• mindbleach@sh.itjust.works
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    2 days ago

    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).

    That’s not Multiplication, it’s Distribution, a(b+c)=(ab+ac), a(b)=(axb).

    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.

    • 💡𝚂𝗆𝖺𝗋𝗍𝗆𝖺𝗇 𝙰𝗉𝗉𝗌📱@programming.dev
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      2 days ago

      Dude you’re not even hitting the right reply buttons anymore

      Yes I am

      Is that what you do when you’re drunk?

      Is that why you think I’m hitting the wrong buttons?

      It’d explain leading with ‘nope! I’ve said exactly what you accused me of.’

      I have no idea what you’re talking about. Maybe stop drinking

      You keep pretending distribution is different from multiplication

      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 not Multiplication, it’s Distribution, a(b+c)=(ab+ac), a(b)=(axb).

      That’s right.

      And then posting images that explicitly say the contents of the brackets should be multiplied

      The “contents OF THE BRACKETS”, done in the BRACKETS step , not the MULTIPLICATION step - there you go quoting proof that I’m correct! 😂

      Or that they can be simplified first.

      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

      I am not playing dueling-sources with you

      No, because you haven’t got any 😂

      your own sources call bullshit on what you keep hassling strangers about

      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

      • mindbleach@sh.itjust.works
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        1 day ago

        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.

        • 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 🙄

          E=mc2 is E=(mc)2

          No, it means E=mc² is E=mcc=(mxcxc)

          Throwing other numbers on there

          I have no idea what you’re talking about 🙄

          • mindbleach@sh.itjust.works
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            1 day ago

            Because BRACKETS - ab=(axb) BY DEFINITION

            “Parentheses must be introduced”!

            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.

            • 💡𝚂𝗆𝖺𝗋𝗍𝗆𝖺𝗇 𝙰𝗉𝗉𝗌📱@programming.dev
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              17 hours ago

              But you understand E=mc2 does not mean E=(mxc)2

              I already answered, and I have no idea what your point is.

              This is you acknowledging that distribution and juxtaposition are only multiplication

              Nope. It’s me acknowledging they are both BRACKETS 🙄

              E=mcc=(mxcxc) <== BRACKETS

              a(b+c)=(ab+ac) <== BRACKETS

              and only precede

              everything 😂

              • mindbleach@sh.itjust.works
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                17 hours ago

                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.

                • 💡𝚂𝗆𝖺𝗋𝗍𝗆𝖺𝗇 𝙰𝗉𝗉𝗌📱@programming.dev
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                  17 hours ago

                  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

                  • mindbleach@sh.itjust.works
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                    17 hours ago

                    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.