this post was submitted on 03 Dec 2023
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[–] SmartmanApps 1 points 8 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] 0 points 8 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.

[–] SmartmanApps 1 points 8 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"?