this post was submitted on 27 Nov 2025
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This is your own source - and it says, juxtaposition is just multiplication. It doesn't mean E=mc^2^ 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=mc^2^ 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.