mindbleach comments on I dunno

[–] mindbleach@sh.itjust.works 0 points 10 months ago (21 children)

Because BRACKETS - ab=(axb) BY DEFINITION

“Parentheses must be introduced”!

But you understand E=mc^2^ does not mean E=(mxc)^2^.

This is you acknowledging that distribution and juxtaposition are only multiplication - and only precede other multiplication.

In your chosen Introduction To Algebra, Chrystal 1817, on page 80 (page 100 of the PDF you used), under Exercises XII, question 24 reads (x+1)(x-1)+2(x+2)(x+3)=3(x+1)^2^. The answer on page 433 of the PDF reads -2. If 3(x+1)^2^ worked the way you pretend it does, that would mean 3=9.

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[–] SmartmanApps@programming.dev 0 points 10 months ago* (last edited 10 months ago) (20 children)

But you understand E=mc2 does not mean E=(mxc)2

I already answered, and I have no idea what your point is.

This is you acknowledging that distribution and juxtaposition are only multiplication

Nope. It's me acknowledging they are both BRACKETS 🙄

E=mcc=(mxcxc) <== BRACKETS

a(b+c)=(ab+ac) <== BRACKETS

and only precede

everything 😂

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[–] mindbleach@sh.itjust.works 0 points 10 months ago* (19 children)

Then why doesn't the juxtaposition of mc precede the square?

In your chosen book is the example you're pestering moriquende for, and you can't say shit about it.

Another: Keys To Algebra 1-4's answer booklet, page 19, upper right: "book 2, page 9" expands 6(ab)^3^ to 6(ab)(ab)(ab), and immediately after that, expands (6ab)^3^ to (6ab)(6ab)(6ab). The bullshit you made up says they should be equal.

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[–] SmartmanApps@programming.dev 0 points 10 months ago* (last edited 10 months ago) (18 children)

Then why doesn’t the juxtaposition of mc precede the square?

For starters stop calling it "juxtaposition" - it's a Product/Term. Second, as I already told you, c²=cc, so I don't know why you're still going on about it. I have no idea what your point is.

In your chosen book

You know I've quoted dozens of books, right?

you can’t say shit about it

Again I have no idea what you're talking about.

expands 6(ab)3 to 6(ab)(ab)(ab)

Ah, ok, NOW I see where you're getting confused. 6ab²=6abb, but 6(ab)²=6abab. Now spot the difference between 6ab and 6(a+b). Spoiler alert - the latter is a Factorised Term, where separate Terms have been Factorised into 1 term, the former isn't. 2 different scenario's, 2 different rules relating to Brackets, the former being a special case to differentiate between 6ab² and 6a²b²=6(ab)²

P.S.

is like arguing 1+2 is different from 2+1 because 8/1+2 is different from 8/2+1

this is correct - 2+1 is different from 1+2, but (1+2) is identically equal to (2+1) (notice how Brackets affect how it's evaluated? 😂) - but I had no idea what you meant by "throwing other numbers on there", so, again, I have no idea what your point is

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[–] mindbleach@sh.itjust.works 0 points 10 months ago (17 children)

Juxtaposition is key to the bullshit you made up, you infuriating sieve. You made a hundred comments in this thread about how 2*(8)^2^ is different from 2(8)^2^. Here is a Maths textbook saying, you're fucking wrong.

Here's another: First Steps In Algebra, Wentworth 1904, on page 143 (as in the Gutenberg PDF), in exercise 54, question 9 reads (x-a)(2x-a)=2(x-b)^2^. The answer on page 247 is x=(2b^2^-a^2^)/(4b-3a). If a=1, b=0, the question and answer get 1/3, and the bullshit you've made up does not.

You have harassed a dozen people specifically to insist that 6(ab)^2^ does not equal 6a^2^b^2^. You've sassed me specifically to say a variable can be zero, so 6(a+b) can be 6(a+0) can just be 6(a). There is no out for you. This is what you've been saying, and you're just fucking wrong, about algebra, for children.

