Activity Questions 1.3

The statement that is false is:

d) The product of two irrational numbers is always an irrational number.

Here’s an explanation for each statement:

• a) The sum of two natural numbers is always a natural number.

  • True. Natural numbers are defined as the set {0, 1, 2, 3, 4, 5, ….}. When discussing arithmetic operations on natural numbers, the sources indicate that addition always results in a natural number, unlike subtraction which can lead to numbers outside the set of natural numbers (requiring the introduction of integers).

• b) The difference between two integers is always an integer.

  • True. Integers are defined as the set {…,−4,−3,−2,−1, 0, 1, 2, 3, 4, …}. The concept of integers was introduced to account for operations like subtraction that would take results “below 0” when starting with natural numbers. This means that the set of integers is closed under subtraction, ensuring that the difference between any two integers will always be another integer. For instance, subtracting 6 from 5 (5 - 6) results in -1, which is an integer.

• c) The product of two rational numbers is always a real number.

  • True. Rational numbers are defined as numbers that can be written in the form p/q, where p and q are both integers. Real numbers are defined as all rational numbers combined with irrational numbers. The number systems are hierarchical, with natural numbers being a subset of integers, integers a subset of rational numbers, and rational numbers a subset of real numbers. Therefore, if the product of two rational numbers yields another rational number, that result will inherently also be a real number. For example, if you multiply (p1/q1) by (p2/q2), the result is (p1p2)/(q1q2), where p1p2 and q1q2 are also integers, making the product a rational number, and by extension, a real number.

• d) The product of two irrational numbers is always an irrational number.

  • False. Irrational numbers are numbers that cannot be expressed as a simple fraction p/q, where p and q are integers. Examples include $\sqrt{2}$, $\sqrt{3}$, $\pi$, and $e$. While the product of two irrational numbers can be irrational (e.g., $\sqrt{2} \times \sqrt{3} = \sqrt{6}$, and $\sqrt{6}$ is an irrational number), it is not always irrational. For example, if you multiply $\sqrt{2}$ by $\sqrt{2}$, both of which are irrational numbers, the product is 2. The number 2 is a natural number, an integer, and a rational number (as it can be written as 2/1), thus it is not an irrational number. The sources explicitly state that the square root of any integer that is not a perfect square is an irrational number.

Let’s evaluate each statement based on the provided sources:

• $-\sqrt{-2}$ is a complex number.

  • The concept of complex numbers is introduced to address the square root of a negative number. Specifically, if one wishes to find the square root of negative numbers, a new class of numbers called complex numbers must be created. The set of complex numbers includes the square root of -1 and implicitly, the square roots of all other negative numbers. Since $\sqrt{-2}$ involves the square root of a negative number, it falls into the category of complex numbers. The presence of a negative sign in front of $\sqrt{-2}$ (i.e., $-\sqrt{-2}$) does not change its classification as a complex number.
  • Therefore, this statement is true.

• Real numbers extend rational numbers.

  • The sources explicitly state this relationship multiple times. For instance, it is mentioned that real numbers extend rational numbers by incorporating “irrational numbers which are very much on the number line, but which cannot be written on the form p by q”. This means that every rational number is inherently a real number because the set of real numbers encompasses all rational numbers, in addition to irrational numbers. The hierarchical relationship is also visualised through Venn diagrams, showing natural numbers as a subset of integers, integers as a subset of rationals, and rationals as a subset of reals.
  • Therefore, this statement is true.

Since both of the provided statements are true, the option “None of these” is false.