Question

show that any positive odd integer is of the
form 4q+1 or 4q +3 for
positive integer​

Answer 1

Step-by-step explanation:

a=bq + r

(b=4)So, a=4q+r

0≤r>b

0≤r>4

r=0,r=1,r=2,r=3.

a=bq+r

case 1 (r=0) :-

a=4q+0

=4q ans.even

case 2 (r=1):-

a=bq+r

=4q+1 ans.odd

case 3 (r=2):-

a=bq+r

=4q+2 ans . even

case 4 (r=3):-

a=bq+r

=4q+3 ans. odd

Answer 2

Step-by-step explanation:

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 .