Question

show that any positiveodd intiger is of the form 6q+1,or6q+3,or6q+5, where q is some intiger​

Answer 1

Correct Question:

Show that any positive odd integers in in the form of 6q + 1, 6q + 3 & 6q + 5 where q is some integer.

AnswEr :

Let us Consider that a & b are two positive integers

$\large\bold{\underline{\sf{By\: Using\: Euclid's\: Division\: lemma}}}$

$:\implies\sf\: a = bq + r$

Here, 0 < r < b & b = 6

• r can be 0,1,2,3,4 & 5

$\rule{150}2$

If r = 0

$:\implies\sf\: a = 6q$

If r = 1

$:\implies\sf\: a = 6q + 1$

If r = 2

$:\implies\sf\: a = 6q + 2$

If r = 3

$:\implies\sf\: a = 6q + 3$

If r = 4

$:\implies\sf\: a = 6q + 4$

If r = 5

$:\implies\sf\: a = 6q + 5$

Here, we can see that 6q, 6q +2 & 6q + 4 are even Numbers and 6q + 1, 6q + 3 & 6q + 5 are odd numbers.

Hence, any positive odd integers is in the form of 6q + 1, 6q + 3 & 6q + 5.

Answer 2

$\huge \mathfrak{answer}$

__________________

❁Question:❁

show that any positiveodd intiger is of the form 6q+1,or6q+3,or6q+5, where q is some integer

✾step to step explanation✾

$\bf{ \: { \boxed{ \red{ \tt{euclid \: division \: lemma \: }}}}}$

$\sf{if \: a \: and \: b \: are \: positive \: integer}$

then

$\rm \underline \red{a = bq + r}$

where 0<r<b

let positive integer be a and b=6

so

$\sf{a = 6q + r}$

where 0<r<6

r is integer greater than or equal to 0 andess than 6

so it can be 0,1,2,3,4,5

case 1➷

$\sf{if \: r = 1}$

$\sf{a = 6q + r}$

$\sf{a = 6q + 1}$

case 2➷

$\sf{if \: r = 3}$

$\sf{a = 6q + r}$

$\sf{a = 6q + 3}$

case 3➷

$\sf{if \: r = 5}$

$\sf{a = 6q + r}$

$\sf{a = 6q + 5}$

so any odd integer is in the the form

$\sf{6q + 1 ,\: 6q + 3 ,\: 6q + 5}$

hence proved