Question

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

Answer 1

Hi !

So, let us see,

Let's first check for 4m.

4m = 2(2m)

= 2q (q = 2m)

It is an even number.

Now, 4m + 1

= 2(2m) + 1

= 2q + 1 (q = 2m)

It is an odd number.

Again, 4m + 2

= 2(2m + 1)

= 2q (q = 2m + 1)

It is an even number.

Now, 4m + 3

= 2(2m + 1) + 1

= 2q + 1 (q = 2m + 1)

It is an odd number.

Also, 4m + 4

= 2(2m + 2)

=2q (q = 2m + 2)

It is an even number.

So, from the above discussion, it is clear that integers like 4m, 4m + 2 and 4m + 4 are even. Also, integers like 4m + 1 and 4m + 3 are odd.

So, we can conclude that any positive integer of the form 4m + 1 or 4m + 3 where m is some integer.

Answer 2

Step-by-step explanation:

Note :- 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 .

°•° 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

THANKS

#BeBrainly.