Verify that the following sets form rings under the indicated operations: Z_e (t
ID: 3118967 • Letter: V
Question
Verify that the following sets form rings under the indicated operations: Z_e (the even integers), with ordinary addition and multiplication. Q[i] = {a + bi|a.b Element Q;i^2 = -1}, with complex addition and multiplication. M_2(Z) = { [a b c d] |a, b, c, d Element, Z }, with matrix addition and multiplication. Z_4 = {0.1.2. 3}. with addition and multiplication modulo 4. In which of the rings of Problem 1 is multiplication commutative? Which of them have an identity element for multiplication (that is, an element e satisfying a e = e a = a for all a)?Explanation / Answer
1. To verify that the given sets form rings, we must verify that the sets follow properties of rings. Now we verify properties of rings for each subquestion one by one.
(a) The set of even integers with respect to (w.r.t.) usual addition & multiplication of integers.
Therefore we can write x , y as : x=2*k1 and y=2*k2 , where k1, k2 are integers.
x+y= 2k1+ 2k2
= 2( k1 + k2)
= 2k where k = k1 + k2 is again an integer.
Therefore x + y = 2k is an even integer.
Therefore the operation addition is closed.
Therefore we can write x , y , z as : x=2*k1 , y=2*k2 , z=2*k3 where k1, k2, k3 are integers.
we have to verify the equality : x+( y + z ) = ( x + y ) + z
Consider Left Hand Side (L.H.S.) = x+( y + z )
= 2k1 +( 2k2 + 2k3 )
= 2*[ k1 + ( k2 + k3 ) ]
= 2*[ ( k1 + k2 ) + k3 ] ...( Since addition of integers is associative ).
= ( 2k1 + 2k2 ) +2k3
= R.H.S. ( Right Hand Side )
Therefore addition of even integers is associative.
Let x be any even integer. Therefore we can write x as : x = 2k1 where k1 is an integer .
Consider x + 0 = 2k1 + 0
= 2k1
= x
Therefore we get that x + 0 = x
i.e. The identity element is 0 .
We have to show that there exists an element x' in the set of even integers such that : x + x' = 0 .
For any even integer x , -x is also an even integer.
But we know that , x + (-x) = 0
Therefore, x' = -x is additive inverse of x .
Therefore every even integer has an inverse element in the set of even integers.
Therefore we can write x , y as : x=2*k1 and y=2*k2 , where k1, k2 are integers. ...(1)
We have to show that : x + y = y + x
Consider L.H.S. = x + y
= 2k1 +2k2 ... From(1)
= 2( k1 + k2 )
= 2( k2 + k1 ) ...(Addition of integers is commutative.)
= 2k2 + 2k1
= y + x
= R.H.S.
Therefore Addition of even integers is commutative .
Let x, y, z be any three even integers.
Therefore we can write x , y , z as : x=2*k1 , y=2*k2 , z=2*k3 where k1, k2, k3 are integers.
Consider L.H.S. = x * ( y * z )
= 2k1 * ( 2k2 * 2k )
= 2k1 * 2k2 * 2k3
= ( 2k1 * 2k2 ) * 2k3
= ( x * y ) * z
= R.H.S.
Therefore multiplication is associative .
R.D. : ( y + z ) * x = ( y * x ) + ( z * x )
We prove L.D. and similarly R.D. can be proved.
Let x, y, z be any three even integers.
Therefore we can write x , y , z as : x=2*k1 , y=2*k2 , z=2*k3 where k1, k2, k3 are integers.
Consider L.H.S. = x * ( y + z )
= 2k1 * ( 2k2 +2k3 )
= ( 2k1 * 2k2 ) + ( 2k1 * 2k3 ) ...( Multiplication distributes over Addition of integers .)
= ( x * y ) + ( x * z )
= R.H.S.
Therefore the set of even integers satisfy distributivity property .
Hence the set of even integers satisfy all the properties of rings.
Therefore the set of even integers is a ring.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.