Answer:
In explanation
Please let me know if something doesn't make sense.
Step-by-step explanation:
a)
*This relation is not reflexive.
0 is an integer and (0,0) is not in the relation because 0(0)>0 is not true.
*This relation is symmetric because if a(b)>0 then b(a)>0 since multiplication is commutative.
*This relation is transitive.
Assume a(b)>0 and b(c)>0.
Note: This means not a,b, or c can be zero.
Therefore we have abbc>0.
Since b^2 is positive then ac is positive.
Since a(c)>0, then (a,c) is in R provided (a,b) and (b,c) is in R.
*The relation is not antisymmretric.
(3,2) and (2,3) are in R but 3 doesn't equal 2.
b)
*This relation is reflective.
Since a^2=a^2 for any a, then (a,a) is in R.
*The relation is symmetric.
If a^2=b^2, then b^2=a^2.
*The relation is transitive.
If a^2=b^2 and b^2=c^2, then a^2=c^2.
*The relation is not antisymmretric.
(1,-1) and (-1,1) is in the relation but-1 doesn't equal 1.
c)
*The relation is reflexive.
a/a=1 for any a in the naturals.
*The relation is not symmetric.
Wile 4/2 is an integer, 2/4 is not.
*The relation is transitive.
If a/b=z and b/c=y where z and y are integers, then a=bz and b=cy.
This means a=cyz. This implies a/c=yz.
Since the product of integers is an integer, then (a,c) is in the relation provided (a,b) and (b,c) are in the relation.
*The relation is antisymmretric.
Assume (a,b) is an R. (Note: a,b are natural numbers.) This means a/b is an integer. This also means a is either greater than or equal to b. If b is less than a, then (b,a) is not in R. If a=b, then (b,a) is in R. (Note: b/a=1 since b=a)
