Given: The number $\sqrt{5}$.

To Prove: $\sqrt{5}$ is an irrational number.

Step 1: Assumption for Contradiction
Let us assume, to the contrary, that $\sqrt{5}$ is a rational number. By the definition of rational numbers, if $\sqrt{5}$ is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (i.e., their highest common factor, $\text{HCF}(a, b) = 1$).

Step 2: Algebraic Manipulation
We have:
$\sqrt{5} = \frac{a}{b}$
Squaring both sides, we get:
$(\sqrt{5})^2 = (\frac{a}{b})^2$
$5 = \frac{a^2}{b^2}$
$a^2 = 5b^2$ --- (Equation 1)

Step 3: Applying the Fundamental Theorem of Arithmetic
From Equation 1, it is evident that $a^2$ is divisible by $5$.
[Theorem: If a prime number $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.]
Therefore, $a$ must also be divisible by $5$.
We can write $a = 5k$ for some integer $k$.

Step 4: Substitution and Further Deduction
Substitute $a = 5k$ into Equation 1:
$(5k)^2 = 5b^2$
$25k^2 = 5b^2$
Divide both sides by $5$:
$b^2 = 5k^2$
This implies that $b^2$ is divisible by $5$.
Following the same theorem as in Step 3, if $b^2$ is divisible by $5$, then $b$ must also be divisible by $5$.

Step 5: Analysis of the Contradiction
From Step 3 and Step 4, we have concluded that both $a$ and $b$ have at least $5$ as a common factor. This contradicts our initial assumption that $a$ and $b$ are coprime (i.e., $\text{HCF}(a, b) = 1$).

Step 6: Conclusion
The contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is a rational number. Therefore, we conclude that $\sqrt{5}$ must be irrational.

Final Answer: Since the assumption that $\sqrt{5}$ is rational leads to a contradiction, it is proven that $\sqrt{5}$ is an irrational number.

Given: A real number $3 + 2\sqrt{5}$.

To Prove: $3 + 2\sqrt{5}$ is an irrational number.

Step 1: Assumption for Contradiction
Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is a rational number. By the definition of rational numbers, if a number is rational, it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., they have no common factors other than 1).

Step 2: Setting up the Equation
Based on our assumption, we can write:
$3 + 2\sqrt{5} = \frac{p}{q}$

Step 3: Isolating the Irrational Part
We perform algebraic manipulations to isolate the term containing the square root.
Subtract 3 from both sides:
$2\sqrt{5} = \frac{p}{q} - 3$
$2\sqrt{5} = \frac{p - 3q}{q}$

Now, divide both sides by 2:
$\sqrt{5} = \frac{p - 3q}{2q}$

Step 4: Analyzing the Rationality of the Expression
Since $p$ and $q$ are integers, the expression $\frac{p - 3q}{2q}$ must also be a rational number because:
1. The difference of two integers ($p - 3q$) is an integer.
2. The product of two integers ($2q$) is an integer.
3. The quotient of two integers (where the denominator is non-zero) is a rational number.

Step 5: Identifying the Contradiction
From our equation in Step 3, we have:
$\sqrt{5} = \text{a rational number}$

However, we know from the fundamental properties of real numbers that $\sqrt{5}$ is an irrational number. This creates a contradiction because a number cannot be both rational and irrational simultaneously.

Step 6: Conclusion
The contradiction has arisen because of our initial incorrect assumption that $3 + 2\sqrt{5}$ is a rational number. Therefore, we conclude that our assumption is false and $3 + 2\sqrt{5}$ must be irrational.

Final Answer: Since the assumption that $3 + 2\sqrt{5}$ is rational leads to a contradiction, it is proven that $3 + 2\sqrt{5}$ is an irrational number.

Given: A real number $6 + \sqrt{2}$.

To Prove: $6 + \sqrt{2}$ is an irrational number.

Step 1: Assumption for Contradiction
To prove that $6 + \sqrt{2}$ is irrational, we begin by assuming the contrary. Let us assume that $6 + \sqrt{2}$ is a rational number. [By the method of contradiction].

Step 2: Definition of Rational Numbers
If $6 + \sqrt{2}$ is a rational number, then it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., they have no common factors other than 1).
Thus, we can write:
$6 + \sqrt{2} = \frac{p}{q}$

Step 3: Algebraic Rearrangement
We isolate the irrational part ($\sqrt{2}$) on one side of the equation to analyze its nature:
$\sqrt{2} = \frac{p}{q} - 6$
$\sqrt{2} = \frac{p - 6q}{q}$

Step 4: Analyzing the Right-Hand Side (RHS)
Since $p$ and $q$ are integers, $6$ is an integer, and the difference of two integers ($p - 6q$) is also an integer. Similarly, $q$ is an integer. Therefore, $\frac{p - 6q}{q}$ is a quotient of two integers, which satisfies the definition of a rational number. [Since the set of rational numbers is closed under subtraction and division].

