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I am aware that there is a theorem which states the following:

Given real polynomials $P(x), D(x)$ where $\deg[P(x)]\ge \deg[D(x)]$. The quotient $\frac{P(x)}{D(x)}$ has remainder with degree $\lt \deg[D(x)]$.

Please correct me if I'm wrong but I am pretty sure that is correct.

My question is how does one go about proving this?

Michael Hardy
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Isaac Greene
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    Typing $$\rm division\ algorithm\ for\ polynomials$$ into Google probably works. – Gerry Myerson Sep 13 '17 at 02:39
  • So, have you tried it? – Gerry Myerson Sep 15 '17 at 06:44
  • @GerryMyerson Yes but the division algorithm itself is far from a rigorous proof – Isaac Greene Sep 16 '17 at 21:48
  • So what you want is a rigorous proof of the division algorithm? and you haven't been able to find one? – Gerry Myerson Sep 16 '17 at 23:45
  • @GerryMyerson No a proof independent from the division algorithm – Isaac Greene Sep 17 '17 at 03:36
  • Now I'm really confused. You don't want a rogirous proof of the division algorithm, but you do want a proof of the result you quote – but that result is the division algorithm. Can you please clarify, preferably by editing the body of your post, since people shouldn't have to read through all these comments to work out what you want. – Gerry Myerson Sep 17 '17 at 06:42
  • @GerryMyerson Note that the theorem I quoted has nothing to do with the quotient (Q(x)) but only the remainder. I understand that the division algorithm is one way to arrive at the conclusion mentioned in the question but all the division algorithm tells you is that when dividing polynomials you "just stop" when the degree of the divisor > degree of the dividend. But can you see that the division algorithm assumes the theorem above. Specifically, it "just stops" when the degree of the divisor is larger than the degree of the dividend. My question is how do prove such a theorem. – Isaac Greene Sep 18 '17 at 02:49
  • What most textbooks call the Division Algorithm (and what I prefer to call the Division Theorem) is in fact the theorem that if $P$ and $D$ are polynomials with coefficients in a field $F$ (in our case, the reals), and $D$ is not the zero polynomial, then there exist unique polynomials $Q$ and $R$ such that $P=DQ+R$ and the degree of $R$ is less than the degree of $D$. The textbooks prove this theorem. For an online example, see https://math.okstate.edu/people/binegar/3613/3613-l17.pdf – Gerry Myerson Sep 18 '17 at 06:17
  • Are we on the same page now, Iso? – Gerry Myerson Sep 19 '17 at 12:43
  • @GerryMyerson Yes, thank you very much for the pdf – Isaac Greene Sep 19 '17 at 22:33
  • See also https://math.stackexchange.com/questions/48958/proof-of-the-polynomial-division-algorithm – Gerry Myerson Sep 19 '17 at 23:16

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