Question

show that any positive odd integer is of from 4q+1 or 4q+3 where is some integer

Answer 1

Let a be any odd positive integer which when divided by 4 gives q as quotient and r as remainder.

By Euclid's division lemma

a = bq + r

Where

0≥ r <b

that means

r = 0 , 1 , 2 , 3

a = 4q

a = 4q + 1

a = 4q + 2

a = 4q + 3

So,

a = 4q + 1 and a = 4q + 3
is the form because when we multiply any number by 4 it give even and by adding odd number it form odd resultant.

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 .