Assume (1)/(sqrt(2)) is a rational number.

If $\frac{1}{\sqrt{2}}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Squaring both sides:

From $2a^2 = b^2$, it means that $b^2$ is divisible by 2.
If $b^2$ is divisible by 2, then $b$ must also be divisible by 2.
So, we can write $b = 2k$ for some integer $k$.

From $a^2 = 2k^2$, it means that $a^2$ is divisible by 2.
If $a^2$ is divisible by 2, then $a$ must also be divisible by 2.

From Step 3, $b$ is divisible by 2. From Step 5, $a$ is divisible by 2.
This means that $a$ and $b$ have a common factor of 2.
However, in Step 1, we assumed that $a$ and $b$ are coprime integers (meaning they have no common factors other than 1).
This contradicts our initial assumption.

Since our assumption leads to a contradiction, our initial assumption that $\frac{1}{\sqrt{2}}$ is rational must be false.
Therefore, $\frac{1}{\sqrt{2}}$ is an irrational number.

If $7\sqrt{5}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Since $a$ and $b$ are integers, $a$ is an integer and $7b$ is a non-zero integer.
Therefore, $\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.

However, we know that $\sqrt{5}$ is an irrational number.
This contradicts our conclusion from Step 3.

Since our assumption leads to a contradiction, our initial assumption that $7\sqrt{5}$ is rational must be false.
Therefore, $7\sqrt{5}$ is an irrational number.

If $6 + \sqrt{2}$ is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Since $a$ and $b$ are integers, $a - 6b$ is an integer and $b$ is a non-zero integer.
Therefore, $\frac{a - 6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.

However, we know that $\sqrt{2}$ is an irrational number.
This contradicts our conclusion from Step 3.

Since our assumption leads to a contradiction, our initial assumption that $6 + \sqrt{2}$ is rational must be false.
Therefore, $6 + \sqrt{2}$ is an irrational number.