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I disagree that this proof is better pedagogically; it assumes the quadratic case of the Fundamental Theorem of Algebra (the correspondence between factorizatio
by dooglius 7y ago
I disagree that this proof is better pedagogically; it assumes the quadratic case of the Fundamental Theorem of Algebra (the correspondence between factorization and solutions), which at this stage would have to be taken on faith by students, whereas completing the square is fully justified.
- boxfire 7y agoNot sure why you think it's easier to take that on faith than complete the square, but I'm sure you never tutored new students to mathematics. At this point we are taking all of abstract algebra and some number theory for granted in anyone's education. Pedagogically it's easier to understand, the thesis of the article.
- dooglius 7y agoGranted, I have not tutored new students. I am making the assumption that someone encountering this would be familiar with basic algebraic manipulations, and the solutions to x^2=a. This is all that is necessary to justify completing the square. The assertion that quadratic equations must have at most two solutions R and S, and that it can be equivalently written (x-R)(x-S) is what has to be taken on faith.
- pdonis 7y ago> it assumes the quadratic case of the Fundamental Theorem of Algebra No, it doesn't; it proves that case. The key step in the proof is the fact that, over the complex numbers, every number has a square root: from that it follows that the factorization used in the proof must always exist.