Question

Show any positive odd integer is of the form 4m+1 or 4m+3. Where m is some integer??

Answer 1

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.

Answer 2

hey

here is answer
let a be any positive integer

then

b=8

0≤r<b

0≤r<4

r=0,1,2, 3


i am taking q instead of m
case 1.

r=0

a=bq+r

4q+0

4q

case 2.
r=1
a=bq+r

4q+1



case 3.

r=2

4q+2



case4.

r=3
4q+3



from above it is proved.



hope it helps

thanks