Yes, they do if it's 2x, but not if it's 2(2+2) - despite them mathematically being the same thing - leading to wrong answers to expressions such as the OP. In fact, that's true of every e-calculator I've ever seen, except for MathGPT (Desmos used to handle it correctly, but then they made a change to make it easier to enter fractions, and consequently broke evaluating divisions correctly).

how do you calculate 2(1+1)? It’s 4. It’s called implicit multiplication

No, it's not called implicit multiplication. It's distribution.

We could write this up as +2*(+1+(+1))

No, you can't. Adding that multiplication has broken it up into 2 terms. You either need to not add the multiply, or add another set of brackets if you do, to keep it as 1 term.

I can’t think of a situation where it’s incorrect

If a=2 and b=3, then...

1/axb=3/2

1/ab=(1/6)

If there is a letter or number or anything next to a bracket, it’s multiplication

No, it's distribution. Multiplication refers literally to multiplication signs, of which there aren't any in this expression.

2x is the same as 2*x

No, 2A is the same as (2xA). i.e. it's a single Term. 2xA is 2 Terms (multiplied).

If a=2 and b=3, then...

axb=2x3 (2 terms)

ab=6 (1 term)

This guy talks in hashtags.

Only in the first post in each thread, so that people following those hashtags will see the first post, and can then click on it if they want to see the rest of the thread. Also "this guy" is me. :-)

Grab a calculator, look at wolfram docs, ask a professor or teacher

I'm a Maths teacher with a calculator and many textbooks - I'm good. :-) Also, if you'd clicked on the thread you would've found textbook references, historical Maths documents, proofs, the works. :-)

8/2(2+2), let’s remove the confusion

8/2*(2+2), brackets

8/2*(4), mult & div, left -> right

4*(4), let go

2 mistakes here. Adding the multiplication sign in the 2nd step has broken up the term in the denominator, thus sending the (2+2) into the numerator, hence the wrong answer (and thus why we have a rule about Terms). Then you did division when there was still unsolved brackets left, thus violating order of operations rules.

it’s more like 8/(2x), but it’s just harder to read, so we don’t

But that's exactly what we do (but no extra brackets needed around 2x nor 2(2+2) - each is a single term).

you could argue that it should be handled like the scenario with the x

Which is what the rules of Maths tells us to do - treat a single term as a single term. :-)

there are no rules of this

Yeah, there is. :-)

you use fractions instead of inline division

No, never. A fraction is a single term (grouped by a fraction bar) but division is 2 terms (separated by the division operator). Again it's the definition of Terms.

And by all means, correct me if I’m wrong

Have done, and appreciate the proper conversation (as opposed to those who call me names for simply pointing out the actual rules of Maths).

link something that isn’t an unreadable

No problem. I t doesn't go into as much detail as the Mastodon thread though, but it's a shorter read (overall - with the Mastodon thread I can just link to specific parts though, which makes it handier to use for specific points), just covering the main issues.

as long as we are civilised

Thanks, appreciated.

[–] [email protected] 1 points 7 months ago (1 children)

Idk where you teach, but I'm thankful you didn't teach me.

Let me quizz you, how do you solve 2(2+2)^2? because acording to your linked picture, because brackets are leftmost you do them first. If I were to believe you:

(2*2+2*2)^2
(4+4)^2, = 64

but it's just simply incorrect.

2(4)^2, wow we're at a 2x^2
2*16 = 32

The thing that pisses me off most, is the fact that, yes. Terms exists, yes they have all sorts of properties. But they are not rules, they are properties. And they only apply when we have unknows and we're at the most simplified form. For example your last link, the dude told us that those terms get prio because they are terms!? There are no mention of term prio in the book. It just simply said that when we have a simplified expression like: 2x^2+3x+5 we call 2x^2 and 3x and 5 terms. And yes they get priority, not because we named them those, but because they are multiplications. These help us at functions the most. Where we can assume that the highest power takes the sign at infinity. Maybe if the numbers look right, we can guess where it'd switch sign.

I don't even want to waste energy proofreading this, or telling you the obvious that when we have a div. and a mult. and no x's there really is no point in using terms, as we just get a single number.

But again, I totally understand why someone would use this, it's easier. But it's not the rule still. That's why at some places this is the default. I forgot the name/keywords but if you read a calculator's manual there must be a chapter or something regarding this exact issue.

So yeah, use it. It's good. Especially if you teach physics. But please don't go around making up rules.

