Let S = {1, 2, 3, ...}, A relation R on S × S is defined by xRy if log ax > log ay when a \(\rm = \frac 1 2.\) Then the relation is:
transitive only
The question asks us to determine the properties (reflexive, symmetric, transitive) of a relation R defined on the set S × S, where S = {1, 2, 3, ...}. The relation is given by xRy if \(\log_a x > \log_a y\) with \(a = \frac{1}{2}\).
The given condition for xRy is \(\log_{1/2} x > \log_{1/2} y\). To understand this inequality, we need to consider the base of the logarithm. The base is \(a = \frac{1}{2}\), which is between 0 and 1 (\(0 < \frac{1}{2} < 1\)).
For a logarithm with a base \(b\) such that \(0 < b < 1\), the logarithmic function \(f(z) = \log_b z\) is a decreasing function. This means that if \(\log_b u > \log_b v\), then it must be true that \(u < v\).
Applying this to our relation, xRy holds if and only if \(x < y\), given that \(x\) and \(y\) are elements of S = {1, 2, 3, ...}, so they are positive integers.
Thus, the relation R is simply the "less than" relation defined on the set of natural numbers S.
Now let's check if this relation R (where xRy means x < y) is reflexive, symmetric, or transitive on S.
A relation R on a set S is reflexive if for every element \(x \in S\), \(xRx\) holds.
For our relation R, \(xRx\) would mean \(x < x\). Is this true for any \(x \in S\)?
Therefore, the relation R is not reflexive.
A relation R on a set S is symmetric if for every \(x, y \in S\), whenever \(xRy\) holds, \(yRx\) also holds.
For our relation R, if \(xRy\) holds, it means \(x < y\). If R were symmetric, then \(yRx\) would also have to hold, meaning \(y < x\).
Therefore, the relation R is not symmetric.
A relation R on a set S is transitive if for every \(x, y, z \in S\), whenever \(xRy\) holds and \(yRz\) holds, \(xRz\) also holds.
For our relation R, if \(xRy\) holds, it means \(x < y\). If \(yRz\) holds, it means \(y < z\). If R is transitive, then \(xRz\) must hold, meaning \(x < z\).
This property holds true for all elements in S. Therefore, the relation R is transitive.
Based on our analysis:
So, the relation is transitive only.
Let's look at the given options:
| Property | Definition | Does R satisfy it? | Explanation |
|---|---|---|---|
| Reflexive | \(xRx\) for all \(x \in S\) | No | \(x < x\) is false |
| Symmetric | If \(xRy\), then \(yRx\) for all \(x, y \in S\) | No | If \(x < y\), then \(y < x\) is false (for \(x \ne y\)) |
| Transitive | If \(xRy\) and \(yRz\), then \(xRz\) for all \(x, y, z \in S\) | Yes | If \(x < y\) and \(y < z\), then \(x < z\) is true |
| Relation Property | Condition for R on set A | Example (xRy iff x < y on S) |
|---|---|---|
| Reflexive | For all \(a \in A\), \(aRa\). | Is \(x < x\) true for all \(x \in \{1, 2, 3, ...\}\)? No. |
| Symmetric | For all \(a, b \in A\), if \(aRb\) then \(bRa\). | If \(x < y\) (for \(x, y \in \{1, 2, 3, ...\}\)), does \(y < x\) follow? No. |
| Transitive | For all \(a, b, c \in A\), if \(aRb\) and \(bRc\), then \(aRc\). | If \(x < y\) and \(y < z\) (for \(x, y, z \in \{1, 2, 3, ...\}\)), does \(x < z\) follow? Yes. |
Understanding how the base of a logarithm affects inequalities is crucial here. For \(\log_b u > \log_b v\):
Relations are fundamental concepts in mathematics used to describe how elements within a set or between sets are connected. Properties like reflexivity, symmetry, and transitivity help classify relations and are important in various areas like set theory, graph theory, and database theory. An equivalence relation, for instance, must be reflexive, symmetric, and transitive. A partial order relation must be reflexive, antisymmetric, and transitive.
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