in the sets z of integers define mrn if m-n is a multiple of 12 prove that a R is an equivalence relation
Answers
Answer:Given:
Relation mRn defined on set of integers Z
m-n is a multiple of 12.
To prove:
The relation R is an equivalence relation.
Solution:
First of all, let us have a look at the definition of equivalence relation.
A relation R defined on set Z is known as equivalence relation if it is:
1. Reflexive:
if aRa, \forall a\in Z
2. Symmetric
if aRb\Rightarrow bRA \ \forall a,b\in Z
3. Transitive:
aRb, bRc \Rightarrow aRc \ \forall a,b,c \in Z
The given relation is:
R = \{(m,n) : m-n \text{ is a multiple of 12}; m,n\in Z\}
1. Reflexive:
mRm = m-m=0
0 is a multiple of 12, \therefore \bold{reflexive}.
2. Symmetric:
mRn\Rightarrow m-n=12a
nRM\Rightarrow n-m=-12a
Both are multiple of 12, \therefore \bold{symmetric}.
3. Transitive:
mRn\Rightarrow m-n=12a .... (1)
nRo\Rightarrow n-o=12b .... (2)
Adding (1) and (2):
m-o=12(a+b)
Hence, mRo. \therefore \bold{transitive}.
Hence proved that the relation R is equivalence relation.
Step-by-step explanation:
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ANSWER
We know that a number is a factor as well as multiple of itself.
Since 'n' will be a factor of 'n' for any natural number, hence (n,n) will be a subset of the relation. In other words nRn for a natural number 'n' holds true. This makes the relation reflexive.
If 'n' is a factor of 'm' then simultaneously 'm' cannot be a factor of 'n' since n≤m.
Thus mRn→nRm is not true. Thus the relation is not symmetric.
Now, if 'm' is multiple of 'n' and 'p' be a multiple of 'm', (all being distinct natural numbers), then 'n' will be a common factor of 'm' and 'p'. Thus
pRm→mRn→pRn is true.
Hence the relation is transitive.