Question

show that any positive even integer is of the form 4q or 4q +2 ,where q is any integer​

Answer 1

Let n be any positive even number,

so,

n = QB + r, where 0 ≤ r < b.

Case I, r = 0

n = 4q + 0

n = 4q ... Which is clearly even.

Case II, r = 1

n = 4q + 1. (Here, 4q is even but 1 is odd, therefore, in this case, it's odd)

Case III, r = 2

n = 4q + 2 ( Clearly it even )

Case IV, r = 3

n = 4q + 3. ( It is odd because 4q is even but 3 is odd)

Hence, Every positive even integer is of the form 4q, 4q + 2

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 even positive integer, so a can't be 4q + 1 , or 4q + 3 [ As these are odd ] .

∴ a can be of the form 4q or 4q + 2 for some integer q .

Hence , it is solved