To prove that 3 + 2√5 is an irrational number, we will use the method of contradiction. We assume, for the sake of contradiction, that 3 + 2√5 is a rational number. If it is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a, b are coprime (have no common factors other than 1).

To prove that 3 + 2√5 is an irrational number, we will use the method of contradiction. We assume, for the sake of contradiction, that 3 + 2√5 is a rational number. If it is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a, b are coprime (have no common factors other than 1).

Now, we rearrange the equation to isolate the term involving √5. We subtract 3 from both sides of the equation.

Since a and b are integers, a - 3b will also be an integer, and 2b will be a non-zero integer. Therefore, the expression (a - 3b)/(2b) is a rational number. This implies that √5 is a rational number. However, we are given that √5 is an irrational number. This creates a contradiction, as an irrational number cannot be equal to a rational number.

Since a and b are integers, a - 3b will also be an integer, and 2b will be a non-zero integer. Therefore, the expression (a - 3b)/(2b) is a rational number. This implies that √5 is a rational number. However, we are given that √5 is an irrational number. This creates a contradiction, as an irrational number cannot be equal to a rational number.

Prove that $(3 + 2\sqrt{5})$ is an irrational number, given that $\sqrt{5}$ is an irrational number.

Answer: Therefore, $3 + 2\sqrt{5}$ is an irrational number.

Step-by-step solution

Step 1: Assume the opposite

To prove that $3 + 2\sqrt{5}$ is an irrational number, we will use the method of contradiction. We assume, for the sake of contradiction, that $3 + 2\sqrt{5}$ is a rational number. If it is rational, it can be expressed as a fraction $\frac{a}{b}$ , 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 term involving $\sqrt{5}$ . We subtract 3 from both sides of the equation.

Step 3: Simplify the rational side

We combine the terms on the right-hand side by finding a common denominator. This simplifies the expression to a single fraction.

Step 4: Further isolate the irrational term

To completely isolate $\sqrt{5}$ , we divide both sides of the equation by 2. This gives us an expression for $\sqrt{5}$ in terms of $a$ and $b$ .

Step 5: Reach a contradiction

Since $a$ and $b$ are integers, $a - 3b$ will also be an integer, and $2b$ will be a non-zero integer. Therefore, the expression $\frac{a - 3b}{2b}$ is a rational number. This implies that $\sqrt{5}$ is a rational number. However, we are given that $\sqrt{5}$ is an irrational number. This creates a contradiction, as an irrational number cannot be equal to a rational number.

Step 6: Conclusion

Our initial assumption that $3 + 2\sqrt{5}$ is rational led to a contradiction. Therefore, our assumption must be false, and $3 + 2\sqrt{5}$ must be an irrational number.

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