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SmartmanApps@programming.dev ⁨23⁩ ⁨hours⁩ ago

This is your own source - and it says, juxtaposition is just multiplication

inside brackets. Don’t leave out the inside brackets that they have specifically said you must use - “Parentheses must be introduced”! 🤣 BTW, this is a 19th Century textbook, from before they started calling them PRODUCTS 🙄

E=mc2 is E=(mc)2

No, it means E=mc² is E=mcc=(mxcxc)

Throwing other numbers on there

I have no idea what you’re talking about 🙄

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mindbleach@sh.itjust.works ⁨21⁩ ⁨hours⁩ ago

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 ⁨13⁩ ⁨hours⁩ ago
  
  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 ⁨13⁩ ⁨hours⁩ ago
    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: 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.
    
    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 ⁨13⁩ ⁨hours⁩ ago
    
    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 latter 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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