Polynomial deg Q

Oct 2009
273
2
How does my proof look?

We show that \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\). Let \(\displaystyle \mathbb{Q}\) be a polynomial ring. It will be the set of all polynomial functions \(\displaystyle f(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n\), \(\displaystyle a_i\in \mathbb{Q}\). If \(\displaystyle a_n\neq 0\) then we say \(\displaystyle n=deg(f(x))\). When adding polynomials we see that \(\displaystyle
\sum_{i=0}^h a_ix^i + \sum_{i=0}^k a_ix^i\leq \sum_{i=0}^{max(h,k)} (a_i+b_i)x^i\). Thus, \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\) as desired.
 
Oct 2009
4,261
1,836
How does my proof look?

We show that \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\). Let \(\displaystyle \mathbb{Q}\) be a polynomial ring.

Not a good idea to use the same symbol for two different things within the same 4-5 lines. Why not \(\displaystyle \mathbb{Q}[x]\)?


It will be the set of all polynomial functions \(\displaystyle f(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n\), \(\displaystyle a_i\in \mathbb{Q}\). If \(\displaystyle a_n\neq 0\) then we say \(\displaystyle n=deg(f(x))\). When adding polynomials we see that \(\displaystyle
\sum_{i=0}^h a_ix^i + \sum_{i=0}^k a_ix^i\leq \sum_{i=0}^{max(h,k)} (a_i+b_i)x^i\).

What does it mean that a polynomial is less or equal than other polynomial? This is meaningless unless properly defined and checked it works as a linear order...and again you use the same symbol (in this case, \(\displaystyle a_i\) ) to denote two different things: the coefficients of polynomial \(\displaystyle f(x)\) and those of pol. \(\displaystyle g(x)\)...!


Thus, \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\) as desired.

You haven't yet proved anything: a good idea would be to show that in \(\displaystyle f(x)+g(x)\) we have the monomials \(\displaystyle a_hx^h\,,\,b_kx^k\) of \(\displaystyle f(x)\,,\,g(x)\) resp., and thus...
.
 
Jun 2009
641
161
Africa
here's a nice proof,
Let \(\displaystyle P\) and \(\displaystyle Q\) two polynomials with degrees \(\displaystyle p \)and \(\displaystyle q\),
\(\displaystyle P=(a_0,a_1X,a_2X^2,..,a_pX^p);a_p \neq 0\) and \(\displaystyle Q=(b_0,b_1X,b_2X^2,..,b_qX^q);b_q \neq 0\)
now suppose \(\displaystyle p> q\).
we have,
\(\displaystyle P+Q= (a_0+b_0)+(a_1+b_1)X+\)\(\displaystyle ...+(a_q+b_q)X^{q}+(a_{q+1}+0)X^{q+1}+(a_{q+2}+0)X^{q+2}+...+(a_p+0)X^p+(0+0)+...\)
which means,
\(\displaystyle P+Q=\)\(\displaystyle (a_0+b_0)+(a_1+b_1)X+...+(a_q+b_q)X^{q}+a_{q+1}X^{q+1}+a_{q+2}X^{q+2}+...+a_pX^p+0+...\)
the last coefficient \(\displaystyle (\neq 0)\) of \(\displaystyle P+Q \) is \(\displaystyle a_p\) therefore,