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I've seen an answer (How can I prove that a polynomial with degree $n$ is continuous everywhere in $\mathbb{R}$ using definitions?) on the question to prove that all polynomials are continuous, so I tried to follow his steps. Could you please tell me if that is alright and if not tell me what I could consider? Also the 1,2, and 4 are fairly obvious(from my lecture notes) so I did not write them out. P.S How can I (start) prove that if r=p/q is a ratio of two polynomials then it is continuous at every point of R where q≠0.

1) $f(x)=x$ is continuous everywhere

2) If $f(x)$ and $g(x)$ are continuous in $D$ then $f(x)\cdot g(x)$ in continuous on $D$.

3) Using $2$ and $1$ show that $x^n$ is continuous for every $n \in \mathbb N$

4) If $f(x)$ and $g(x)$ are continuous on $D$ then $f(x)+g(x)$ is continous on $D$

5) Now use $3$ and $4$.

So here is how I followed the steps:

1) elementary proof

2) algebraic property of coninuous functions proof

3) Proof: If $f_1(x)=x$ and $f_2(x)=x$ are both continuous on $D$ then from 2) we know that $f_1(x)\cdot f_2(x)=x^2$ is continuous. Suppose now that $f_1(x)=x$ $f_2(x)=x^2$ $f_3(x)=x^3 ... f_m(x)=x^m$. Hence we can conclude that $f_1(x)\cdot ....f_m(x)=x^n$ is continuous on $D$, for $n=m(m+1)/2$.

4) algebraic property of continuous functions

5) proof: from $3)$ we know that $x^n$ is contionuous on $D$ and hence if we add a polynomial of degree $n-1$ or smaller then by $4)$ we can conclude that $x^n+x^{n-1}+...+1$ is continuous.

Hence all polynomials are continuous.

  • There is one little matter with 5, since not every polynomial are in form of $x^n+x^{n-1}+...+1$, should prove for any $a_nx_n+a_{n-1}x^{n-1}+...+a_0$. –  Mar 30 '19 at 12:51
  • For the second question regarding rational function, if $q(x)$ is not $0$ in some point $c$, then it is nonzero in a neighbourhood of $c$, then just apply $\epsilon - \delta$ defintion as usual. –  Mar 30 '19 at 12:53
  • Also, in step 3, $n=m$, not $m(m+1)/2$. –  Mar 30 '19 at 12:58
  • Well on step 3 I said that n=m(m+1)/2 since x.x^2.x^3...x^m=x^n, isnt that right? or that's an unnecessary thing? –  Mar 30 '19 at 13:04
  • But your definition seems to be $f_1(x)=f_2(x)=...=f_m(x)=x$ –  Mar 30 '19 at 13:06
  • yeah sorry edited it. –  Mar 30 '19 at 14:04
  • All you need to do is two easy induction: $\lim_{x\to a} x^n= a^n$ and $\lim_{x\to a} \sum^n_{k=1}f_k(x)= \sum ^n_{k=1}f_k(a)$ – Matematleta Mar 30 '19 at 15:30

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