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 😂
False equivalence is you arguing about brackets and exponents by pointing to equations without exponents.
This entire thing is about your lone-fool campaign to insist 2(8)2doesn’t mean 2*82, despite multiple textbook examples that only work because a(b)c is a*bc and not acbc.
I found four examples, across two centuries, of your certain circumstances: addition in brackets, factor without multiply symbol, exponent on the bracket. You can’t pivot to pretending this is a division syntax issue, when you’ve explicitly said 2(8)2 is (2*8)2. Do you have a single example that matches that, or are you just full of shit?
No it isn’t! 😂 8 is a single numerical factor. ab is a Product of 2 algebraic factors.
False equivalence is you arguing about brackets and exponents
Nope. I was talking about 1/a(b+c) the whole time, as the reason the Distributive Law exists, until you lot decided to drag exponents into it in a False Equivalence argument. I even posted a textbook that showed more than a century ago they were still writing the first set of Brackets. i.e. 1/(a)(b+c). i.e. It’s the FOIL rule, (a+b)(c+d)=(ac+ad+bc+bd) where b=0, and these days we don’t write (a)(b+c) anymore, we just write a(b+c), which is already a single Term, same as (a+b)(c+d) is a single term, thus doesn’t need the brackets around the a to show it’s a single Term.
3(x-y) is a single term…
This entire thing is about your lone-fool campaign
Hilarious that all Maths textbooks, Maths teachers, and most calculators agree with me then, isn’t it 😂
insist 2(8)2 doesn’t mean 2*82,
Again, you lot were the ones who dragged exponents into it in a False Equivalence argument to 1/a(b+c)
I found four examples, across two centuries
None of which relate to the actual original argument about 1/a(b+c)=1/(ab+ac) and not (b+c)/a
You can’t pivot to pretending this is a division syntax issue
I’m not pivoting, that was the original argument. 😂 The most popular memes are 8/2(2+2) and 6/2(1+2), and in this case they removed the Division to throw a curve-ball in there (note the people who failed to notice the difference initially). We know a(b+c)=(ab+ac), because it has to work when it follows a Division, 1/a(b+c)=1/(ab+ac). It’s the same reason that 1/a²=1/(axa), and not 1/axa=a/a=1. It’s the reason for the brackets in (ab+ac) and (axa), hence why it’s done in the Brackets step (not the MULTIPLY step). It’s you lot trying to pivot to arguments about exponents, because you are desperately trying to separate the a from a(b+c), so that it can be ax(b+c), and thus fall to the Multiply step instead of the Brackets step, but you cannot find any textbooks that say a(b+c)=ax(b+c) - they all say a(b+c)=(ab+ac) - so you’re trying this desperate False Equivalence argument to separate the a by dragging exponents into it and invoking the special Brackets rule which only applies in certain circumstances, none of which apply to a(b+c) 😂
2(8)2 is (2*8)2.
That’s right
are you just full of shit?
says the person making False Equivalence arguments. 🙄 Let me know when you find a textbook that says a(b+c)=ax(b+c), otherwise I’ll take that as an admission of being wrong that you keep avoiding the actual original point that a(b+c) is a single Term, as per Maths textbooks
So is 3xy, according to that textbook. That doesn’t mean 3xy2 is 9*y2*x2. The power only applies to the last element… like how (8)22 only squares the 2.
Four separate textbooks explicitly demonstrate that that’s how a(b)c works. 6(ab)3 is 6(ab)(ab)(ab), not (6ab)(6ab)(6ab). 3(x+1)2 for x=-2 is 3, not 9. 15(a-b)3x2 doesn’t involve coefficients of 3375. 2(x-b)2 has a 2b2 term, not 4b2. If any textbook anywhere shows a(b)c producing (ab)c, or x(a-b)c producing (xa-xb)c, then reveal it, or shut the fuck up.
2(ab)2 is 2(ab)(ab) the same way 6(ab)3 is 6(ab)(ab)(ab). For a=8, b=1, that’s 2*(8*1)*(8*1).
Factor yes, hence the special rule about Brackets and Exponents that only applies in that context
like how (8)22 only squares the 2
It doesn’t do anything, being an invalid syntax to follow brackets immediately with a number. You can do ab, a(b), but not (a)b
not (6ab)(6ab)(6ab)
Yep, as opposed to 6(a+b), which is (6a+6b)
3(x+1)2 for x=-2 is 3, not 9
No it isn’t. See previous point. Do we have an a(b+c), yes we do. Do we have an a(bc)²? No we don’t.
