Make a C++ program Theorem: if n is composite (not prime), then n has an integer
ID: 3933922 • Letter: M
Question
Make a C++ program
Theorem: if n is composite (not prime), then n has an integer divisor n. Proof: Assume n is composite, then There is an integer a, 1 < a < n such that a is a factor of n; then There is an integer b > 1 such that n = ab; then Prove: Either a n or b n. This is the “then” part of the theorem. Proof of 4 by indirect proof: Assume the negation of 4): a > n and b > n; then ab > (n)2 ; then ab > n, contradicting 3) (where ab = n).
Make a C++ program
Theorem: if n is composite (not prime), then n has an integer divisor n. Proof: Assume n is composite, then There is an integer a, 1 < a < n such that a is a factor of n; then There is an integer b > 1 such that n = ab; then Prove: Either a n or b n. This is the “then” part of the theorem. Proof of 4 by indirect proof: Assume the negation of 4): a > n and b > n; then ab > (n)2 ; then ab > n, contradicting 3) (where ab = n).
Make a C++ program
Theorem: if n is composite (not prime), then n has an integer divisor n. Proof: Assume n is composite, then There is an integer a, 1 < a < n such that a is a factor of n; then There is an integer b > 1 such that n = ab; then Prove: Either a n or b n. This is the “then” part of the theorem. Proof of 4 by indirect proof: Assume the negation of 4): a > n and b > n; then ab > (n)2 ; then ab > n, contradicting 3) (where ab = n).
Explanation / Answer
**********************Theorem: if n is composite (not prime), then n has an integer divisor n.*****************
#include <iostream>
#include<math.h>
using namespace std;
int main()
{
int a, b, n, i, j, factor;
int m=0;
cout<<"Enter number n: ";
cin>>n;
for(a = 2; a <n; a++){
for(b = 2; b <n; b++){
if(n%a == 0){
factor=a;
}
if (n == a * a){//if n=a*a then n>a
m=a; // if n is perfect square of a
}
if(n ==a*b){ //if n=a*b
i=a;
j=b;
}
}
}
if(m !=0){ //if number n is perfect square
if(m>=i){//for either a<= root of n
cout<<"n has integer divisor <= root of n : ";
}
if(m >=j){//for or b<= root of n
cout<<"n has integer divisor <= root of n : ";
}
}
return 0;
}
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.