Comment on I dunno
SmartmanApps@programming.dev 13 hours agoThen 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
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
mindbleach@sh.itjust.works 10 hours ago
…
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.
SmartmanApps@programming.dev 10 hours ago
No it isn’t! 😂 Spot the difference 1/2(8)²=1/256 vs.
Nope. Never said that either 🙄
Because that isn’t what I said. See previous point 😂
From 2(8)², which isn’t the same thing as 2(ab)² 🙄
Because you’re on a completely different page and making False Equivalence arguments.
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.
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)…
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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.
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 the is multiplication only, in the latter there is Division, hence False Equivalence in trying to say what applies to Multiplication also applies to Division
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 😂