Question
Show that every positive odd integer is of the form $(4 q+1)$ or $(4 q+3)$ for some integer $q$.
✓
Answer
Let 'a' be a given positive od integer.
On dividing ' $a$ ' by $4$ , let $q$ be the quotient and $r$ be the remainder.
Then, by euclid's algorithm, we have
$a = 4q + r$, where $0 ≤ r < 4$
$⇒ a = 4q + r$, where$ r = 0, 1, 2, 3$
$⇒ a = 4q$ or$ a = 4q + 1$ or $a = 4q + 2$ or $a = 4q + 3$
But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$
On dividing ' $a$ ' by $4$ , let $q$ be the quotient and $r$ be the remainder.
Then, by euclid's algorithm, we have
$a = 4q + r$, where $0 ≤ r < 4$
$⇒ a = 4q + r$, where$ r = 0, 1, 2, 3$
$⇒ a = 4q$ or$ a = 4q + 1$ or $a = 4q + 2$ or $a = 4q + 3$
But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$
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