Question

Show that any positive odd integer is of the form $(6m + 1)$ or $(6m + 3)$ or $(6m + 5)$ where $m$ is some integer.

Answer 1

Hint: Firstly, to show that any positive odd integer is of the form $(6m + 1)$ or $(6m + 3)$ or $(6m + 5)$ where $m$ is some integer, we assume any two positive integers. Then to divide them we use Euclid’s Division Lemma which is mentioned below and by simplifying it we can prove it further.

Formula used: Euclid’s Division Lemma: If $a$ and $b$ be any two positive integers. Then, there exist unique integers $q$ and $r$ such that $a = bq + r,0 \leqslant r < b$ .

Complete step-by-step solution:
Firstly, we assume that $a$ and $b$ are two positive integers.
Where $a$ be any positive integer and we take $b = 6$ .
Then, by Euclid’s Division Lemma there exist two integers $m$ and $r$ , now by Division algorithm we have
$a = 6m + r,$ where $0 \leqslant r < 6$ - - - - - - - - - - $(1.)$
Now, we know that the value of $r$ lies from $0$ to less than $6$ . And we also know that $r$ is an integer. So an integer less than $6$ is $5$ .
So, by substituting all values of $r$ in $(1.)$ , we get
$a = 6m$ or $6m + 1$ or $6m + 2$ or $6m + 3$ or $6m + 4$ or $6m + 5$ .
But, we know that, here, $6\;m$ , $6m + 2$ and $6m + 4$ are even integers. Because $6$ is an even integer and if we add an even integer in this the result will be an even integer again.
Thus, we say that, $a = 6m + 1$ or $6m + 3$ or $6m + 5$ , are positive odd integers where $m$ is some integer.

Note: Euclid’s division lemma is another form of the long division method we have been doing from early grades. In Euclid’s division lemma, $q$ and $r$ represents the quotient and remainder, respectively. Euclid’s division lemma is described only for positive integers, but it also can be extended for all integers.