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[–] SmartmanApps@programming.dev 0 points 10 months ago* (last edited 10 months ago) (16 children)

Juxtaposition is key to the bullshit you made up

Terms/Products is mathematical fact, as is The Distributive Law. Maths textbooks never use the word "juxtaposition".

You made a hundred comments in this thread about how 2*(8)2 is different from 2(8)2

That's right. 1/2(8)²=1/256, 1/2x8²=32, same difference as 8/2(1+3)=1 but 8/2x(1+3)=16

Here is a Maths textbook saying, you’re fucking wrong

Nope! It doesn't say that 1/a(b+c)=1/ax(b+c). You're making a false equivalence argument

Here’s another:

Question about solving an equation and not about solving an expression. False equivalence again.

You have harassed a dozen people specifically to insist that 6(ab)2 does not equal 6a2b2

Nope! I have never said that, which is why you're unable to quote me saying that. I said 6(a+b)² doesn't equal 6x(a+b)², same difference as 8/2(1+3)=1 but 8/2x(1+3)=16

You’ve sassed me specifically to say a variable can be zero, so 6(a+b) can be 6(a+0) can just be 6(a).

That's right

There is no out for you

Got no idea what you're talking about

This is what you’ve been saying

Yes

you’re just fucking wrong

No, you've come up with nothing other than False Equivalence arguments. You're taking an equation with exponents and no division, and trying to say the same rules apply to an expression with division and no exponents, even though we know that exponent rule is a special case anyway, even if there was an exponent in the expression, which there isn't. 🙄

about algebra, for children

For teenagers, who are taught The Distributive Law in Year 7

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[–] mindbleach@sh.itjust.works 0 points 10 months ago (15 children)

I have never said that, which is why you’re unable to quote me saying that.

...

1/2(8)²=1/256

That's you saying it. You are unambiguously saying a(b)^c^ somehow means (ab)^c^=a^c^b^c^ instead of ab^c^, except when you try to nuh-uh at anyone pointing out that's what you said. Where the fuck did 256 come from if that's not exactly what you're doing?

You're allegedly an algebra teacher, snipping about terms I am quoting from a textbook you posted, and you wanna pretend 2(x-b)^2^ isn't precisely what you insist you're talking about? Fine, here's yet another example:

A First Book In Algebra, Boyden 1895, on page 47 (49 in the Gutenberg PDF), in exercise 24, question 18 reads, divide 15(a-b)^3^x^2^ by 3(a-b)x. The answer on page 141 of the PDF is 5(a-b)^2^x. For a=2, b=1, the question and answer get 5x, while the bullshit you've made up gets 375x.

Show me any book where the equations agree with you. Not words, not acronyms - an answer key, or a worked example. Show me one time that published math has said x(b+c)^n^ gets an x^n^ term. I've posted four examples to the contrary and all you've got is pretending not to see x(b+c)^n^ right fuckin' there in each one.

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[–] SmartmanApps@programming.dev 0 points 10 months ago* (14 children)

That’s you saying it

No it isn't! 😂 Spot the difference 1/2(8)²=1/256 vs.

6(ab)2 does not equal 6a2b2

You are unambiguously saying a(b)^c somehow means (ab)^c=a^c b^c

Nope. Never said that either 🙄

except when you try to nuh-uh at anyone pointing out that’s what you said

Because that isn't what I said. See previous point 😂

Where the fuck did 256 come from if that’s not exactly what you’re doing?

From 2(8)², which isn't the same thing as 2(ab)² 🙄 The thing you want it to mean is 2(8²)

snipping about terms I am quoting from a textbook you posted,

Because you're on a completely different page and making False Equivalence arguments.

you wanna pretend 2(x-b)2 isn’t precisely what you insist you’re talking about?

No idea what you're talking about, again. 2(x-b)2 is most certainly different to 2(xb)2, no pretense needed. you're sure hung up on making these False Equivalence arguments.

Show me any book where the equations agree with you

Easy. You could've started with that and saved all this trouble. (you also would've found this if you'd bothered to read my thread that I linked to)...

Thus, x(x-1) is a single term which is entirely in the denominator, consistent with what is taught in the early chapters of the book, which I have posted screenshots of several times.

