Question

Prove that 1 is not a linear combination of the polynomials 2 and x in , that is, prove it is impossible to find such that .

Short answer

It is proved that 1 is not a linear combination of the polynomials 2 and x in $\mathbb{Z}\left[x\right]$.

Step-by-step solution

Defining linear combination of functions

It is an addition of functions, which is constructed by a set of terms, and every term is multiplied by a constant and adds the result.

Proving that 1 is not a linear combination of the polynomials 2 and x in ℤ[x]

Let’s assume, 1 can be written as a linear combination of 2 and x in $\mathbb{Z}\left[x\right]$such that for any$f\left(x\right),g\left(x\right)\in \mathbb{Z}\left[x\right]$ ,$2f\left(x\right)+xg\left(x\right)=1$ …….$\left(i\right)$

We know $f\left(x\right)$and $g\left(x\right)$are functions in $\mathbb{Z}\left[x\right]$.

Therefore, we can write$f\left(x\right)=\sum _{i=0}^{\infty }{p}_i{x}^i$ and $g\left(x\right)=\sum _{i=0}^{\infty }{q}_i{x}^i$where, $p_i,q_i\in \mathbb{Z}$.

Putting the values of $f\left(x\right)$and$g\left(x\right)$ in the equation$\left(i\right)$ as:

$\begin{array}{l}2\sum _{i=0}^{\infty }{p}_i{x}^i+x\sum _{i=0}^{\infty }{q}_i{x}^i=1\\ \sum _{i=0}^{\infty }2p_i{x}^i+\sum _{i=0}^{\infty }{q}_i{x}^{i+1}=1\end{array}$

By computing the above equation, we get, $2p_0=1\Rightarrow p_0=1/2$.

Since $p_0\in \mathbb{Z}$, $p_0=1/2$is not possible.

Therefore, our assumption is wrong.

Hence, it is proved that 1 is not a linear combination of the polynomials 2 and x in $\mathbb{Z}\left[x\right]$.