2(x-b)2 has a 2b2 term
No, it has a a(b-c) term, squared
shut the fuck up
says someone still trying to make the special case of Exponents and Brackets apply to a Factorised Term when it doesn’t. 😂 I’ll take that as your admission of being wrong about a(b+c)=ax(b+c) then. Thanks for playing
For a=8, b=1, that’s 2*(81)(8*1).
Only if you had defined it as such to begin with, otherwise the Brackets Exponents rule doesn’t apply if you started out with 2(8)², which is different to 2(8²) and 2(ab)²
…and there’s still no exponent in a(b+c) anyway, Mr. False Equivalence
There is no special case. You made it up by confusing yourself about “dismissing a bracket.” To everyone else in the world, brackets are just another term. Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after, because it makes no difference.
Here’s yet another example, PDF page 27: (6+5)x+(-2+10)y. And that’s as factorization. This Maths textbook you plainly didn’t read was published this decade. Still waiting on any book ever that demonstrates your special bullshit.
7bx with b=(m+n) becomes 7(m+n)x and it’s the same damn thing. Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7. Brackets only happen first because they have to be reduced to a single term. A bracket with one number is not “unsolved” - it’s one number. Squaring a bracket with one number is squaring that number.
The base of an exponent is whatever’s in the symbols of inclusion. Hence: 6(ab)3 = 6(ab)(ab)(ab).
No, it has a a(b-c) term, squared
It has a (b-c) term, squared. The base of an exponent is whatever’s in the symbols of inclusion. See page 121 of 696, in the Modern Algebra: Structure And Method PDF you plainly got from Archive.org. “In an expression such as 3a2, the 2 is the exponent of the base a. In an expression such as (3a)2, the 2 is the exponent of the base 3a, because you enclosed the expression in a symbol of inclusion.” You will never find a published example that makes an exception for distribution first.
On the page before your screenshot - 116 of 696 - this specific Maths textbook refers to both 8x7 and 8(7) as “symbols of multiplication.” It’s just multiplication. It’s only ever multiplication. It’s not special, you crank. 8(7) is a product identical to 8x7. Squaring either factor only squares that factor.
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 😂
a=8, b=1, it’s the same thing.
False equivalence is you arguing about brackets and exponents by pointing to equations without exponents.
This entire thing is about your lone-fool campaign to insist 2(8)2 doesn’t mean 2*82, despite multiple textbook examples that only work because a(b)c is a*bc and not acbc.
I found four examples, across two centuries, of your certain circumstances: addition in brackets, factor without multiply symbol, exponent on the bracket. You can’t pivot to pretending this is a division syntax issue, when you’ve explicitly said 2(8)2 is (2*8)2. Do you have a single example that matches that, or are you just full of shit?
No it isn’t! 😂 8 is a single numerical factor. ab is a Product of 2 algebraic factors.
Nope. I was talking about 1/a(b+c) the whole time, as the reason the Distributive Law exists, until you lot decided to drag exponents into it in a False Equivalence argument. I even posted a textbook that showed more than a century ago they were still writing the first set of Brackets. i.e. 1/(a)(b+c). i.e. It’s the FOIL rule, (a+b)(c+d)=(ac+ad+bc+bd) where b=0, and these days we don’t write (a)(b+c) anymore, we just write a(b+c), which is already a single Term, same as (a+b)(c+d) is a single term, thus doesn’t need the brackets around the a to show it’s a single Term.