I’ve posted four examples to the contrary

You've posted 4 False Equivalence arguments 🙄 If you don't understand what that means, it means proving that ab=axb does not prove that 1/ab=1/axb. In the former there is multiplication only, in the latter there is Division, hence False Equivalence in trying to say what applies to Multiplication also applies to Division

all you’ve got is

Pointing out that you're making a False Equivalence argument. You're taking examples where the special Exponent rule of Brackets applies, and trying to say that applies to expressions with no Exponents. It doesn't. 🙄 The Distributive Law always applies. The special exponent rule with Brackets only applies in certain circumstances. I already said this several posts back, and you're pretending to not know it's a special case, and make a False Equivalence argument to an expression that doesn't even have any exponents in it 😂

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[–] mindbleach@sh.itjust.works 0 points 10 months ago (13 children)

From 2(8)², which isn’t the same thing as 2(ab)²

a=8, b=1, it's the same thing.

False equivalence is you arguing about brackets and exponents by pointing to equations without exponents.

This entire thing is about your lone-fool campaign to insist 2(8)^2^ doesn't mean 2*8^2^, despite multiple textbook examples that only work because a(b)^c^ is a*b^c^ and not a^c^b^c^.

I found four examples, across two centuries, of your certain circumstances: addition in brackets, factor without multiply symbol, exponent on the bracket. You can't pivot to pretending this is a division syntax issue, when you've explicitly said 2(8)^2^ is (2*8)^2^. Do you have a single example that matches that, or are you just full of shit?

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[–] SmartmanApps@programming.dev 1 point 10 months ago* (last edited 10 months ago) (12 children)

a=8, b=1, it’s the same thing

No it isn't! 😂 8 is a single numerical factor. ab is a Product of 2 algebraic factors.

False equivalence is you arguing about brackets and exponents

Nope. I was talking about 1/a(b+c) the whole time, as the reason the Distributive Law exists, until you lot decided to drag exponents into it in a False Equivalence argument. I even posted a textbook that showed more than a century ago they were still writing the first set of Brackets. i.e. 1/(a)(b+c). i.e. It's the FOIL rule, (a+b)(c+d)=(ac+ad+bc+bd) where b=0, and these days we don't write (a)(b+c) anymore, we just write a(b+c), which is already a single Term, same as (a+b)(c+d) is a single term, thus doesn't need the brackets around the a to show it's a single Term.

3(x-y) is a single term...

This entire thing is about your lone-fool campaign

Hilarious that all Maths textbooks, Maths teachers, and most calculators agree with me then, isn't it 😂

insist 2(8)2 doesn’t mean 2*82,

Again, you lot were the ones who dragged exponents into it in a False Equivalence argument to 1/a(b+c)

I found four examples, across two centuries

None of which relate to the actual original argument about 1/a(b+c)=1/(ab+ac) and not (b+c)/a

You can’t pivot to pretending this is a division syntax issue

I'm not pivoting, that was the original argument. 😂 The most popular memes are 8/2(2+2) and 6/2(1+2), and in this case they removed the Division to throw a curve-ball in there (note the people who failed to notice the difference initially). We know a(b+c)=(ab+ac), because it has to work when it follows a Division, 1/a(b+c)=1/(ab+ac). It's the same reason that 1/a²=1/(axa), and not 1/axa=a/a=1. It's the reason for the brackets in (ab+ac) and (axa), hence why it's done in the Brackets step (not the MULTIPLY step). It's you lot trying to pivot to arguments about exponents, because you are desperately trying to separate the a from a(b+c), so that it can be ax(b+c), and thus fall to the Multiply step instead of the Brackets step, but you cannot find any textbooks that say a(b+c)=ax(b+c) - they all say a(b+c)=(ab+ac) - so you're trying this desperate False Equivalence argument to separate the a by dragging exponents into it and invoking the special Brackets rule which only applies in certain circumstances, none of which apply to a(b+c) 😂

2(8)2 is (2*8)2.

