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My question is concerning polynomials and the following axiom for real vector spaces

  • Associativity of Addition: (Axiom 3) $\displaystyle \mathbf{u} +(\mathbf{v} +\mathbf{w}) =(\mathbf{u} \ +\ \mathbf{v}) +\mathbf{w}$

Below is problem 10, chapter 4.1 of Anton's book elementary linear algebra.

  • Show that the set of polynomials of the form $\displaystyle a_{0} +a_{1} x$ with the operations below is a vector space. \begin{gather} ( a_{0} +a_{1} x) +( b_{0} +b_{1} x) =( a_{0} +b_{0}) +( a_{1} +b_{1}) x\\ k( a_{0} +a_{1} x) =( ka_{0}) +( ka_{1}) x \end{gather}

  • Please note: I realize this is a subspace, so it is only enough to show that it is nonempty and closed under multiplication and addition. However, that's not what the question is asking for.

Below are two ways I show the axiom for associativity of addition.

Axiom 3: Solution 1 \begin{gather*} ( a_{0} +a_{1} x) +(( b_{0} +b_{1} x) +( c_{0} +c_{1} x)) \ =\ \\ ( a_{0} +( b_{0} +c_{0})) +( a_{1} +( b_{1} +c_{1})) x\ =\\ (( a_{0} +b_{0}) +c_{0}) +(( a_{1} +b_{1}) +c_{1}) x=\\ (( a_{0} +a_{1} x) \ +\ ( b_{0} +b_{1} x)) \ +\ ( c_{0} +c_{1} x) \ \end{gather*} Axiom 3: Solution 2 \begin{gather*} ( a_{0} +a_{1} x) +(( b_{0} +b_{1} x) +( c_{0} +c_{1} x)) \ =\ \\ ( a_{0} +a_{1} x) +(( b_{0} +c_{0}) +( b_{1} +c_{1}) x) =\\ ( a_{0} +( b_{0} +c_{0})) +( a_{1} +( b_{1} +c_{1})) x\ =\\ (( a_{0} +b_{0}) +c_{0}) +(( a_{1} +b_{1}) +c_{1}) x=\\ (( a_{0} +b_{0}) +( a_{1} +b_{1}) x) +( c_{0} +c_{1} x) =\\ (( a_{0} +a_{1} x) \ +\ ( b_{0} +b_{1} x)) \ +\ ( c_{0} +c_{1} x) \ \end{gather*}

  • My thoughts.
    Solution 2 seems more correct because I feel like Solution 1 is imposing operations on the set V then expecting the vector space axioms to hold. However, they could both be wrong.

  • Closing comments. The bulk of my confusion stems from how these proofs are demonstrated depending on the book I look at. Elementary Linear Algebra is a fantastic book, but the way they go about some of these proofs gives me a very loose understanding of the underpinning logic; my other book, linear algebra done right by axler, is verbose.

I thank you in advance for your help.

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    $\begingroup$ So Solution $1$ is, basically, a sketch of Solution $2$, with some steps omitted. That is fine. A major part of writing proofs is not about being rigorous (we are always rigorous) but about communicating ideas. As such, a shorter proof, even if it skips some steps, may be more insightful, easier to read/teach, quicker to comprehend. Just as the centipede froze in horror when it started to think about which leg to move, too much detail may be counterproductive to understanding proofs. You need to tailor your proofs to your audience. $\endgroup$ Jan 14 at 18:58
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    $\begingroup$ Solution (1) seems to skip a step. It is essentially obvious, so not a sin in a proof, but, at least for those new to proofs, (2), with that step filled in, is probably better. $\endgroup$ Jan 14 at 19:20
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    $\begingroup$ Essentially, mathematicians learn over time what kind of “obvious” steps they can skip. Most proofs have the goal of not just being correct, but communicating that logic. Indeed, in the real world, we rarely write “formal proof,” and there are often steps we assume the reader can fill in. $\endgroup$ Jan 14 at 19:24
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    $\begingroup$ I do wonder if, given non-print-media, we could use an approach to proofs that allows the reader to dig down into “missing steps.” So, the reader starts with almost an outline of the proof, but then each section can be opened up and we find more details. I’ve certainly seen such an outline presented in a class, first, before going into the details, but I’m thinking a more interactive approach. $\endgroup$ Jan 14 at 19:27
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    $\begingroup$ @ThomasAndrews Once you drill all the way down, you see ZF axioms or something like that? Um ... yep, it would be amazing. I suspect, however, the reality will be - click on a step and it asks you to pay money to the journal where that step's proof was published... $\endgroup$ Jan 14 at 19:33

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