Comment on I dunno
mindbleach@sh.itjust.works 3 weeks agoFrom 2(8)², which isn’t the same thing as 2(ab)²
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*8^2^, despite multiple textbook examples that only work because a(b)^c^ is a*b^c^ and not a^c^b^c^.
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?
SmartmanApps@programming.dev 3 weeks ago
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, thus doesn’t need the brackets around the a to show it’s a single Term.
3(x-y) is a single term…
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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 pople 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 (cxc), 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), 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
mindbleach@sh.itjust.works 3 weeks ago
So is 3xy, according to that textbook. That doesn’t mean 3xy^2^ is 9*y^2^*x^2^. The power only applies to the last element… like how (8)2^2^ 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. 2(x-b)^2^ has a 2b^2^ term, not 4b^2^. 15(a-b)^3^x^2^ is not (3375a-3375b)x^2^. 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).
SmartmanApps@programming.dev 3 weeks ago
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
mindbleach@sh.itjust.works 3 weeks ago
We can see the Acrobat window in those scans you found online.
You think 2(8)^2^ is 128 if that’s simplified from 2(8+0)^2^… but 256 if it’s simplified from 2(8*1)^2^. In short: no.
I think you’re about fifteen years old. You had an unpleasant teacher who belittled you, and you’ve identified with the aggressor. Your whole online persona is posturing to always be smarterer than everybody else, even if that means saying Wolfram fucking Alpha is wrong about basic algebra.
Faced with a contradiction that requires you to insist (8*1) ≠ (8+0), you’re going to type laughter and spam emojis as if that inspires any reaction besides pity. The word you should be looking for is, “oops.”
mindbleach@sh.itjust.works 3 weeks ago
PDFs found online. From which you are ignoring counterexamples using a(b+c)^n^. Fraud.
Your own spammed screenshots say 3 gets multiplied.
There is no special case. You made it up. 8+0 equals 8 (or sorry, does it just mean 8?) so 2(8+0)^2^ is the same as 2(8)^2^. The latter is the next step in simplifying the former. You’ve admitted simplifying first is valid, when your nose was rubbed in your own found PDFs doing exactly that.
You don’t have an opinion. You make no claim, anymore. All you have left are derision and emojis. You’ve admitted 2(8*1)^2^ means 2(8)(8), and insist that’s different from 2(8)^2^ because… ibid. You cannot explain it even now.
mindbleach@sh.itjust.works 3 weeks ago
Your bullshit hit max comment depth.
So when you said 2(8)^2^ is 256, you were wrong.
Otherwise - walk me through how 2(8*1)^2^, 2(8+0)^2^, and 2(8)^2^ aren’t equal, alleged math teacher.
“3(x+y) means 3*(x+y).”
“It depends on what the definition of is, is,” says someone definitely not trapped in a contradiction.
mindbleach@sh.itjust.works 3 weeks ago
That would mean 2(8*1)^2^ is 128. You are the one saying it’s not 2a^2^b^2^, because you think it’s 2^2^a^2^b^2^, and that 2(8*1)^2^ is 256. I’m not touching anything without exponents because exponents are where you are blatantly full of shit.
Source: your ass. Every published example disagrees, and you just go, nuh-uh, that up-to-date Maths textbook must be wrong. You alone are correct on this accursed Earth.
Hey look, another one of the textbooks you insist I read says you’re completely wrong: “The multiplication sign is often not included between letters, e.g. 3ab means 3 * a * b.” Page 31 of the PDF… right above where you’ve dishonestly twisted the “expanding brackets” text. Next page: “3(x+y) means 3*(x+y).”
Page 129 of that PDF, exercise 5, question 14: simplify 2(e^4^)^2^. The answer on PDF page 414 is 2e^8^. Your bullshit would say 4e^8^.
Right below that, exercise 5*, question 4: 4(4^4^)^4^. The answer on PDF page 414 is 1.72x10^10^. The bullshit you’ve made up would be 1.10x10^11^. 5* questions 7, 9, 10, and 11 also have the same a(b)^c^ format as 2(8)^2^, if you somehow need further proof of how this actually works.
PDF page 134, exam practice question 10a, simplify 3(q^2^)^2^. PDF page 415 says 3q^6^. Your bullshit says 9q^6^.
Damn dude, that’s five textbooks you chose saying you’re full of shit, and zero backing you up. One more and I get a free hoagie. Your bullshit has brought us to max comment depth.
mindbleach@sh.itjust.works 3 weeks ago
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 PDF you plainly got from Archive.org. “In an expression such as 3a^2^, 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.
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.