Analyzing the Relation R = {(a, b) : a ≤ b²}

The given relation is $R = \{(a, b) : a \le b^2\}$ defined on the set of real numbers, $\mathbb{R}$. A relation is considered an equivalence relation if it satisfies three properties: reflexivity, symmetry, and transitivity. We will check each of these properties for the given relation $R$.

Understanding Equivalence Relation Properties

For a relation R on a set A:

• Reflexive: For every element $a \in A$, $(a, a) \in R$.
• Symmetric: For every $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$.
• Transitive: For every $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Checking for Reflexivity

For the relation $R$ to be reflexive on $\mathbb{R}$, it must be true that for every $a \in \mathbb{R}$, $(a, a) \in R$. This means $a \le a^2$ must hold for all real numbers $a$.

Let's test some values:

• If $a = 2$, is $2 \le 2^2$? Yes, $2 \le 4$. So $(2, 2) \in R$.
• If $a = -3$, is $-3 \le (-3)^2$? Yes, $-3 \le 9$. So $(-3, -3) \in R$.
• If $a = 1$, is $1 \le 1^2$? Yes, $1 \le 1$. So $(1, 1) \in R$.

However, let's consider values of $a$ between 0 and 1.

• If $a = 0.5$, is $0.5 \le (0.5)^2$? This means $0.5 \le 0.25$. This is false. So $(0.5, 0.5) \notin R$.

Since we found a real number $(0.5)$ for which $(a, a) \notin R$, the relation $R$ is not reflexive.

Checking for Symmetry

For the relation $R$ to be symmetric, it must be true that for every $a, b \in \mathbb{R}$, if $(a, b) \in R$ (i.e., $a \le b^2$), then $(b, a) \in R$ (i.e., $b \le a^2$).

Let's test some values:

• Consider $a = 1$ and $b = 2$. Is $(1, 2) \in R$? Yes, because $1 \le 2^2$ ($1 \le 4$) is true.
• Now, is $(2, 1) \in R$? This requires $2 \le 1^2$ ($2 \le 1$). This is false.

Since we found $a=1, b=2$ such that $(1, 2) \in R$ but $(2, 1) \notin R$, the relation $R$ is not symmetric.

Checking for Transitivity

For the relation $R$ to be transitive, it must be true that for every $a, b, c \in \mathbb{R}$, if $(a, b) \in R$ (i.e., $a \le b^2$) and $(b, c) \in R$ (i.e., $b \le c^2$), then $(a, c) \in R$ (i.e., $a \le c^2$).

Let's test some values where $a \le b^2$ and $b \le c^2$ hold, but $a \le c^2$ might fail.

• Consider $a = 50$, $b = 7$, and $c = 3$.
• Is $(50, 7) \in R$? Yes, because $50 \le 7^2$ ($50 \le 49$). This is false. Need better examples.

Let's try $a=5, b=3, c=2$.

• Is $(5, 3) \in R$? Yes, because $5 \le 3^2$ ($5 \le 9$). This is true.
• Is $(3, 2) \in R$? Yes, because $3 \le 2^2$ ($3 \le 4$). This is true.
• Now, is $(5, 2) \in R$? This requires $5 \le 2^2$ ($5 \le 4$). This is false.

Since we found $a=5, b=3, c=2$ such that $(5, 3) \in R$ and $(3, 2) \in R$, but $(5, 2) \notin R$, the relation $R$ is not transitive.

Conclusion about the Relation R

We have shown that the relation $R = \{(a, b) : a \le b^2\}$ defined on the set of real numbers fails to satisfy reflexivity, symmetry, and transitivity. A relation must satisfy all three properties to be an equivalence relation.

Therefore, the relation $R$ is not an equivalence relation.

Based on the analysis, the correct statement is that the relation $R = \{(a, b) : a \le b^2\}$ is Not an Equivalence Relation.