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 🙄
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.
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.
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
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=(2b2-a2)/(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 6a2b2. 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.
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
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=acbc instead of abc, 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)3x2 by 3(a-b)x. The answer on page 141 of the PDF is 5(a-b)2x. 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 xn term. I’ve posted four examples to the contrary and all you’ve got is pretending not to see x(b+c)nright fuckin’ there in each one.
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=ac 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 😂
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 🙄
No, it means E=mc² is E=mcc=(mxcxc)
I have no idea what you’re talking about 🙄
But you understand E=mc2 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.
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 😂
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.
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 former 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
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=(2b2-a2)/(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 6a2b2. 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.
Terms/Products is mathematical fact, as is The Distributive Law. Maths textbooks never use the word “juxtaposition”.
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 1/a(b+c)=1/ax(b+c). You’re making a false equivalence argument
Question about solving an equation and not about solving an expression. False equivalence again.
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, 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. 🙄
For teenagers, who are taught The Distributive Law in Year 7
…
That’s you saying it. You are unambiguously saying a(b)c somehow means (ab)c=acbc instead of abc, 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)3x2 by 3(a-b)x. The answer on page 141 of the PDF is 5(a-b)2x. 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 xn 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.
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 🙄 The thing you want it to mean is 2(8²)
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)…
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 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
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 😂