That's right

are you just full of shit?

says the person making False Equivalence arguments. 🙄 Let me know when you find a textbook that says a(b+c)=ax(b+c), otherwise I'll take that as an admission of being wrong that you keep avoiding the actual original point that a(b+c) is a single Term, as per Maths textbooks

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[–] mindbleach@sh.itjust.works 0 points 10 months ago* (11 children)

3(x-y) is a single term…

So is 3xy, according to that textbook. That doesn't mean 3xy^2^ is 9*y^2^*x^2^. The power only applies to the last element... like how (8)2^2^ only squares the 2.

Four separate textbooks explicitly demonstrate that that's how a(b)^c^ works. 6(ab)^3^ is 6(ab)(ab)(ab), not (6ab)(6ab)(6ab). 3(x+1)^2^ for x=-2 is 3, not 9. 15(a-b)^3^x^2^ doesn't involve coefficients of 3375. 2(x-b)^2^ has a 2b^2^ term, not 4b^2^. If any textbook anywhere shows a(b)^c^ producing (ab)^c^, or x(a-b)^c^ producing (xa-xb)^c^, then reveal it, or shut the fuck up.

2(ab)^2^ is 2(ab)(ab) the same way 6(ab)^3^ is 6(ab)(ab)(ab). For a=8, b=1, that's 2*(8*1)*(8*1).

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[–] SmartmanApps@programming.dev -1 points 10 months ago* (last edited 10 months ago) (10 children)

So is 3xy

That's right

That doesn’t mean 3xy2 is 9y2x2.

That's right. It means 3abb=(3xaxbxb)

The power only applies to the last element

Factor yes, hence the special rule about Brackets and Exponents that only applies in that context

like how (8)22 only squares the 2

It doesn't do anything, being an invalid syntax to follow brackets immediately with a number. You can do ab, a(b), but not (a)b

not (6ab)(6ab)(6ab)

Yep, as opposed to 6(a+b), which is (6a+6b)

3(x+1)2 for x=-2 is 3, not 9

No it isn't. See previous point. Do we have an a(b+c), yes we do. Do we have an a(bc)²? No we don't.

2(x-b)2 has a 2b2 term

No, it has a a(b-c) term, squared

shut the fuck up

says someone still trying to make the special case of Exponents and Brackets apply to a Factorised Term when it doesn't. 😂 I'll take that as your admission of being wrong about a(b+c)=ax(b+c) then. Thanks for playing

For a=8, b=1, that’s 2*(81)(8*1).

Only if you had defined it as such to begin with, otherwise the Brackets Exponents rule doesn't apply if you started out with 2(8)², which is different to 2(8²) and 2(ab)²

..and there's still no exponent in a(b+c) anyway, Mr. False Equivalence

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[–] mindbleach@sh.itjust.works 0 points 10 months ago (1 child)

We can see the Acrobat window in those scans you found online.

You think 2(8)^2^ is 128 if that's simplified from 2(8+0)^2^... but 256 if it's simplified from 2(8*1)^2^. In short: no.

I think you're about fifteen years old. You had an unpleasant teacher who belittled you, and you've identified with the aggressor. Your whole online persona is posturing to always be smarterer than everybody else, even if that means saying Wolfram fucking Alpha is wrong about basic algebra.

Faced with a contradiction that requires you to insist (8*1) ≠ (8+0), you're going to type laughter and spam emojis as if that inspires any reaction besides pity. The word you should be looking for is, "oops."

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[–] SmartmanApps@programming.dev 0 points 10 months ago

We can see the Acrobat window in those scans you found online.

...and the ones that came with the textbook, and not in the photo's 😂

You think 2(8)2 is 128 if that’s simplified from 2(8+0)2… but 256 if it’s simplified from 2(8*1)2. In short: no

says person who can't tell the difference between a(b+c) and a(bc) 😂

I think you’re about fifteen years old. You had an unpleasant teacher who belittled you, and you’ve identified with the aggressor. Your whole online persona is posturing to always be smarterer than everybody else

#EveryAccusationIsAConfession

Wolfram fucking Alpha is wrong about basic algebra

Which is an established fact

Faced with a contradiction that requires you to insist (8*1) ≠ (8+0)

again says person who can't tell the difference between a(b+c) and a(bc) 😂

The word you should be looking for is, “oops.”

yet again says person who can't tell the difference between a(b+c) and a(bc) 😂

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[–] mindbleach@sh.itjust.works 0 points 10 months ago (1 child)

who mysteriously owns dozens of Maths textbooks

PDFs found online. From which you are ignoring counterexamples using a(b+c)^n^. Fraud.

