Question

show that positive even integer is of the form 4m and every positive odd integer is ofvthe form 4m+1 for some integer m
​

Answer 1

Answer:

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 .

Therefore 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 ] .

Therefore  a can be of the form 4q + 1 or 4q + 3 for some integer q .

HOPE IT HELPS :)