To prove that $5 - \frac{2}{7} \sqrt{3}$ is irrational, we will use the method of contradiction. We assume that the number is rational and then show that this assumption leads to a contradiction.

Step 1: Assume the number is rational.
Let $x = 5 - \frac{2}{7} \sqrt{3}$ be a rational number. Since rational numbers are closed under subtraction and multiplication, if $\sqrt{3}$ were rational, then $x$ would be rational. However, we know that $\sqrt{3}$ is irrational. Let's proceed with the assumption.

Step 2: Rearrange the equation to isolate the irrational part.
$x = 5 - \frac{2}{7} \sqrt{3}$
$\frac{2}{7} \sqrt{3} = 5 - x$
$\sqrt{3} = \frac{7}{2} (5 - x)$

Step 3: Analyze the right-hand side (RHS).
Since we assumed $x$ is rational, and 5 is rational, $5 - x$ is rational (rational numbers are closed under subtraction). The product of a rational number ( $\frac{7}{2}$ ) and a rational number ( $5 - x$ ) is rational. Therefore, the entire RHS, $\frac{7}{2} (5 - x)$ , is rational.

Step 4: Identify the contradiction.
The equation states that $\sqrt{3}$ is equal to a rational number. However, it is a proven fact that $\sqrt{3}$ is irrational. A number cannot be both rational and irrational. This is a contradiction.

Step 5: Conclude.
Since our initial assumption that $x$ is rational leads to a contradiction, the assumption must be false. Therefore, $5 - \frac{2}{7} \sqrt{3}$ is irrational.

Related Topics and Formulae

Irrational Numbers: A number is called irrational if it cannot be expressed in the form $\frac{p}{q}$ , where $p$ and $q$ are integers and $q \neq 0$ . Examples include $\sqrt{2}$ , $\sqrt{3}$ , $π$ .

Proof by Contradiction: A common method to prove that a number is irrational. You assume the opposite (that it is rational) and show that this assumption leads to an impossible situation or a statement that contradicts a known fact.

Closure Properties of Rational Numbers: The set of rational numbers is closed under addition, subtraction, multiplication, and division (by a non-zero rational). This means that performing these operations on rational numbers always yields another rational number. This property is crucial for constructing the contradiction in the proof.