this post was submitted on 03 Dec 2023
420 points (99.8% liked)

196

16597 readers
1800 users here now

Be sure to follow the rule before you head out.

Rule: You must post before you leave.

^other^ ^rules^

founded 1 year ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 76 points 1 year ago (5 children)

In some countries we're taught to treat implicit multiplications as a block, as if it was surrounded by parenthesis. Not sure what exactly this convention is called, but afaic this shit was never ambiguous here. It is a convention thing, there is no right or wrong as the convention needs to be given first. It is like arguing the spelling of color vs colour.

[–] [email protected] 57 points 1 year ago (3 children)

This is exactly right. It's not a law of maths in the way that 1+1=2 is a law. It's a convention of notation.

The vast majority of the time, mathematicians use implicit multiplication (aka multiplication indicated by juxtaposition) at a higher priority than division. This makes sense when you consider something like 1/2x. It's an extremely common thing to want to write, and it would be a pain in the arse to have to write brackets there every single time. So 1/2x is universally interpreted as 1/(2x), and not (1/2)x, which would be x/2.

The same logic is what's used here when people arrive at an answer of 1.

If you were to survey a bunch of mathematicians—and I mean people doing academic research in maths, not primary school teachers—you would find the vast majority of them would get to 1. However, you would first have to give a way to do that survey such that they don't realise the reason they're being surveyed, because if they realise it's over a question like this they'll probably end up saying "it's deliberately ambiguous in an attempt to start arguments".

[–] [email protected] 26 points 1 year ago (4 children)

The real answer is that anyone who deals with math a lot would never write it this way, but use fractions instead

[–] [email protected] 6 points 1 year ago* (last edited 1 year ago) (2 children)

So are you suggesting that Richard Feynman didn't "deal with maths a lot", then? Because there definitely exist examples where he worked within the limitations of the medium he was writing in (namely: writing in places where using bar fractions was not an option) and used juxtaposition for multiplication bound more tightly than division.

Here's another example, from an advanced mathematics textbook:

Both show the use of juxtaposition taking precedence over division.

I should note that these screenshots are both taken from this video where you can see them with greater context and discussion on the subject.

[–] [email protected] 10 points 1 year ago (2 children)

Mind you, Feynmann clearly states this is a fraction, and denotes it with "/" likely to make sure you treat it as a fraction.

[–] [email protected] 10 points 1 year ago (13 children)

Yep with pen and paper you always write fractions as actual fractions to not confuse yourself, never a division in sight, while with papers you have a page limit to observe. Length of the bars disambiguates precedence which is important because division is not associative; a/(b/c) /= (a/b)/c. "calculate from left to right" type of rules are awkward because they prevent you from arranging stuff freely for readability. None of what he writes there has more than one division in it, chances are that if you see two divisions anywhere in his work he's using fractional notation.

Multiplication by juxtaposition not binding tightest is something I have only ever heard from Americans citing strange abbreviations as if they were mathematical laws. We were never taught any such procedural stuff in school: If you understand the underlying laws you know what you can do with an expression and what not, it's the difference between teaching calculation and teaching algebra.

load more comments (13 replies)
load more comments (1 replies)
load more comments (1 replies)
load more comments (3 replies)
[–] [email protected] 4 points 1 year ago (2 children)

So 1/2x is universally interpreted as 1/(2x), and not (1/2)x, which would be x/2.

Sorry but both my phone calculator and TI-84 calculate 1/2X to be the same thing as X/2. It's simply evaluating the equation left to right since multiplication and division have equal priorities.

X = 5

Y = 1/2X => (1/2) * X => X/2

Y = 2.5

If you want to see Y = 0.1 you must explicitly add parentheses around the 2X.

Before this thread I have never heard of implicit operations having higher priority than explicit operations, which honestly sounds like 100% bogus anyway.

You are saying that an implied operation has higher priority than one which I am defining as part of the equation with an operator? Bogus. I don't buy it. Seriously when was this decided?

