To show that $3 \sqrt{7}$ is an irrational number, we will use a proof by contradiction. An irrational number is a number that cannot be expressed as a ratio of two integers (i.e., in the form $\frac{p}{q}$ , where $p$ and $q$ are integers and $q \neq 0$ ).

Step 1: Assume the opposite.
Let us assume that $3 \sqrt{7}$ is rational. Therefore, it can be written as a fraction in its simplest form: $3 \sqrt{7} = \frac{a}{b}$ , where $a$ and $b$ are co-prime integers (they have no common factor other than 1) and $b \neq 0$ .

Step 2: Rearrange the equation.
We can rearrange the equation to isolate the square root term. $\sqrt{7} = \frac{a}{3 b}$

Step 3: Analyze the new equation.
The left side of the equation, $\sqrt{7}$ , is irrational. For the equation to hold true, the right side, $\frac{a}{3 b}$ , must also be irrational. However, since $a$ and $b$ are integers, $3 b$ is also an integer. This means $\frac{a}{3 b}$ is a rational number (a ratio of two integers). We now have a rational number equal to an irrational number ( $\sqrt{7}$ ). This is a contradiction.

Conclusion:
Our initial assumption that $3 \sqrt{7}$ is rational has led to a contradiction. Therefore, $3 \sqrt{7}$ must be an irrational number.

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