Question

If $n\in Z$and $n\ne 0$, prove that $n$ can be written uniquely in the form $n=2^{k}m$, where $k\ge 0$ and $m$ is odd.

Short answer

It is proved that $n$can be written uniquely in the form $n=2^{k}m$, where $k\ge 0$and $m$ is odd.

, where is odd.

Step-by-step solution

Define  n

Assume that $2n$, then$n=2^{0}n$ . Here, $n$is odd.

When $2n$ , then by Fundamental theorem of Arithmetic

$\begin{array}{rcl}n& =& 2^{r_1}{p_2}^{r_2}{p_3}^{r_3}.........{p_k}^{r_k}\\ & =& 2^{k}m\end{array}$

Here, $k=r_1$ and ${p_2}^{r_2}{p_3}^{r_3}......{p_k}^{r_k}=m$.

Remember that $2{p_2}^{r_2}{p_3}^{r_3}.......{p_k}^{r_k}$ ; therefore it is odd.

Unique representation 

Let’s assume that

$n=2^{k_1}{m}_1=2^{k_2}{m}_2$with $k_2>k_1$

This gives$m_1=2^{k_2-k_1}{m}_2$

Here, the above equation shows that $2m_1$and $m_1$is odd.

Hence, $k_1=k_2$ can also be written as $m_1=m_2$. Hence it is proved.