Q1(i):

State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number.

Step 1: Theoretical Foundation of the Real Number System

To evaluate the truth value of the statement, we must first establish the formal definitions of the sets of numbers involved in the real number system.

• Rational Numbers ($\mathbb{Q}$): Any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
• Irrational Numbers ($\mathbb{I}$): Any number that cannot be expressed in the form $\frac{p}{q}$. Their decimal expansions are non-terminating and non-repeating (e.g., $\sqrt{2}$, $\pi$, $e$).
• Real Numbers ($\mathbb{R}$): The set of real numbers is defined as the continuous collection of all rational and irrational numbers.

[By the axioms of set theory], the set of Real Numbers is the union of Rational and Irrational numbers. Mathematically, this is expressed as:

$\mathbb{R} = \mathbb{Q} \cup \mathbb{I}$

Step 2: Logical Deduction and Set Inclusion

Given the union $\mathbb{R} = \mathbb{Q} \cup \mathbb{I}$, it logically follows that the set of irrational numbers is a proper subset of the set of real numbers.

$\mathbb{I} \subset \mathbb{R}$

[Per the definition of a subset], if an element $x$ belongs to the set of irrational numbers ($x \in \mathbb{I}$), it must necessarily belong to the set of real numbers ($x \in \mathbb{R}$). Therefore, every single irrational number is, by definition, a real number.

Step 3: Visualizing the Number System

The following Euler diagram illustrates the partition of the Real Number system, demonstrating that the Irrational Numbers are entirely contained within the boundaries of the Real Numbers.

Step 4: Final Evaluation

Because the collection of all rational numbers and irrational numbers together forms the complete collection of real numbers, there is no irrational number that exists outside the domain of real numbers. Every point on the real number line corresponds to either a rational or an irrational number.

Final Solution: True. Every irrational number is a real number because the set of real numbers is composed of the union of all rational and irrational numbers ($\mathbb{R} = \mathbb{Q} \cup \mathbb{I}$).