Show that $5 - \sqrt{3}$ is irrational.

Step 1: Assume for contradiction it is rational

To prove that $5 - \sqrt{3}$ is irrational, we will use the method of contradiction. We start by assuming that $5 - \sqrt{3}$ is a rational number. If it is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$, $b$ are coprime (have no common factors other than 1).

Step 2: Isolate the irrational term

Now we rearrange the equation to isolate the irrational term, $\sqrt{3}$, on one side. We subtract $\frac{a}{b}$ from both sides and add $\sqrt{3}$ to both sides to achieve this.

Step 3: Simplify the rational expression

We combine the terms on the right-hand side of the equation into a single fraction. Since $a$ and $b$ are integers and $b \neq 0$, $5b - a$ will be an integer and $b$ will be a non-zero integer. Therefore, the expression $\frac{5b - a}{b}$ represents a rational number.

Step 4: State the contradiction

From reference-1, we know that $\sqrt{3}$ is an irrational number. However, the right-hand side of our equation, $\frac{5b - a}{b}$, is a rational number because it is expressed as a ratio of two integers. This leads to a contradiction: an irrational number cannot be equal to a rational number.

Step 5: Conclude the proof

Since our initial assumption that $5 - \sqrt{3}$ is rational leads to a contradiction, this assumption must be false. Therefore, $5 - \sqrt{3}$ must be an irrational number.

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