Comment on I dunno

• mindbleach@sh.itjust.works ⁨10⁩ ⁨hours⁩ ago

  From 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?

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  • SmartmanApps@programming.dev ⁨9⁩ ⁨hours⁩ ago

    a=8, b=1, it’s the same thing

    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, thus doesn’t need the brackets around the a to show it’s a single Term.

    3(x-y) is a single term…

    Image

    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 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) 😂

    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

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    • mindbleach@sh.itjust.works ⁨8⁩ ⁨hours⁩ ago

      3(x-y) is a single term…

      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).

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      • SmartmanApps@programming.dev ⁨8⁩ ⁨hours⁩ ago

        So is 3xy

        That’s right

        That doesn’t mean 3xy2 is 9y2x2.

        That’s right. It means 3abb=(3xaxbxb)

        The power only applies to the last element

        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

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