Go ahead and tell me how you would explain what 3(x+y) means without referring to Multiplication?

Your own spammed screenshots say 3 gets multiplied.

There is no special case. You made it up. 8+0 equals 8 (or sorry, does it just mean 8?) so 2(8+0)^2^ is the same as 2(8)^2^. The latter is the next step in simplifying the former. You've admitted simplifying first is valid, when your nose was rubbed in your own found PDFs doing exactly that.

You don't have an opinion. You make no claim, anymore. All you have left are derision and emojis. You've admitted 2(8*1)^2^ means 2(8)(8), and insist that's different from 2(8)^2^ because... ibid. You cannot explain it even now.

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[–] SmartmanApps@programming.dev 0 points 10 months ago* (last edited 10 months ago)

PDFs found online

Nope! If you looked more carefully you'll find some of them are photo's and scans. You also seem to be forgetting that every modern textbook comes with a PDF as well 🙄

From which you are ignoring counterexamples using a(b+c)n. Fraud

says the actual fraud who keeps ignoring that there is no exponent in a(b+c) 😂

Go ahead and tell me how you would explain what 3(x+y) means without referring to Multiplication?

Your own spammed screenshots say 3 gets multiplied

So in other words, you weren't able to. Also, it doesn't say that - well done on missing the point for a third time in a row 😂

There is no special case

So you think 2(3x4x5)=(2x3x2x4x2x5) is totally fine? BWAHAHAHAHAHAHAAHA 😂😂😂

You made it up

Weird then that it's in Maths textbooks isn't it, that 2(3x4x5) is in fact only equal to 2(60)? 😂

does it just mean 8?

says person showing they don't know the difference in meaning between "means" and "equals" 😂

so 2(8+0)2 is the same as 2(8)2

Yep.

The latter is the next step in simplifying the former

Yep.

You’ve admitted simplifying first is valid

Yep.

when your nose was rubbed in your own found PDFs doing exactly that

Nope! When you finally discovered that they were both valid, even though only a couple of textbooks I posted specifically said to Distribute first. We in fact teach students to simplify before Distributing - less working out, less mistakes with signs.

You don’t have an opinion

That's right, just facts, as per Maths textbooks 😂

You make no claim, anymore

a(b+c)=(ab+ac), same thing I've been saying the whole time

All you have left are derision and emojis

and facts

You’ve admitted 2(8*1)2 means 2(8)(8)

Nope! Never said anything of the sort, liar. I have said the whole time that Multiplication is a special case, to which you claimed there was no special case.

insist that’s different from 2(8)2 because…

No Multiplication. It's not complicated 🙄

You cannot explain it even now

I already did. Not my fault you don't understand the difference between Addition and Multiplication 😂

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[–] mindbleach@sh.itjust.works 0 points 10 months ago* (1 child)

Your bullshit hit max comment depth.

That would mean 2(8*1)^2^ is 128

That’s right,

So when you said 2(8)^2^ is 256, you were wrong.

Otherwise - walk me through how 2(8*1)^2^, 2(8+0)^2^, and 2(8)^2^ aren't equal, alleged math teacher.

Tell me how a(b)^n^ gets a different answer when you know the values. Gimme the primo bullshit.

None of them have said a(b+c)=ax(b+c)

“3(x+y) means 3*(x+y).”

means not equals

"It depends on what the definition of is, is," says someone definitely not trapped in a contradiction.

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[–] SmartmanApps@programming.dev 0 points 10 months ago

Your bullshit hit max comment depth.

That's hilarious that you're calling textbooks "bullshit" 🤣🤣🤣 BTW there's nothing preventing you from addressing comments made in a different post to the one you're replying to, 🙄 and yet, yet again, you didn't. Did you work out yet why we don't write (a+b)c? It's all in the post you're avoiding.

