Let mand n be positive integers, then m Z ∩ n Z = k Z where ∘ k = l c m ( m , n ) ∘ k = g c d ( m , n ) ∘ k = m , n

Question

Let mand n be positive integers, then   m  Z  ∩  n  Z  =  k  Z where  ∘  k  =  l  c  m  (  m  ,  n  )  ∘  k  =  g  c  d  (  m  ,  n  )  ∘  k  =  m  ,  n

Ragazzonibw · Accepted Answer

We know that   n  Z refers to all the number belonging to integers which are multiples of   n. For example   2  Z is the set of all all numbers which are multiples of   2 which is essentially the set of even numbers.Suppose we have   m  Z  ∩  n  Z  =  k  Z. These means that   k  Z  ⊆  m  Z and   k  Z  ⊆  n  Z from the properties of set. What this implies is that the set of all numbers which are multiples of   k is a subset of set of all numbers which are multiples of   m. Hence   k needs to be a multiple of   m. Similarly,   k needs to be a multiple of   n. Hence we have proved that   k is a common multiple of   m and   n.Now consider another common multiple of   m and   n which is   x. Hence we have   x  Z  ⊆   m  Z and   x  Z subseteq   n  Z. As   x  Z is a subset of both   m  Z and   n  Z, it must be a subset of intersection of   m  Z and   n  Z. Hence   x  Z  ⊆  m  Z  ∩  n  Z. We know that   m  Z  ∩  n  Z  =  k  Z. Hence we have   x  Z  ⊆  k  Z. We know that   x  Z can be a subset of   k  Z only if   k      |    x that is   k divides x completely or   x is a multiple of   k. Thus we can see that   k is the least common multiple. Hence the correct answer is   k  =  L  C  M  (  m  ,  n  )Answer :   l  c  m  (  m  ,  n  )

Let mand n be positive integers, then mZ∩nZ=kZ where ∘k=lcm(m,n) ∘k=gcd(m,n) ∘k=m,n

Answered question

Answer & Explanation

New Questions in Pre-Algebra