Question

Use the Division Algorithm to prove that every odd integer is either of the form $4k+1$ or of the form$4k+3$ for some integer k.

Short answer

It is proved that every odd integer is either of the form $4k+1$ or of the form $4k+3$ for some integer k.

Step-by-step solution

The Division Algorithm

Theorem 1.1 states that consider a, b as an integer with $b>0$. Then q and r areunique integersin which $a=bq+r$ and $0\le r\le b$.

Show that every odd integer is either of the form 4k+1 or the form 4k+3 for some integer k

Each integer n could be divided by 4 with a remainder r equal to 0, 1, 2, or 3.

In this case, either $n=4k,4k+1,4k+2\mathrm{or}4k+3$, with quotient k.

When $n=4k\mathrm{or}n=4k+2$ then n is even.

When n would be odd then $n=4k+1\mathrm{or}n=4k+3$.

Hence, it is proved that every odd integer is either of the form $4k+1$or of the form $4k+3$ for some integer k.