Real Numbers - Exercise 1.2

Assume to the contrary that $\sqrt{5}$ is a rational number.
Therefore, $\sqrt{5}$ can be expressed in the form: $$\sqrt{5} = \frac{a}{b}$$ where $a$ and $b$ are co-prime integers (they have no common factor other than 1) and $b \neq 0$.

Squaring both sides of the equation: $$5 = \frac{a^2}{b^2} \implies 5b^2 = a^2 \quad \text{--- (Equation 1)}$$

Since $5b^2 = a^2$, it means $5$ divides $a^2$.
By Theorem (if a prime $p$ divides $a^2$, then $p$ divides $a$), $5$ divides $a$.
Hence, we can write: $$a = 5k \quad \text{for some integer } k$$

Substitute $a = 5k$ into Equation 1: $$5b^2 = (5k)^2$$ $$5b^2 = 25k^2 \implies b^2 = 5k^2$$ This means $5$ divides $b^2$, and therefore $5$ divides $b$.

From Steps 3 and 4, both $a$ and $b$ have at least 5 as a common factor.
This contradicts our initial statement that $a$ and $b$ are co-prime integers.
This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

Let us assume to the contrary that $3 + 2\sqrt{5}$ is rational.
Then there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that: $$3 + 2\sqrt{5} = \frac{a}{b}$$

Subtract 3 from both sides: $$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$$ Divide both sides by 2: $$\sqrt{5} = \frac{a - 3b}{2b}$$

Since $a$ and $b$ are integers, $\frac{a - 3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is rational.
However, this contradicts the known fact that $\sqrt{5}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.

Assume to the contrary that $\dfrac{1}{\sqrt{2}}$ is rational.
Then $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$, where $a, b \in \mathbb{Z}$, co-prime, and $a \neq 0, b \neq 0$.
Taking reciprocal on both sides: $$\sqrt{2} = \frac{b}{a}$$ Since $a$ and $b$ are integers, $\dfrac{b}{a}$ is rational, which means $\sqrt{2}$ is rational.
This contradicts the fact that $\sqrt{2}$ is irrational.
$\therefore \dfrac{1}{\sqrt{2}}$ is irrational.

Assume to the contrary that $7\sqrt{5}$ is rational.
Then $7\sqrt{5} = \dfrac{a}{b}$, where $a$ and $b$ are co-prime integers and $b \neq 0$.
Dividing both sides by 7: $$\sqrt{5} = \frac{a}{7b}$$ Since $a$ and $b$ are integers, $\dfrac{a}{7b}$ is a rational number.
This implies $\sqrt{5}$ is rational, contradicting the fact that $\sqrt{5}$ is irrational.
$\therefore 7\sqrt{5}$ is irrational.

Assume to the contrary that $6 + \sqrt{2}$ is rational.
Then $6 + \sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are co-prime integers and $b \neq 0$.
Subtract 6 from both sides: $$\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$$ Since $a$ and $b$ are integers, $\dfrac{a - 6b}{b}$ is rational, implying $\sqrt{2}$ is rational.
This contradicts the fact that $\sqrt{2}$ is irrational.
$\therefore 6 + \sqrt{2}$ is irrational.