Question

Show that any positive odd integer is of the form (4m+1) or (4m+3),where m is some integer.

Answer 1

Step-by-step explanation:

Note :- I am taking q as some integer.

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

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#BeBrainly.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let n be an arbitrary odd positive integer.

On dividing n by 4,

Assume m be the quotient,

And r be the remainder.

So, by Euclid's division lemma, we have

n = 4m + r,

Where, 0 ≤ r < 4.

n = 4m or (4m + 1) or (4m + 2) or (4m + 3)

Clearly, 4m and (4m + 2) are even and since n is odd,

So, n ≠ 4m and n ≠ (4m + 2)

Therefore,

n = (4m + 1) or (4m + 3), for some integer m.

Hence,  any positive odd integer is of the form (4m + 1) or (4m + 3),where m is some integer.