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Show that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.
Given:
Statement "Any positive odd integer is of the form $6q\ +\ 1$ or, $6q\ +\ 3$ or, $6q\ +\ 5$, where q is some integer".
To prove:
We have to show that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.
Solution:
According to Euclid's Division Lemma,
If $a$ and $b$ are two positive integers, then
$a\ =\ bq\ +\ r$ where, $0\ \underline{< }\ r\ <\ b$.
Let $a$ be the positive integer which when divided by 6 gives $q$ as quotient and $r$ as remainder.
$a\ =\ 6q\ +\ r$ where, $0\ \underline{< }\ r\ <\ 6$.
So, $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$
Now,
When, $\mathbf{r\ =\ 0}$:
$a\ =\ 6q\ +\ 0\ =\ 6q$, which is divisible by 2, so it is an even number.
When, $\mathbf{r\ =\ 1}$:
$a\ =\ 6q\ +\ 1$, which is not divisible by 2, so it is an odd number.
When, $\mathbf{r\ =\ 2}$:
$a\ =\ 6q\ +\ 2\ =\ 2(3q\ +\ 1)$, which is divisible by 2, so it is an even number.
When, $\mathbf{r\ =\ 3}$:
$a\ =\ 6q\ +\ 3$, which is not divisible by 2, so it is an odd number.
When, $\mathbf{r\ =\ 4}$:
$a\ =\ 6q\ +\ 4\ =\ 2(3q\ +\ 2)$, which is divisible by 2, so it is an even number.
When, $\mathbf{r\ =\ 5}$:
$a\ =\ 6q\ +\ 5$, which is not divisible by 2, so it is an odd number.
Therefore, any odd integer is of the form $6q\ +\ 1$ or $6q\ +\ 3$ or $6q\ +\ 5$.
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