Question

by using Euclid division algorithm prove that for any real number "p",4p + 1 is always odd natural number​

Answer 1

Step-by-step explanation:

'p' is a real number. we need to prove that 4p + 1 is always an odd natural number. Let b = 4 be the other integer,

applying Euclid's division lemma to 'p' and b = 4 we get,

p = bq+ r where, b > r ≥ 0,    

therefore p = 4q or 4q+1 or 4q+2 or 4q +3

Now substituting the value of 'p' in the given equation '4p +1'

Case 1 :  when p = 4q

4p + 1 = 4 (4q) +1 = 16q + 1

Case 2: where p = 4q + 1

4p + 1 = 4 (4q + 1) +1 = 16q + 5

Case 3: where p = 4q + 2

4p + 1 = 4 (4q + 2) +1 = 16q+ 9

Case 4: where p = 4q + 3

4p + 1 = 4 (4q + 3) +1 = 16q+ 13

We know that 16q is in the form of an even number and even + odd results in an odd number. Hence it can be said that for any real number "p",4p + 1 is always an odd natural number​.