Number Sets (ℕ, ℤ, ℚ, ℝ): ℚ, ℝ, ℂ — Rational, Real & Complex

The Number Sets:
• Natural Numbers $\mathbb{N}$: Counting numbers $\{0, 1, 2, 3, \dots\}$
• Integers $\mathbb{Z}$: Whole numbers $\{\dots, -3, -2, -1, 0, 1, 2, \dots\}$
• Rational Numbers $\mathbb{Q}$: Numbers that can be written as a fraction $\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$. Examples: $\frac{1}{2}$, $-\frac{3}{4}$, $0.5$, $0.333\dots$
• Real Numbers $\mathbb{R}$: All rational and irrational numbers (numbers with non-repeating, infinite decimal expansions). Examples: $\pi$, $e$, $\sqrt{2}$.
• Complex Numbers $\mathbb{C}$: Numbers containing the imaginary unit $i$ (where $i^2 = -1$). Examples: $i$, $2 + 3i$.

Powers (Exponents) and Order of Operations:
A power (exponent) shows how many times a number is multiplied by itself. Let's look at different powers and how negative signs behave:
• Squared Numbers ($x^2$): A number multiplied by itself. For example, $3^2 = 3 \times 3 = 9$.
• Cubed Numbers ($x^3$): A number multiplied by itself three times. For example, $2^3 = 2 \times 2 \times 2 = 8$.
• Be careful with negative signs and parentheses!
If the negative sign is inside the parentheses, the sign of the result depends on whether the exponent is even or odd:
- Even exponents result in a positive number: $(-2)^2 = (-2) \times (-2) = 4$, and $(-2)^4 = 16$.
- Odd exponents result in a negative number: $(-2)^3 = (-2) \times (-2) \times (-2) = -8$, and $(-2)^5 = -32$.
If there are no parentheses, the negative sign is applied after the exponent: $-2^3 = -(2^3) = -8$, and $-2^4 = -(2^4) = -16$.
• Fractions:
$(\frac{1}{2})^2 = \frac{1}{4}$, and $(\frac{1}{2})^3 = \frac{1}{8}$, since we apply the power to both the numerator and the denominator: $(\frac{1}{2})^2 = \frac{1^2}{2^2} = \frac{1}{4}$ and $(\frac{1}{2})^3 = \frac{1^3}{2^3} = \frac{1}{8}$.

Square Roots:
The square root $\sqrt{x}$ is the number that, when squared, gives $x$. Examples: $\sqrt{4} = 2$, $\sqrt{9} = 3$.
• Irrational Numbers:
$\sqrt{2}$ cannot be written as a fraction. It is an irrational number. This means $\sqrt{2} \in \mathbb{R}$ but $\sqrt{2} \notin \mathbb{Q}$.
• A neat derivation:
$\sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{2}\sqrt{2}$
Since $\sqrt{2}$ is irrational, $\frac{1}{2}\sqrt{2}$ is also irrational, so it belongs to $\mathbb{R}$.
• Non-real Roots (Complex Numbers):
The square root of a negative number, like $\sqrt{-1}$, is not solvable in the real numbers. We define $\sqrt{-1} = i$, where $i$ is the imaginary unit. Since it is not a real number, $i \notin \mathbb{R}$, but it belongs to the complex numbers: $i \in \mathbb{C}$.

Optional: Proof that $\sqrt{2}$ is Irrational (Not necessary to memorize or fully understand):
Suppose $\sqrt{2}$ is rational. Then we can write it in simplest terms as $\sqrt{2} = \frac{a}{b}$ (where $a$ and $b$ have no common factors).
Squaring both sides gives $2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$. This means $a^2$ is even, so $a$ must be even (say $a = 2k$).
Substitute $a = 2k$ into the equation: $(2k)^2 = 2b^2 \Rightarrow 4k^2 = 2b^2 \Rightarrow b^2 = 2k^2$. This means $b^2$ is even, so $b$ must be even.
Since both $a$ and $b$ are even, they share a common factor of $2$, contradicting our assumption that $\frac{a}{b}$ was in simplest terms. Thus, $\sqrt{2}$ must be irrational.