Question

show that any positive odd integer is the form 4q+1 or 4 q+3

Answer 1

let a be any positive integer.
here b= 4
a=4q+r; 0<r<4
remainder =0,1,2,3
If r=0; a=4q (even) -(1) equations
lf r=1; a = 4q+1 ( odd) - (2) equations
lf r= 2 ; a= 4q+2 (even) -(3) equations
lf r=3 ; a= 4q+3 (odd) - (4) equations

from equations 2 and 4 it is clear that the positive odd integers is the form 4q+1 or 4q+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 .