So when you said 2(8)2 is 256, you were wrong

Nope. 2(8*1)² has a Multiplication inside the Brackets, so The Distributive Law does not apply, 2(8)² doesn't have Multiplication in it, so The Distributive Law does apply. As I've already said repeatedly, if you wanted 2x8², then you could've just written 2x8². If you've written 2(8) rather then 2x8, then you are saying this is a Product, not a Multiplication.

Otherwise - walk me through how 2(8*1)2, 2(8+0)2, and 2(8)2 aren’t equal, alleged math teacher.

I already did multiple times. The first one has Multiplication in it, the other two don't. Multiplication (and Division) is the special case where The Distributive Law does not apply, because you cannot Distribute over Multiplication, only Addition (and Subtraction)

alleged math teacher.

who mysteriously owns dozens of Maths textbooks, many of which quoted in the post you're avoiding 😂

None of them have said a(b+c)=ax(b+c)

“3(x+y) means 3*(x+y).”

Yep, doesn't say equals, exactly as I said 🙄 Congratulations on missing the point a second time in a row. You wanna go for three?

“It depends on what the definition of is, is,”

You think "means" and "equals" are the same word?? BWAHAHAHAHAHAHA! 🤣🤣🤣 You know the language has to get dumbed down to Year 7 level, right? And you're still missing the point, right? 😂 Go ahead and tell me how you would explain what 3(x+y) means without referring to Multiplication? I'll wait. BTW I'll point out yet again That the questions on Page 282, answers on Page 577, prove I am the one interpreting this right. Maths teacher understands Maths textbook language better than someone who isn't a Maths teacher. Who woulda thought?? 😂

says someone definitely not trapped in a contradiction

Yep, I'm definitely not trapped in a contradiction. 🤣🤣🤣 Look at the questions on Page 282, answers on Page 577, and then ask yourself what you think they meant when they said means, 😂and not equals. There is definitely a specific reason they did not say equals

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[–] mindbleach@sh.itjust.works 0 points 10 months ago* (1 child)

So you’re saying there no such rule as 2(ab)²=2a²b².

That would mean 2(8*1)^2^ is 128. You are the one saying it's not 2a^2^b^2^, because you think it's 2^2^a^2^b^2^, and that 2(8*1)^2^ is 256. I'm not touching anything without exponents because exponents are where you are blatantly full of shit.

if it has been written as 2(ab)², not if it has been written a(b+c)²

Source: your ass. Every published example disagrees, and you just go, nuh-uh, that up-to-date Maths textbook must be wrong. You alone are correct on this accursed Earth.

Hey look, another one of the textbooks you insist I read says you're completely wrong: "The multiplication sign is often not included between letters, e.g. 3ab means 3 * a * b." Page 31 of the PDF... right above where you've dishonestly twisted the "expanding brackets" text. Next page: "3(x+y) means 3*(x+y)."

Page 129 of that PDF, exercise 5, question 14: simplify 2(e^4^)^2^. The answer on PDF page 414 is 2e^8^. Your bullshit would say 4e^8^.

Right below that, exercise 5*, question 4: 4(4^4^)^4^. The answer on PDF page 414 is 1.72x10^10^. The bullshit you've made up would be 1.10x10^11^. 5* questions 7, 9, 10, and 11 also have the same a(b)^c^ format as 2(8)^2^, if you somehow need further proof of how this actually works.

PDF page 134, exam practice question 10a, simplify 3(q^2^)^2^. PDF page 415 says 3q^6^. Your bullshit says 9q^6^.

Damn dude, that's five textbooks you chose saying you're full of shit, and zero backing you up. One more and I get a free hoagie. Your bullshit has brought us to max comment depth.