3(x-y) is a single term…
Hilarious that all Maths textbooks, Maths teachers, and most calculators agree with me then, isn’t it 😂
Again, you lot were the ones who dragged exponents into it in a False Equivalence argument to 1/a(b+c)
None of which relate to the actual original argument about 1/a(b+c)=1/(ab+ac) and not (b+c)/a
I’m not pivoting, that was the original argument. 😂 The most popular memes are 8/2(2+2) and 6/2(1+2), and in this case they removed the Division to throw a curve-ball in there (note the people who failed to notice the difference initially). We know a(b+c)=(ab+ac), because it has to work when it follows a Division, 1/a(b+c)=1/(ab+ac). It’s the same reason that 1/a²=1/(axa), and not 1/axa=a/a=1. It’s the reason for the brackets in (ab+ac) and (axa), hence why it’s done in the Brackets step (not the MULTIPLY step). It’s you lot trying to pivot to arguments about exponents, because you are desperately trying to separate the a from a(b+c), so that it can be ax(b+c), and thus fall to the Multiply step instead of the Brackets step, but you cannot find any textbooks that say a(b+c)=ax(b+c) - they all say a(b+c)=(ab+ac) - so you’re trying this desperate False Equivalence argument to separate the a by dragging exponents into it and invoking the special Brackets rule which only applies in certain circumstances, none of which apply to a(b+c) 😂
That’s right
says the person making False Equivalence arguments. 🙄 Let me know when you find a textbook that says a(b+c)=ax(b+c), otherwise I’ll take that as an admission of being wrong that you keep avoiding the actual original point that a(b+c) is a single Term, as per Maths textbooks
So is 3xy, according to that textbook. That doesn’t mean 3xy2 is 9*y2*x2. The power only applies to the last element… like how (8)22 only squares the 2.
Four separate textbooks explicitly demonstrate that that’s how a(b)c works. 6(ab)3 is 6(ab)(ab)(ab), not (6ab)(6ab)(6ab). 3(x+1)2 for x=-2 is 3, not 9. 15(a-b)3x2 doesn’t involve coefficients of 3375. 2(x-b)2 has a 2b2 term, not 4b2. If any textbook anywhere shows a(b)c producing (ab)c, or x(a-b)c producing (xa-xb)c, then reveal it, or shut the fuck up.
2(ab)2 is 2(ab)(ab) the same way 6(ab)3 is 6(ab)(ab)(ab). For a=8, b=1, that’s 2*(8*1)*(8*1).
That’s right
That’s right. It means 3abb=(3xaxbxb)
Factor yes, hence the special rule about Brackets and Exponents that only applies in that context
It doesn’t do anything, being an invalid syntax to follow brackets immediately with a number. You can do ab, a(b), but not (a)b
Yep, as opposed to 6(a+b), which is (6a+6b)
No it isn’t. See previous point. Do we have an a(b+c), yes we do. Do we have an a(bc)²? No we don’t.
No, it has a a(b-c) term, squared
says someone still trying to make the special case of Exponents and Brackets apply to a Factorised Term when it doesn’t. 😂 I’ll take that as your admission of being wrong about a(b+c)=ax(b+c) then. Thanks for playing
Only if you had defined it as such to begin with, otherwise the Brackets Exponents rule doesn’t apply if you started out with 2(8)², which is different to 2(8²) and 2(ab)²
…and there’s still no exponent in a(b+c) anyway, Mr. False Equivalence
There is no special case. You made it up by confusing yourself about “dismissing a bracket.” To everyone else in the world, brackets are just another term. Several of the textbooks I’ve linked will freely juxtapose brackets and variables before or after, because it makes no difference.
Here’s yet another example, PDF page 27: (6+5)x+(-2+10)y. And that’s as factorization. This Maths textbook you plainly didn’t read was published this decade. Still waiting on any book ever that demonstrates your special bullshit.
7bx with b=(m+n) becomes 7(m+n)x and it’s the same damn thing. Splitting it like 7xm+7xn is no different from splitting (m+n)/7 into m/7+n/7. Brackets only happen first because they have to be reduced to a single term. A bracket with one number is not “unsolved” - it’s one number. Squaring a bracket with one number is squaring that number.
The base of an exponent is whatever’s in the symbols of inclusion. Hence: 6(ab)3 = 6(ab)(ab)(ab).
It has a (b-c) term, squared. The base of an exponent is whatever’s in the symbols of inclusion. See page 121 of 696, in the Modern Algebra: Structure And Method PDF you plainly got from Archive.org. “In an expression such as 3a2, the 2 is the exponent of the base a. In an expression such as (3a)2, the 2 is the exponent of the base 3a, because you enclosed the expression in a symbol of inclusion.” You will never find a published example that makes an exception for distribution first.
On the page before your screenshot - 116 of 696 - this specific Maths textbook refers to both 8x7 and 8(7) as “symbols of multiplication.” It’s just multiplication. It’s only ever multiplication. It’s not special, you crank. 8(7) is a product identical to 8x7. Squaring either factor only squares that factor.
Variables don’t work differently when you know what they are. b=1 is not somehow an exception that isn’t allowed, remember?
There’s an exponent in 2(8)2 and it concisely demonstrates to anyone who passed high school that you can’t do algebra.