Question

State that any positive odd integers is of the form 4q+1, or 4a+3 where q is a positive integer

Answer 1

Hey mate
Here is your answer
Let a is any positive even integer

Since we know by Euclid algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying

a = bq + r where 0 <= r < b

Here b = 4, then

a = 4q + r where 0 <= r < 4

Since 0 <= r < 4, then possible remainder are 0, 1, 2 and 3.

Now possible values of a can be 4q, 4q + 1, 4q + 2, 4q + 3

Since a is even, a cannot be 4q+1 or 4q + 3 as they are both not divisible by 2.

Hense, any even integer is of the form 4q or 4q + 2.
if it will help you pls mark it as brainlist

Thanks

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 .