Question

Show that any positive odd integer is of the form $4q+1$ or $4q+3$ where $q$is some integer.

Answer 1

$1$ is the first odd positive number but it does not leave a remainder $1$.

Euclid's division algorithm:

Given positive integers $a$ $\&b$, there exist unique $q\&r$ satisfying $a=bq+r,0\le r<b$

If $a\&b$ are two integers, then as per Euclid's division algorithm $a=bq+r,0\le r<b$

Let the positive integers be $a\&b$and let $b$be equal to $4$

$0\le r<4$ , $r$ is an integer it is greater than $0$ and less than $4$

Case 1:

When $r=1$

$$\begin{gathered} a=bq+r \\ a=4q+1 \end{gathered}$$

It is an odd integer.

Case 2:

When

$$\begin{gathered} r=3 \\ a=bq+r \\ a=4q+3 \end{gathered}$$

It is also an odd integer.

In the case of $r=2$, it will be an even number.

Therefore, Positive odd integers are in the form of $4q+1or4q+3$.

Hence, Proved.