Question

Show that any positive odd integer is of the form 6q+ 1,or 6q +3,or 6q +5), where q is
some integer.​

Answer 1

According to Euclid division lemma method,.

If a and b are two positive integers,then

$\color{red} \boxed{a = bq + r}$

where

$0 \leqslant r < \: b$

b=>6

it means

$0 \leqslant r < 6$

$\color{red} \boxed{a = 6q + r}$

r is the interger greater than or equal to 0 or less than 6

hence r will be0, 1,2,3,4,5,

If r=>1

Our equation will be

a=>6q+1

This will always be in form of odd integer

______________________

If r=>3

Our equation will be

a=>6q+3

This will always be in form of odd interger

______________________

If r=>5

Our equation will be

a=>6q+5

This will always be in form of odd interger

_______________________

Therefore, all odd integer are in form of 6q+1,6q+3,and 6q+5..

Hence proved ...☺️

Answer 2

HEY FRIEND HERE IS UR ANSWER,

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

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