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[–] SmartmanApps@programming.dev 0 points 10 months ago

Firstly, it's hilarious that you've gone back to a previous comment, thus ignoring the dozen textbook references I posted 😂

That would mean 2(8*1)2 is 128

That's right, because we don't Distribute over Multiplication (and Division), only Addition and Subtraction (it's right there in the Property's name - The Distributive Property of Multiplication over Addition). Welcome to you proving why a(bc)² is a special case 😂 I've been telling you this whole time that a(b+c) and a(bc) aren't the same, and you finally stumbled on why they aren't the same 😂

You are the one saying it’s not 2a2b2,

No I'm not. I never said that, liar. I've been telling you the whole time that it is a special case 🙄 (upon which you claimed there was no special case)

because you think it’s 22a2b2,

No I don't. That's why you can't quote me ever saying that 🙄

exponents are where you are blatantly full of shit

and there are no exponents in a(b+c) and all this stuff about exponents is you being blatantly full of shit 🙄

Source: your ass.

No, this meme

Notice that there are no exponents? 😂

Every published example disagrees

says person who came back to this post to avoid this post which is full of published examples that agree with me - weird that 😂

that up-to-date Maths textbook must be wrong

And I also pointed out why that was wrong here. i.e. the post that you have avoided replying to 😂

You alone are correct on this accursed Earth

No, all textbooks as well, except those which are using the old-fashioned and wrong syntax of (a+b)c, not to mention most calculators as well (only Texas Instruments is still doing it wrongly).

Page 31 of the PDF… right

Before the pages I already posted in the post that you are avoiding replying to 😂

where you’ve dishonestly twisted the “expanding brackets” text. Next page: “3(x+y) means 3*(x+y).”

means not equals, Mr. Person Who Is Actually Dishonestly Twisting The Words, as proven by the exercises on Page 282, answers on Page 577, which are also in the post that you are avoiding replying to 😂

Page 129 of that PDF, exercise 5, question 14: simplify 2(e4)2. The answer on PDF page 414 is 2e8

That's right

Your bullshit would say 4e8.

Nope. Been telling you the whole time that is a special case, upon which you claimed there was no such special case 😂

if you somehow need further proof of how this actually works

No, I don't, it's still a False Equivalence argument 🙄 But if you wanna waste your time on an irrelevant point (which you seem determined to do), go ahead, don't let me stop you, but that's an admission that you are wrong about a(b+c)

Damn dude, that’s five textbooks you chose saying you’re full of shit

Nope! None of them have said a(b+c)=ax(b+c), they have all said a(b+c)=(ab+ac), which is why you're avoiding replying to the post of mine which quotes them all 🙄

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[–] mindbleach@sh.itjust.works 0 points 10 months ago* (1 child)

There is no special case. You made it up by confusing yourself about "dismissing a bracket." To everyone else in the world, brackets are just another term. Several of the textbooks I've linked will freely juxtapose brackets and variables before or after, because it makes no difference.

Here's yet another example, PDF page 27: (6+5)x+(-2+10)y. And that's as factorization. This Maths textbook you plainly didn't read was published this decade. Still waiting on any book ever that demonstrates your special bullshit.

7bx with b=(m+n) becomes 7(m+n)x and it's the same damn thing. Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7. Brackets only happen first because they have to be reduced to a single term. A bracket with one number is not "unsolved" - it's one number. Squaring a bracket with one number is squaring that number.

The base of an exponent is whatever's in the symbols of inclusion. Hence: 6(ab)^3^ = 6(ab)(ab)(ab).

No, it has a a(b-c) term, squared

It has a (b-c) term, squared. The base of an exponent is whatever's in the symbols of inclusion. See page 121 of 696, in the Modern Algebra: Structure And Method PDF you plainly got from Archive.org. "In an expression such as 3a^2^, the 2 is the exponent of the base a. In an expression such as (3a)^2^, the 2 is the exponent of the base 3a, because you enclosed the expression in a symbol of inclusion." You will never find a published example that makes an exception for distribution first.

On the page before your screenshot - 116 of 696 - this specific Maths textbook refers to both 8x7 and 8(7) as "symbols of multiplication." It's just multiplication. It's only ever multiplication. It's not special, you crank. 8(7) is a product identical to 8x7. Squaring either factor only squares that factor.

Only if you had defined it as such to begin with

Variables don't work differently when you know what they are. b=1 is not somehow an exception that isn't allowed, remember?

