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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Answer 2

Let us start with taking a, where is a positive odd integer. We apply the division algorithm with a and b=4.

Since, $0 \leqslant r < 4$ the possible remainders are 0, 1, 2, and 3.

That is, a can be 4m, or 4m+1, or 4m+2, or 4m+3, where q is the quotient. However, since a is odd, a cannot be 4q or 4q+2.

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

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