Comment on I dunno
mindbleach@sh.itjust.works 23 hours agoBecause 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.
SmartmanApps@programming.dev 15 hours ago
I already answered, and I have no idea what your point is.
Nope. It’s me acknowledging they are both BRACKETS 🙄
E=mcc=(mxcxc) <== BRACKETS
a(b+c)=(ab+ac) <== BRACKETS
everything 😂
mindbleach@sh.itjust.works 15 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.
SmartmanApps@programming.dev 14 hours ago
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.
You know I’ve quoted dozens of books, right?
Again I have no idea what you’re talking about.
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.
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
mindbleach@sh.itjust.works 14 hours ago
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.