Question

show that any positive odd integer is of the form (4m+1) or(4m+3) where m is some integer
Why are you taking b=4 and not b=2? 2m+1 and 2m+3 are also odd numbers!!

Answer 1

let a= bq + r
where r= 0(=or<)r > 4
1 case
where r=1
a = 4q+0 = 4q ( if you will simplify it it will become(2q) which is even
2 case
where r = 1
a= 4q+1= 4q+1 [ which cannot be simplified, hence it is odd}
3 case
where r=2
a=4q+ 2= 2(2q+1) which is even when simplified
4 case
where r= 3
a=4q+3= 4q+3 which is odd
hence, any positive odd integer is of the form (4m+1) or ( 4m+3)
you can take q as m
you cannot take 2 because then there will be only two cases i.e. 
a= 2m+0= 2m and
2m+1= 2m+1 which is odd, but the question also mention that it has to be in form of 4m+3. thus if you will take 2 then you will not have enough cases to support your answer.
 hope it helped you.

Answer 2

Step-by-step explanation:

Note :- I am taking q as some integer.

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved

THANKS

#BeBrainly.