Question

show that any opposite integer is of the form 4Q + 1 or 4Q+ 3 where Q is some positive integer

Answer 1

I think it should be odd integer instead of opposite integer.
Ok I will consider it odd integer.

Solution :-
Let the odd integer be 'a'
and, 'b' be another integer.
And b = 4
now, by remainder theorem
a = b . q + r {where, q is some integer and r is the remainder and 0 ≤ r < b}

So, a = 4.q + r
if r = 0
a = 4.q + 0
.•. a is even number as any integer multiplied by any even ( here 4) number is always even.

So, 4q + 1 will be odd as even + odd = odd
Similarly 4q + 2 is even
And, 4q + 3 is odd.

.•. We proved that any odd number is in the form 4.q +1 and 4.q + 3.

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 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 .