Comment on I dunno
SmartmanApps@programming.dev 13 hours agoBut 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 😂
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.
SmartmanApps@programming.dev 13 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 12 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.
SmartmanApps@programming.dev 11 hours ago
Terms/Products is mathematical fact, as is The Distributive Law. Maths textbooks never use the word “juxtaposition”.
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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
Nope! It doesn’t say that a(b+c)=ax(b+c)
Question about solving an equation and not about solving an expression
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
That’s right
Got no idea what you’re talking about
Yes
No
For teenagers, who are taught The Distributive Law in Year 7