Every composite number has at least one prime factor less than or equal to square root of itself.
Answers
Answered by
0
consider composite is even
then 2 is one prime for all even numbers
hence proved
consider a number n² is a perfect square, hence there is at least 1 prime≤n
hence proved
consider an odd number and not perfect square, n²+m, where m,n are odd too
such that √(n²+m) = n+k where k<1
since n²+m is composite it has at least one factor <n
least value of n²+m=15
√15=√(9+6)=3+0.8729 (k<1)
prime less than √15 is 3
hence proved
then 2 is one prime for all even numbers
hence proved
consider a number n² is a perfect square, hence there is at least 1 prime≤n
hence proved
consider an odd number and not perfect square, n²+m, where m,n are odd too
such that √(n²+m) = n+k where k<1
since n²+m is composite it has at least one factor <n
least value of n²+m=15
√15=√(9+6)=3+0.8729 (k<1)
prime less than √15 is 3
hence proved
Similar questions