second, parentheses/brackets are always first. What order you do your exponents in is another ambiguity though.

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

What order you do your exponents in is another ambiguity though

No it isn't - top down.

[–] Tlaloc_Temporal@lemmy.ca 1 points 8 months ago (1 children)

2^3^4 is ambiguous. 2^(3^4) is standard practice, but some calculators aren't that smart and will do (2^3)^4.

It's ambiguous because it works both ways, not because we don't have a standard. Confusion is possible.

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

The only confusion I can see is if you intended for the 4 to be an exponent of the 3 and didn't know how to do that inline, or if you did actually intend for the 4 to be a separate numeral in the same term? And I'm confused because you haven't used inline notation in a place that doesn't support exponents of exponents without using inline notation (or a screenshot of it).

As written, which inline would be written as (2^3)4, then it's 32. If you intended for the 4 to be an exponent, which would be written inline as 2^3^4, then it's 2^81 (which is equal to whatever that is equal to - my calculator batteries are nearly dead).

we don’t have a standard

We do have a standard, and I told you what it was. The only confusion here is whether you didn't know how to write that inline or not.

[–] Tlaloc_Temporal@lemmy.ca 1 points 8 months ago (1 children)

It's ambiguous because it works both ways, not because we don't have a standard.

Try reading the whole sentence. There is a standard, I'm not claiming there isn't. Confusion exists because operating against the standard doesn't immediately break everything like ignoring brackets would.

Just to make sure we're on the same page (because different clients render text differently, more ambiguous standards...), what does this text say?

2^3^4

It should say 2^3^4; "Two to the power of three to the power of four". The proper answer is 2⁸¹, but many math interpreters (including Excel, MATLAB, and many students) will instead compute 8⁴, which is quite different.

We have a standard because it's ambiguous. If there was only one way to do it, we'd just do that, no standard needed. You'd need to go pretty deep into kettle math or group theory to find atypical addition for example.

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

There is a standard, I’m not claiming there isn’t

Ah ok. Sorry, got caught out by a double negative in your sentence.

Confusion exists because operating against the standard doesn’t immediately break everything like ignoring brackets would

Ah but that's exactly the original issue in this thread - the e-calc is ignoring the rules pertaining to brackets. i.e. The Distributive Law.

Ah ok. Well that was my only confusion was what you had actually intended to write, not how to interpret it (depending on what you had intended). Yes should be interpreted 2^81.

including Excel

Yeah, but Excel won't let you put in a factorised term either. It's just severely broken because the people who wrote it didn't bother checking the rules of Maths first. Programmers not knowing the rules of Maths doesn't mean Maths is ambiguous (it certainly creates a lot of confusion though!).

We have a standard because it’s ambiguous. If there was only one way to do it, we’d just do that,

Disagree. There is one way to do it - follow the rules of Maths. That's why they exist. The order of operations rules are at least 400 years old, and make it not ambiguous. If people aren't obeying the rules then they're just wrong - that doesn't make it ambiguous. It's like saying if e-calcs started saying 1+1=3 then that must mean 1+1 is ambiguous. It might create confusion, but it doesn't mean the Maths is ambiguous.

[–] SmartmanApps@programming.dev 2 points 8 months ago

Brackets are ALWAYS first.

[–] CheesyFox@lemmy.world 9 points 11 months ago (3 children)

afair, multiplication was always before division, also as addition was before subtraction

[–] Pipoca@lemmy.world 9 points 11 months ago (2 children)

It's BE(D=M)(A=S). Different places have slightly different acronyms - B for bracket vs P for parenthesis, for example.

But multiplication and division are whichever comes first right to left in the expression, and likewise with subtraction.

Although implicit multiplication is often treated as binding tighter than explicit. 1/2x is usually interpreted as 1/(2x), not (1/2)x.

[–] unoriginalsin@lemmy.world 1 points 11 months ago (1 children)

It's BE(D=M)(A=S). Different places have slightly different acronyms - B for bracket vs P for parenthesis, for example.

But, since your rule has the D&M as well as the A&S in brackets does that mean your rule means you have to do D&M as well as the A&S in the formula before you do the exponents that are not in brackets?

But seriously. Only grade school arithmetic textbooks have formulas written in this ambiguous manner. Real mathematicians write their formulas clearly so that there isn't any ambiguity.

[–] Pipoca@lemmy.world 2 points 11 months ago* (last edited 11 months ago) (1 children)

That's not really true.

You'll regularly see textbooks where 3x/2y is written to mean 3x/(2y) rather than (3x/2)*y because they don't want to format

3x
----
2y

properly because that's a terrible waste of space in many contexts.

[–] unoriginalsin@lemmy.world 1 points 11 months ago (1 children)

You'll regularly see textbooks

That's what I said.

[–] Pipoca@lemmy.world 1 points 11 months ago (1 children)

You generally don't see algebra in grade school textbooks, though.

[–] unoriginalsin@lemmy.world 1 points 11 months ago (1 children)

12 is a grade. I took algebra in the 7th grade.

