Question

show that any positive odd intezer is of the form 4q+1or4q+3,where q is some intezer

Answer 1

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
Take b = 4

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

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

That is, a can be 4q,or 4q + 1, 4q + 2, 4q + 3 where q is the quotient.

Since a is odd, a 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.

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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 .