Math Problem Statement

Determine whether the relation R on the set of all real numbers is reflexive, symmetric,

antisymmetric, and/or transitive, where (π‘₯, 𝑦) ∈ 𝑅 if and only if

(d) π‘₯ is a multiple of 𝑦

Solution

Let's analyze the relation RR on the set of all real numbers R\mathbb{R}, where (x,y)∈R(x, y) \in R if and only if xx is a multiple of yy. Specifically, xx is a multiple of yy if there exists a real number kk such that x=kβ‹…yx = k \cdot y.

Reflexive

A relation is reflexive if every element is related to itself, i.e., (x,x)∈R(x, x) \in R for all x∈Rx \in \mathbb{R}.

  • Analysis: For any real number xx, x=1β‹…xx = 1 \cdot x. Hence, xx is always a multiple of itself, so (x,x)∈R(x, x) \in R for all x∈Rx \in \mathbb{R}.

Conclusion: The relation RR is reflexive.

Symmetric

A relation is symmetric if whenever (x,y)∈R(x, y) \in R, then (y,x)∈R(y, x) \in R as well.

  • Analysis: If (x,y)∈R(x, y) \in R, then x=kβ‹…yx = k \cdot y for some real number kk. For (y,x)(y, x) to be in RR, yy must be a multiple of xx, meaning there must exist a real number mm such that y=mβ‹…xy = m \cdot x.

    • If k=0k = 0, then x=0x = 0, and (0,y)(0, y) is in RR for any yy, but (y,0)(y, 0) will only be in RR if y=0y = 0, so symmetry may not hold.
    • If kβ‰ 0k \neq 0, then symmetry does not necessarily hold because there is no guarantee that y=mβ‹…xy = m \cdot x (i.e., that yy is a multiple of xx).

Conclusion: The relation RR is not symmetric.

Antisymmetric

A relation is antisymmetric if whenever (x,y)∈R(x, y) \in R and (y,x)∈R(y, x) \in R, then x=yx = y.

  • Analysis: If (x,y)∈R(x, y) \in R and (y,x)∈R(y, x) \in R, then xx is a multiple of yy and yy is a multiple of xx. This implies that x=kβ‹…yx = k \cdot y and y=mβ‹…xy = m \cdot x, leading to x=kβ‹…mβ‹…xx = k \cdot m \cdot x. For this to be true for all xx, kβ‹…m=1k \cdot m = 1. The only real numbers that satisfy this condition for all xx are k=1k = 1 and m=1m = 1, which implies x=yx = y.

Conclusion: The relation RR is antisymmetric.

Transitive

A relation is transitive if whenever (x,y)∈R(x, y) \in R and (y,z)∈R(y, z) \in R, then (x,z)∈R(x, z) \in R.

  • Analysis: If (x,y)∈R(x, y) \in R, then x=kβ‹…yx = k \cdot y for some kk, and if (y,z)∈R(y, z) \in R, then y=mβ‹…zy = m \cdot z for some mm. Substituting, x=kβ‹…(mβ‹…z)=(kβ‹…m)β‹…zx = k \cdot (m \cdot z) = (k \cdot m) \cdot z, which implies xx is a multiple of zz, so (x,z)∈R(x, z) \in R.

Conclusion: The relation RR is transitive.

Final Summary:

  • Reflexive: Yes
  • Symmetric: No
  • Antisymmetric: Yes
  • Transitive: Yes

Would you like more details on any of these properties, or do you have any other questions?

Here are five related questions you might find interesting:

  1. How would the properties of this relation change if we restricted the set to positive integers?
  2. Can you think of a real-world scenario where a similar relation might apply?
  3. What happens if we consider the relation where yy is a multiple of xx instead?
  4. How do these properties differ when considering the set of nonzero real numbers?
  5. What are some examples of relations that are symmetric but not antisymmetric?

Tip: When analyzing the symmetry of a relation, always consider edge cases, like when the elements involved might be zero, as they can often break symmetry.