Step 5: Analyzing the Left-Hand Side (LHS) and Identifying Contradiction
From Step 4, we have concluded that the RHS ($\frac{p - 6q}{q}$) is rational. This implies that the LHS ($\sqrt{2}$) must also be rational.
However, we know from the fundamental theorem of real numbers that $\sqrt{2}$ is an irrational number. [A well-established mathematical fact].

Step 6: Conclusion
Our assumption that $6 + \sqrt{2}$ is rational has led to a contradiction, as it implies $\sqrt{2}$ is rational, which is false. Therefore, our initial assumption must be incorrect.
Hence, it is proved that $6 + \sqrt{2}$ is an irrational number.

Final Answer: Since the assumption that $6 + \sqrt{2}$ is rational leads to a contradiction, $6 + \sqrt{2}$ is an irrational number.

Given: A number $\frac{1}{\sqrt{2}}$.

To Prove: $\frac{1}{\sqrt{2}}$ is an irrational number.

Step 1: Assumption for Proof by Contradiction
Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is a rational number. By the definition of rational numbers, if a number is rational, it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor $\text{gcd}(p, q) = 1$).

Therefore, we can write:
$\frac{1}{\sqrt{2}} = \frac{a}{b}$
where $a$ and $b$ are integers, $b \neq 0$, and $\text{gcd}(a, b) = 1$.

Step 2: Algebraic Manipulation
To isolate the radical term, we rearrange the equation:
$\frac{1}{\sqrt{2}} = \frac{a}{b}$
Taking the reciprocal of both sides:
$\sqrt{2} = \frac{b}{a}$

Step 3: Analyzing the Rationality
Since $a$ and $b$ are integers, the quotient $\frac{b}{a}$ must be a rational number by the definition of rational numbers (the ratio of two integers is rational).
[Since $a, b \in \mathbb{Z}$ and $a \neq 0$, $\frac{b}{a} \in \mathbb{Q}$]

This implies that $\sqrt{2}$ is a rational number.

Step 4: Identifying the Contradiction
We know from the fundamental theorems of real numbers that $\sqrt{2}$ is an irrational number. Our assumption that $\frac{1}{\sqrt{2}}$ is rational has led to the conclusion that $\sqrt{2}$ is rational, which contradicts the established mathematical fact that $\sqrt{2}$ is irrational.

[Since the assumption leads to a contradiction, the initial assumption must be false.]

Step 5: Conclusion
Because the assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction, we must conclude that $\frac{1}{\sqrt{2}}$ is irrational.

Final Answer: Since the assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction, it is proven that $\frac{1}{\sqrt{2}}$ is an irrational number.

Given: A number $7\sqrt{5}$.

To Prove: That $7\sqrt{5}$ is an irrational number.

Step 1: Assumption for the Method of Contradiction
To prove that $7\sqrt{5}$ is irrational, we assume the contrary. Let us assume that $7\sqrt{5}$ is a rational number.

Step 2: Definition of Rational Numbers
If $7\sqrt{5}$ is a rational number, then by definition, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are co-prime (i.e., their highest common factor, $HCF(a, b) = 1$).

So, we can write:
$7\sqrt{5} = \frac{a}{b}$

Step 3: Algebraic Manipulation
We isolate the square root term on one side of the equation to analyze its nature.
Divide both sides of the equation by $7$:
$\sqrt{5} = \frac{a}{7b}$

Step 4: Analyzing the Rationality of the Expression
Since $a$ and $b$ are integers, $7$ is also an integer. Therefore, the expression $\frac{a}{7b}$ is a ratio of two integers. By the definition of rational numbers, any number that can be expressed as a quotient of two integers (where the denominator is non-zero) is a rational number.

Thus, $\frac{a}{7b}$ is a rational number.

Step 5: Identifying the Contradiction
From the equation in Step 3, we have:
$\sqrt{5} = \text{a rational number}$

However, it is a well-established mathematical fact (proven via the Fundamental Theorem of Arithmetic) that $\sqrt{5}$ is an irrational number.

[Since the assumption that $7\sqrt{5}$ is rational leads to the conclusion that $\sqrt{5}$ is rational, which contradicts the known fact that $\sqrt{5}$ is irrational, our initial assumption must be false.]

Step 6: Conclusion
Since the assumption that $7\sqrt{5}$ is rational leads to a contradiction, we conclude that $7\sqrt{5}$ must be irrational.

Final Answer: Since the assumption that $7\sqrt{5}$ is rational leads to a contradiction, it is proven that $7\sqrt{5}$ is an irrational number.