Question

using Euclid division algorithm to show that any positive odd integer is of the form 4q+1 or 4q+3 ​

Answer 1

let a be any positive integer according to Euclid division leema a = ba +r so, let b= 4

now

a= 4q+ r r=0,1,2,3 which is less than b

now putting the value

r=o

we get a=4q+0

a=4q

now when r=1

a= 4q+ 1

it is a odd positive integer

again

when r=2

a= 4q+2

which is an even positive integer

when r=3

a=4q+3

which is a odd positive integer

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 .