I think I know what you’re missing - perhaps intentionally 🙄 - in a(b+c), c can be equal to 0. It can be any number, not just positive and negative, leaving us with a(b)=(axb), which is also what I’ve been saying all along (not sure how you missed it, other than to deliberately ignore it)
You mean I know that, because it disobeys The Distributive Law 🙄 The expression you’re looking for is 2x(3+5)², which is indeed not subject to Distribution, since the 2 is not next to the brackets.
If you’d stuck to one dogmatic answer
Instead I’ve stuck to one actual law of Maths, a(b+c)=(ab+ac).
But you’ve concisely proven
The Distributive Law, including c=0 🙄 Not sure why you would think c=0 is somehow an exception from a law
the harassment is the point
No, the rules of Maths is the point
when you can’t do algebra right
Says person who thinks c=0 is somehow an exception that isn’t allowed,🙄but can’t cite any textbook which says that
Dude you’re not even hitting the right reply buttons anymore. Is that what you do when you’re drunk? It’d explain leading with ‘nope! I’ve said exactly what you accused me of.’
You keep pretending distribution is different from multiplication:
The context is Maths, you have to obey the rules of Maths. a(b+c)=(ab+ac), 5(8-5)=(5x8-5x5).
And then posting images that explicitly say the contents of the brackets should be multiplied. Or that they can be simplified first. I am not playing dueling-sources with you, because your own sources call bullshit on what you keep hassling strangers about.
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.
I think I know what you’re missing - perhaps intentionally 🙄 - in a(b+c), c can be equal to 0. It can be any number, not just positive and negative, leaving us with a(b)=(axb), which is also what I’ve been saying all along (not sure how you missed it, other than to deliberately ignore it)
Nope! I’ve said a(b+c)=(ab+ac) is correct.
You mean I know that, because it disobeys The Distributive Law 🙄 The expression you’re looking for is 2x(3+5)², which is indeed not subject to Distribution, since the 2 is not next to the brackets.
Instead I’ve stuck to one actual law of Maths, a(b+c)=(ab+ac).
The Distributive Law, including c=0 🙄 Not sure why you would think c=0 is somehow an exception from a law
No, the rules of Maths is the point
Says person who thinks c=0 is somehow an exception that isn’t allowed,🙄but can’t cite any textbook which says that
Dude you’re not even hitting the right reply buttons anymore. Is that what you do when you’re drunk? It’d explain leading with ‘nope! I’ve said exactly what you accused me of.’
You keep pretending distribution is different from multiplication:
And then posting images that explicitly say the contents of the brackets should be multiplied. Or that they can be simplified first. I am not playing dueling-sources with you, because your own sources call bullshit on what you keep hassling strangers about.
Yes I am
Is that why you think I’m hitting the wrong buttons?
I have no idea what you’re talking about. Maybe stop drinking
No pretending - is is different - it’s why you get different answers to 8/2(1+3) (Distribution) and 8/2x(1+3) (Multiplication) 😂
B 8/2(1+3)=8/(2+6)=8/8
E
DM 8/8=1
AS
B 8/2x(1+3)=8/2x4
E
DM 8/2x4=4x4=16
AS
That’s right.
The “contents OF THE BRACKETS”, done in the BRACKETS step , not the MULTIPLICATION step - there you go quoting proof that I’m correct! 😂
That’s right, you can simplify then DISTRIBUTE, both part of the BRACKETS step, and your point is?
B 8/2(1+3)=8/2(4)=8/(2x4)=8/8
E
DM 8/8=1 <== same answer
AS
No, because you haven’t got any 😂
says person failing to give a single example of that EVER happenning 😂
I’ll take that as an admission of being wrong then. Thanks for playing
This is your own source - and it says, juxtaposition is just multiplication. It doesn’t mean E=mc2 is E=(mc)2.
Throwing other numbers on there is like arguing 1+2 is different from 2+1 because 8/1+2 is different from 8/2+1.
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.