As for your sources, you still linked a blog post.

[–] [email protected] 1 points 7 months ago (1 children)

because brackets are leftmost you do them first

No, not because leftmost (did I say leftmost? No, I did not), because brackets. Brackets are always first in order of operations.

2(4)^2, wow we’re at a 2x^2

No, we're at x^2, because 2(4) is a bracketed term, and order of operations rules is brackets before exponents, and to solve the brackets we have to distribute the 2, so 2(4)^2=(2x4)^2=8^2=64.

all sorts of properties. But they are not rules

Depends. The Distributive Property is a property, but The Distributive Law is a rule. Properties explain how/why things work, but rules have to be obeyed if you want to get the right answer. Terms is a rule, based on properties (similarly, The Distributive Law is a rule, which makes use of the Distributive Property).

they only apply when we have unknows

Are you referring to pronumerals? Textbooks are quite explicit that the same rules apply to pronumerals as to numerals (since pronumerals literally stand-in for numerals).

terms get prio because they are terms!?

Not priority, they are already fully solved because they are terms. If we have 2a, then there's literally nothing to be done (except substitute a value for a if you've been told what it is). 2xa on the other hand needs to be multiplied (2 terms separated by a multiplication).

Noted that you ignored where I pointed out why it makes a difference

There are no mention of term prio in the book.

Which book? I don't know what you're talking about now.

we have a simplified expression

AKA Terms. And Terms are not expressions. Expressions are defined as being made up of Terms and operators. See previous textbook screenshot. 2a is a Term, 2xa is an Expression. And yes, you are right that a Term is a simplified expression, and being simplified, there is no further simplification to be done.

2x^2+3x+5 we call 2x^2 and 3x and 5 terms. And yes they get priority, not because we named them those, but because they are multiplications

No, they are Terms. There is no multiplication. Multiplication refers literally to multiplication symbols. A Term is a product. i.e. the result of a multiplication. That's why they don't have multiplication symbols in them - it has already been done.

using terms, as we just get a single number

EXACTLY!! When a=2 and b=3, ab=6, a single number. AKA a Term.

I totally understand why someone would use this, it’s easier

We use it because that's how Maths works, and is a rule taught in all the textbooks, and has been for more than a century.

I forgot the name/keywords but if you read a calculator’s manual there must be a chapter or something regarding this exact issue.

The name is Term. You can read about this exact issue in Maths textbooks.

Especially if you teach physics

I teach Maths, on which much of Physics is built.

As for your sources, you still linked a blog post

In other words, you didn't even read it. The sources are in it - there are Maths textbooks in it.

[–] [email protected] 1 points 7 months ago (1 children)

Alright there buddy, I'd like to close this.

It's clear that your a troll. However, on the offchance that you didn't know, I'll tell you where you went wrong on the first one.

2(4)^2=(2x4)^2=8^2=64

You can't distribute into a bracket, that's raised to the power of anything other than 1, like this. To do this you need to raise distributed number to the bracket's power's inverse. in this case 1/2.

2(4)^2=(2^(1/2)*4)^2=(sqrt(2)*4)^2=2*4^2=2*16=32
or y'know 2*16=32

Maybe if we look at it with roots you'd get it. wolfram syntax

2(4)^2=2Surd[4,1/2]
2Surd[4,1/2]= Surd[4*2^(1/2),1/2]= (4*sqrt(2))^2= 4^2*2= 16*2= 32

I hope you don't get scared from this math, you're a teacher afterall. I have no Idea how you could have gotten a degree or not kicked from school on day 1. Unless.. you are trolling me, fuck you for that. If you respond with more bullshiting, I'll block you.

[–] [email protected] 1 points 7 months ago

2(4)^2=(2x4)^2=8^2=64

Yes, that's right. Brackets before Exponents, as per the order of operations rules.

You can’t distribute into a bracket

You know that's literally what The Distributive Law says you must do, right? Unless you have a source somewhere saying there's an exception?

Apparently you didn't bother reading any of the links I gave you, so here's one of the many textbooks which says you must distribute...

In case that's unclear, that means that 2x² and 2(x)² aren't the same thing (since 2(x)=(2x) by definition).

wolfram syntax

You know Wolfram disobeys The Distributive Law, right? I know I'm not the only one who knows this. Is that why you're insisting your way is right? Cos they're known to be wrong about this.

If you respond with more bullshiting,

You call quoting Maths textbooks "bullshiting"?