Cartesian Product Basics and Universal Sets

The core task is to find a universal set that encompasses both the Cartesian product $Q \times N$ and the Cartesian product $N \times Q$. We need to understand what these sets represent and how they relate to potential universal sets.

Number Sets: N, Q, R Explained

First, let's define the sets involved:

• Set of Natural Numbers ($N$): This set contains all positive integers starting from 1. $N = \{1, 2, 3, 4, ...\}$.
• Set of Rational Numbers ($Q$): This set includes numbers that can be written as a fraction $\frac{p}{q}$, where $p$ (numerator) and $q$ (denominator) are integers, and the denominator $q$ is not zero. Examples include $\frac{1}{2}$, $-3$, $0.75 = \frac{3}{4}$. Formally, $Q = \{ \frac{p}{q} | p \in \mathbb{Z}, q \in \mathbb{Z}, q \neq 0 \}$.
• Set of Real Numbers ($R$): This set comprises all rational numbers and irrational numbers (like $\pi$ or $\sqrt{2}$). It represents all points on the continuous number line.

Crucially, these sets have a hierarchical relationship:

• $N \subset Q$ (Every natural number is also a rational number).
• $Q \subset R$ (Every rational number is also a real number).

This means $N$ is a subset of $Q$, and $Q$ is a subset of $R$. Consequently, $N$ is also a subset of $R$. The order is $N \subset Q \subset R$.

Understanding Universal Sets

A set $U$ is defined as a universal set for other sets, say $A$ and $B$, if both $A$ and $B$ are subsets of $U$. This means every element in $A$ must also be in $U$, and every element in $B$ must also be in $U$. For this problem, we are looking for a set $U$ such that:

• $Q \times N \subseteq U$
• $N \times Q \subseteq U$

where $ \subseteq $ denotes the subset relationship.

Analyzing $Q \times R$ as a Universal Set

Let's evaluate the option $Q \times R$. This represents the set of all possible ordered pairs $(x, y)$ where $x$ is an element from the set of rational numbers ($Q$) and $y$ is an element from the set of real numbers ($R$).

The set $Q \times R$ can be written as: $Q \times R = \{ (x, y) | x \in Q, y \in R \}$.

Now, we need to verify if the other two sets, $Q \times N$ and $N \times Q$, are indeed subsets of $Q \times R$.

$Q \times N$ and $N \times Q$ Subset Verification

1. Verifying $Q \times N \subseteq Q \times R$:

Take any arbitrary element $(a, b)$ from the set $Q \times N$. By the definition of a Cartesian product, this means $a \in Q$ and $b \in N$.

We know that $N$ is a subset of $R$ ($N \subset R$). This implies that any element belonging to $N$ must also belong to $R$. Therefore, $b \in R$.

So, for the element $(a, b)$, we have $a \in Q$ and $b \in R$.

According to the definition of $Q \times R$, any ordered pair where the first component is from $Q$ and the second component is from $R$ is an element of $Q \times R$. Since $a \in Q$ and $b \in R$, the pair $(a, b)$ is an element of $Q \times R$.

This demonstrates that every element in $Q \times N$ is also present in $Q \times R$. Hence, $Q \times N$ is a subset of $Q \times R$.

2. Verifying $N \times Q \subseteq Q \times R$:

Now, consider any arbitrary element $(c, d)$ from the set $N \times Q$. This implies $c \in N$ and $d \in Q$.

To check if $(c, d)$ is in $Q \times R$, we need to ensure that its first component $c$ is in $Q$ and its second component $d$ is in $R$.

Let's examine the components:

• For the first component, $c$: We are given $c \in N$. Since $N \subset Q$, it follows directly that $c \in Q$.
• For the second component, $d$: We are given $d \in Q$. Since $Q \subset R$, it follows directly that $d \in R$.

We have established that $c \in Q$ and $d \in R$. Therefore, the ordered pair $(c, d)$ satisfies the conditions for membership in $Q \times R$.

This shows that every element in $N \times Q$ is also present in $Q \times R$. Hence, $N \times Q$ is a subset of $Q \times R$.

Conclusion

Since $Q \times R$ contains all elements from both $Q \times N$ and $N \times Q$, it functions as a universal set for these two Cartesian products.