Question

1. Prove that$\mathit{f}\left(x\right)\mathbf{+}\mathit{f}\left(x\right)\mathbf{=}\mathbf{0}$ for every $\mathit{f}\left(x\right)\mathbf{\in }{\mathit{\mathbb{Z}}}_{\mathbf{2}}\left[x\right]$.
2. Prove that$\mathit{u}\mathbf{+}\mathit{u}\mathbf{=}\mathbf{0}$ for every u in the field K.

Short answer

1. It has been proved$f\left(x\right)+f\left(x\right)=0$ that for every$f\left(x\right)\in {\mathbb{Z}}_2\left[x\right]$.
2. It has been proved that$u+u=0$ for every u in the field K.

Step-by-step solution

Prove that f(x)+f(x)=0

Let $f\left(x\right)\in {\mathbb{Z}}_2\left[x\right]$

Now$f\left(x\right)=a_0+a_{1}x+a_2{x}^2+\mathrm{...}+a_{n-1}{x}^{n-1}$ with all $a_i\in {\mathbb{Z}}_2$

Now, $f\left(x\right)+f\left(x\right)=2a_0+2a_{1}x+2a_2{x}^2+\mathrm{...}+2a_{n-1}{x}^{n-1}$for all$a_i\in {\mathbb{Z}}_2$

In ${\mathbb{Z}}_2$,$2a_i=0$ (since $2\left(0\right)=0$and$2\left(1\right)=2=0$ )

Thus $f\left(x\right)+f\left(x\right)=0$

Hence, $f\left(x\right)+f\left(x\right)=0$for every$f\left(x\right)\in {\mathbb{Z}}_2\left[x\right]$.

Prove that u+u=0

In any field K, for $u\in K$

If $u=\left[f\left(x\right)\right]$, then

$u+u=\left[f\left(x\right)+f\left(x\right)\right]$

Now $\left[f\left(x\right)+f\left(x\right)\right]=\left[0\right]$

Thus $u+u=0$

Hence,$u+u=0$ for every u in the field K.