Question

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

Answer 1

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here is your answers

Let a is any positive even integer

Since we know by Euclid algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying

a = bq + r where 0 <= r < b

Here b = 4, then

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

Since 0 <= r < 4, then possible remainder are 0, 1, 2 and 3.

Now possible values of a can be 4q, 4q + 1, 4q + 2, 4q + 3

Since a is even, a cannot be 4q+1 or 4q + 3 as they are both not divisible by 2.

Hense,  any even integer is of the form 4q or 4q + 2.

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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 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