Yes, irrational. Let p = root 2 and q = root 3. Then (q - p)2 = 5 - 2root6, which is irrational because it is the sum of an integer (5) and an irrational (2root6), and so q - p (which is root3 - root2) is irrational.

We will use the fact that if p prime, a divides p, then a = p or a = 1. Then if p + q = r, for primes p, q, r, then one of p,q,r is even, or all three are (consider mod 2). p = q = r = 2 clearly doesn't work, and p + q = 2 doesn't work for primes p,q >= 3. So without loss of generality p =…

#include<stdio.h> void main() int n,m,q,sum=0; do { pf("enter the value of n"); scanf("%d",&n); q=n%10 sum=sum+(q*q*q) n=n/10 } while(n>0); pf("palindrome"); while(n<0); pf('not palindrome"); } }

IF (1+3+5+....+P) + (1+3+5+...+q)=(1+3+5+....+r) where each set of parentheses contain the sum of consecutive add integer as shown. what is the samellest possible value of (p+q+r) where p>6?

The two vectors are P & Q..Sum of the two vecotors is P+Q=R..R Is called the resultant vector of this two vector..the action of the resultant vector R is equal to the actions of two vectors P & Q..