I am no mathematics expert, but I have taken up to calc 2 and differential equations and never heard this "rule" before.

[–] [email protected] 3 points 1 year ago (1 children)

I can say that this is a common thing in engineering. Pretty much everyone I know would treat 1/2x as 1/(2x).

Which does make it a pain when punched into calculators to remember the way we write it is not necessarily the right way to enter it. So when put into matlab or calculators or what have you the number of brackets can become ridiculous.

[–] [email protected] 3 points 1 year ago (1 children)

I'm an engineer. Writing by hand I would always use a fraction. If I had to write this in an email or something (quickly and informally) either the context would have to be there for someone to know which one I meant or I would use brackets. I certainly wouldn't just wrote 1/2x and expect you to know which one I meant with no additional context or brackets

load more comments (1 replies)
load more comments (1 replies)
load more comments (1 replies)
[–] [email protected] 17 points 1 year ago (3 children)

BDMAS bracket - divide - multiply - add - subtract

[–] [email protected] 15 points 1 year ago (9 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.

load more comments (9 replies)
[–] [email protected] 9 points 1 year ago (3 children)

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

[–] [email protected] 9 points 1 year 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.

[–] [email protected] 1 points 1 year ago (2 children)

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

[–] [email protected] 3 points 1 year ago (1 children)

It’s commutative, not communicative, btw

[–] [email protected] 3 points 1 year ago

whoops, my bad

[–] [email protected] 2 points 1 year ago* (last edited 1 year 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.

[–] [email protected] 1 points 1 year ago

ok, i guess you're right

[–] [email protected] 1 points 1 year 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.

[–] [email protected] 2 points 1 year ago* (last edited 1 year 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.

[–] [email protected] 1 points 1 year ago (1 children)

You'll regularly see textbooks

That's what I said.

[–] [email protected] 1 points 1 year ago (1 children)

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

[–] [email protected] 1 points 1 year ago (1 children)

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

[–] [email protected] 1 points 1 year ago* (last edited 1 year 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.

[–] [email protected] 1 points 1 year 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.

[–] [email protected] 1 points 1 year 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).

[–] [email protected] 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.

[–] [email protected] 1 points 11 months ago

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

[–] [email protected] 7 points 1 year ago* (last edited 1 year 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

[–] [email protected] 7 points 1 year ago* (last edited 1 year 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

[–] [email protected] 3 points 1 year ago

4 would be correct since you go left to right.

[–] [email protected] 3 points 1 year ago

2nd one is correct, divisions first.

[–] [email protected] 2 points 1 year 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.

[–] [email protected] 2 points 1 year ago (2 children)
[–] [email protected] 4 points 1 year ago

BDSM Brackets ... ok

[–] [email protected] 2 points 1 year ago

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

[–] [email protected] 14 points 1 year ago (2 children)

I think when a number or variable is adjacent a bracket or parenthesis then it's distribution to the terms within should always take place before any other multiplication or division outside of it. I think there is a clear right answer and it's 1.

[–] [email protected] 13 points 1 year ago (2 children)

No there is no clear right answer because it is ambiguous. You would never seen it written that way.

Does it mean A÷[(B)(C)] or A÷B*C

[–] [email protected] 2 points 1 year ago* (last edited 1 year ago) (2 children)

It means

A ÷ B(C) which is equivalent to A ÷ (B*C)

I literally just explained this. The Parenthesis takes priority over multiplication and division outright.

Maybe
B*C = B(C)
But
A ÷ B(C) =! A ÷ B * C
load more comments (2 replies)
[–] [email protected] 3 points 1 year ago

It's 16, addition in bracket comes first

[–] SmartmanApps 1 points 8 months ago

Not sure what exactly this convention is called

It's 2 actual rules of Maths - Terms and The Distributive Law.

never ambiguous

Correct.

there is no right or wrong

Yes there is - obeying the rules is right, disobeying the rules is wrong.

load more comments (1 replies)