Question

Show that every positive odd integers is the form(4q+1)or (4q+3) for some integer

Answer 1

Hi there!

Let 'a' 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, a = bq + r

where, 0 ≤ r < b

Taking b = 4

=> a = 4q + r

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

That is, 'a' can be 

♦ 4q
♦ 4q + 1
♦ 4q + 2
♦ 4q + 3

Where, q is the quotient.

Since 'a' is odd, 'a' can't 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. ---( Proved. )

Hope it helps! :)

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 .