[–] Pipoca@lemmy.world 1 points 11 months ago* (last edited 11 months ago) (1 children)

Grade school is a US synonym for primary or elementary school; it doesn't seem to be used as a term in England or Australia. Apparently, they're often K-6 or K-8; my elementary school was K-4; some places have a middle school or junior high between grade school and high school.

[–] unoriginalsin@lemmy.world 1 points 11 months ago (1 children)

I don't know why you're getting lost on the pedantry of defining "grade school", when I was clearly discussing the fact that you only see this kind of sloppy formula construction in arithmetic textbooks where students are learning the basics of how to perform the calculations. Once you get into applied mathematics and specialized fields that use actual mathematics, like engineering, chemistry and physics, you stop seeing this style of formula construction because the ambiguity of the terms leads directly to errors of interpretation.

[–] Pipoca@lemmy.world 1 points 11 months ago (1 children)

Sure, the definition of grade school doesn't really matter too much. Because college texts are written in ways that violate pemdas.

Look, for example, at https://www.feynmanlectures.caltech.edu/I_45.html

For example, if f(x,y)=x2+yx, then (∂f/∂x)y=2x+y, and (∂f/∂y)x=x. We can extend this idea to higher derivatives: ∂^2f/∂y^2 or ∂^2f/∂y∂x. The latter symbol indicates that we first differentiate f with respect to x, treating y as a constant, then differentiate the result with respect to y, treating x as a constant. The actual order of differentiation is immaterial: ∂2f/∂x∂y=∂2f/∂y∂x.

Notice: ∂^2f/∂y∂x is clearly written to mean ∂^2f/(∂y∂x).

[–] unoriginalsin@lemmy.world 1 points 11 months ago* (last edited 11 months ago) (1 children)

What an interesting error to point out in support of pemdas.

Clearly the formatting of a paragraph of text in a textbook full of clearly and unambiguously written formulas discussing the very order of operations itself compared to the formatting of an actual formula diagram is going to be less clear. But here you've chosen to point to a discussion of why the order is irrelevant in the case under question.

Your example is the conclusion of a review of mathematics.

First we shall review some mathematics.

...

The actual order of differentiation is immaterial:

The fact that the example formula is written sloppy is irrelevant, because at no point is this going to be an actual formula meant to be solved, it's merely an illustration of why, in this case, the order of a particular operation is "immaterial".

Even if ∂^2f/∂y∂x is clearly written to mean ∂^2f/(∂y∂x), it doesn't matter because "∂2f/∂x∂y=∂2f/∂y∂x". So long as you're consistently applying pemdas, you're going to get the same answer whether you derive x first or y.

However, when it's time to discuss the actual formulas and equations being taught in the example text, clearly and unambiguously written formulas are illustrated as though copied from Ann illustration on a whiteboard instead of inserted into paragraphs that might have simply been transcribed from a lecture. Which, somewhat coincidentally, is exactly what your citation is.

[–] Pipoca@lemmy.world 1 points 11 months ago

Under PEMDAS, ∂2f/∂x∂y = (∂2f/∂x) * ∂y = ∂2∂y/∂x

[–] CheesyFox@lemmy.world 1 points 11 months ago (2 children)

a fair point, but aren't division and subtraction are non-communicative, hence both operands need to be evaluated first?

[–] EffortlessEffluvium@lemm.ee 3 points 11 months ago (1 children)

It’s commutative, not communicative, btw

[–] CheesyFox@lemmy.world 3 points 11 months ago

whoops, my bad

[–] Pipoca@lemmy.world 2 points 11 months ago* (last edited 11 months ago) (1 children)

1 - 3 + 1 is interpreted as (1 - 3) + 1 = -1

Yes, they're non commutative, and you need to evaluate anything in parens first, but that's basically a red herring here.

[–] CheesyFox@lemmy.world 1 points 11 months ago

ok, i guess you're right

[–] And009@reddthat.com 7 points 11 months ago* (last edited 11 months ago) (1 children)

~~Multiplication VS division doesn't matter just like order of addition and subtraction doesn't matter.. You can do either and get same results.~~

Edit : the order matters as proven below, hence is important

[–] prime_number_314159@lemmy.world 7 points 11 months ago* (last edited 11 months ago) (2 children)

If you do only multiplication first, then 2×3÷3×2 = 6÷6 = 1.

If you do mixed division and multiplication left to right, then 2×3÷3×2 = 6÷3×2 = 2×2 = 4.

Edit: changed whitespace for clarity

[–] And009@reddthat.com 3 points 11 months ago

2nd one is correct, divisions first.

[–] Johanno@feddit.de 2 points 11 months ago

4 would be correct since you go left to right.

[–] Squirrel@thelemmy.club 2 points 11 months ago

I was taught that division is just inverse multiplication, and to be treated as such when it came to the order of operations (i.e. they are treated as the same type of operation). Ditto with addition and subtraction.

[–] SamVergeudetZeit@feddit.de 2 points 11 months ago (2 children)
[–] answersplease77@lemmy.world 3 points 11 months ago

BDSM Brackets ... ok

[–] And009@reddthat.com 2 points 11 months ago

Glad to be of help, I remember it being taughy back in the 4th grade and it stuck well.