Question

show that any positive odd integer is in the form of 4q+1, or 4q+3 where q is some integer

Answer 1

hii mate 

Let  be any positive integer

We know by Euclid's algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying, where.

Take b=4

a=4q+r

Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.

That is,  can be , where q is the quotient.

Since  is odd,  cannot be 4q or 4q + 2 as they are both divisible by 2.

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

hope this will help u

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 .