Question

show that any positive even integer is in the form of 4q , 4q+2 where q is a whole number​

Answer 1

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 ] .

$\therefore$ a can be of the form 4q or 4q + 2 for some integer q .

Hence , it is solved

Answer 2

heya!

Here is ur answer..

______________________________

Let "a" is any positive integer and b = 4

According to Euclid's division postulate,

a = bq+r [ where, 0 ≤r<b]

a = 4q+r [ where, 0 ≤ r<4]

Therefore,

Possible values of "r" are, 0,1,2 and 3

If,

r = 0 => a = 4q

r = 1 => a = 4q+1

r = 2 => a = 4q+2

r = 3 => a = 4a+3

Since, 4q+1 and 4q+3 are the odd integers.

Therefore, any positive integer is of the form 4q, 4q+2.

Hence proved!