Let m and n be positive integers. Prove that mZ nZ = kZ. Where k is the term of m and n.
Solution
kZ is a set consisting of all multiples of k.
Let, a   (mZ Intersection nZ)
So, m divides a, and also n divides a
We need to show that lcm(m,n)=k also divides a.
Lets use Division Algorithm:
a = pk + r ------------- Equation (1)
where p and r are integers such that
0 <= r < k
(by <=Â Â i mean greater than or equal to)
Since a is a multiple of both m and n, so lcm(m,n) divides a.
So, k divides L.H.S. of equation(1)
Also, k divides pk.
So, k must divide r.
But since r <k, So r = 0
Hence, from equation(1),  a = pk
So, k divides a
Now we choosed a as arbitrary element in (mZ Intersection nZ), and we showed that its in kZ,
So mZ Intersection nZ is contained in kZ.
Now to show kZ is contained in mZ Intersection nZ.
Let, b kZ
So, b is a multiple of k
So, b is a multiple of lcm(m.n)
So, b is a multiple of both m and n.
So, b is element of both mZ and nZ.
Hence b lies in (mZ Intersection nZ).
Since b is arbitrary element of kZ.
So kZ is contained in (mZ Intersection nZ).
Hence kZ = (mZ Intersection nZ)
Hence we proved.
:)
.
Let m and n be positive integers- Prove that mZ nZ - kZ- Where k is t.docx
1. Let m and n be positive integers. Prove that mZ nZ = kZ. Where k is the term of m and n.
Solution
kZ is a set consisting of all multiples of k.
Let, a   (mZ Intersection nZ)
So, m divides a, and also n divides a
We need to show that lcm(m,n)=k also divides a.
Lets use Division Algorithm:
a = pk + r ------------- Equation (1)
where p and r are integers such that
0 <= r < k
(by <=Â Â i mean greater than or equal to)
Since a is a multiple of both m and n, so lcm(m,n) divides a.
So, k divides L.H.S. of equation(1)
Also, k divides pk.
So, k must divide r.
But since r <k, So r = 0
Hence, from equation(1),  a = pk
So, k divides a
2. Now we choosed a as arbitrary element in (mZ Intersection nZ), and we showed that its in kZ,
So mZ Intersection nZ is contained in kZ.
Now to show kZ is contained in mZ Intersection nZ.
Let, b kZ
So, b is a multiple of k
So, b is a multiple of lcm(m.n)
So, b is a multiple of both m and n.
So, b is element of both mZ and nZ.
Hence b lies in (mZ Intersection nZ).
Since b is arbitrary element of kZ.
So kZ is contained in (mZ Intersection nZ).
Hence kZ = (mZ Intersection nZ)
Hence we proved.
:)
