# Polynomial deg Q

#### sfspitfire23

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.

#### tonio

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...
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#### Raoh

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,
$image=http://latex.codecogs.com/gif.latex?%5Cfn_cs%20deg%28P+Q%29=p=degP=max%28degP,degQ%29&hash=fecca8fd87612b078ec52dcfb33344d0$

#### smacktalk88

What if a_p=-b_p Raoh??