There's an exponent in 2(8)^2^ and it concisely demonstrates to anyone who passed high school that you can't do algebra.

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[–] SmartmanApps@programming.dev 0 points 10 months ago*

There is no special case

So you're saying there no such rule as 2(ab)²=2a²b². Got it. you're admitting you're wrong then 😂

You made it up by confusing yourself about “dismissing a bracket.”

Says person who just claimed there's no such rule as the one they've been making the basis of their wrong claims 😂

To everyone else in the world, brackets are just another term

That's right. That's why you cannot separate the coefficient from it 😂

Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after

Several of your textbooks are outdated then

because it makes no difference

So is (2+3)-4 equal to -20 or 1? I'll wait

And that’s as factorization.

Wrong Factorisation. ab+ac=a(b+c)

This Maths textbook you plainly didn’t read was published this decade

And yet, is still wrong

Still waiting on any book ever that demonstrates your special bullshit

You were the one who just said there's no such special rule as 2(ab)²=2a²b² 😂

7bx with b=(m+n) becomes

7x(m+n)

and it’s the same damn thing

No, 7(m+n)x is invalid syntax due to ambiguity when x is negative.

Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7

That's right. I never said otherwise.

Brackets only happen first because they have to be reduced to a single term

They happen first because they already are a single Term...

A bracket with one number is not “unsolved”

Yes it is! If you haven't solved Brackets then you cannot progress onto Exponents.

it’s one number

Exactly! That's why you cannot separate the a from (b+c) - it's all ONE NUMBER 😂

Squaring a bracket with one number is squaring that number.

if it has been written as 2(ab)², not if it has been written a(b+c)². Also, again, there is no exponent in a(b+c), so that rule doesn't apply anyway 😂

Hence: 6(ab)3

6(ab)² <== note: not 6(a+b)², nor 6(a+b) for that matter. You're still desperately trying to make a False Equivalence argument 😂

It has a (b-c) term

No it doesn't. a(b-c) is one term

The base of an exponent is whatever’s in the symbols of inclusion.

And there's no exponent in a(b+c) 🙄

See page 121 of 696

See page 37

Also see page 282 and answers on page 577. x(x-1) is one term, as taught on page 37

In an expression such as 3a2,

3a² <== note: not 3(a+b), which has no exponent 🙄 It's hilarious how much effort you're putting into such an obvious False Equivalence argument 😂

You will never find a published example that makes an exception for distribution first

BWAHAHAHAHAHAHAH! I see you haven't read ANY of the sources I've posted so far then 🤣

refers to both 8x7 and 8(7) as “symbols of multiplication.”

Yep, present tense and past tense, since 8(7) is a Product, a number, as per Pages 36 and 37, which you're still conveniently ignoring, despite me having posted it multiple times

It’s just multiplication

Nope. 8(7) is a Product, a single Term, as per Pages 36 and 37. You won't find them writing 8(7)=8x7 anywhere in the whole book, always 8(7)=56, because it's a number

It’s not special, you crank

Yes it is, as per the textbook you're quoting from! 🤣

8(7) is a product identical to 8x7

Nope! 8(7) is a number, as per the textbook you are - selectively - quoting from 😂 8(7) is one term, 8x7 is two terms, as per Page 37

Squaring either factor only squares that factor

Where there are multiple pronumeral factors and you need the brackets to specify which factors the square is applying to, which, again, none of which applies to a(b+c) anyway Mr. False Equivalence 😂

Variables don’t work differently when you know what they are.

They're not variables if you know what they are - they are constants, literally a number as per pages 36 and 37

b=1 is not somehow an exception that isn’t allowed, remember?

That's right. a(1+c)=(a+ac), and you're point is??

There’s an exponent in 2(8)2

and no Pronumerals, and 2(8) is a number as per pages 36 and 37. 🙄And if you had read those pages, you would find it also tells you why you cannot write 2(8) as 28 (in case it's not obvious). if you want 2x8², then you can just write 2x8² 🙄

it concisely demonstrates to anyone who passed high school that you can’t do algebra

says someone who is trying to say that a rule about exponents applies to expressions without exponents 🤣

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