Now, we square both sides of the equation √3 = (a)/(b) to eliminate the square root. This gives us 3 = (a²)/(b²). Rearranging the terms, we get 3b² = a². This equation implies that a² is a multiple of 3, which means a itself must be a multiple of 3.

Now, we square both sides of the equation √3 = (a)/(b) to eliminate the square root. This gives us 3 = (a²)/(b²). Rearranging the terms, we get 3b² = a². This equation implies that a² is a multiple of 3, which means a itself must be a multiple of 3.

Since a is a multiple of 3, we can write a = 3c for some integer c. Substituting this into the equation 3b² = a², we get 3b² = (3c)², which simplifies to 3b² = 9c². Dividing both sides by 3, we find b² = 3c². This implies that b² is a multiple of 3, and therefore b must also be a multiple of 3.

Since a is a multiple of 3, we can write a = 3c for some integer c. Substituting this into the equation 3b² = a², we get 3b² = (3c)², which simplifies to 3b² = 9c². Dividing both sides by 3, we find b² = 3c². This implies that b² is a multiple of 3, and therefore b must also be a multiple of 3.

Prove that √3 is irrational.

Answer: $\sqrt{3}$ is irrational.

Step-by-step solution

Step 1: Assume the opposite

To prove that $\sqrt{3}$ is irrational, we use the method of contradiction. We start by assuming the opposite, that $\sqrt{3}$ is a rational number. A rational number can always be expressed as a fraction $\frac{a}{b}$ , where $a$ and $b$ are integers, $b \neq 0$ , and $a$ and $b$ have no common factors other than 1 (i.e., they are coprime).

Step 2: Square both sides

Now, we square both sides of the equation $\sqrt{3} = \frac{a}{b}$ to eliminate the square root. This gives us $3 = \frac{a^2}{b^2}$ . Rearranging the terms, we get $3b^2 = a^2$ . This equation implies that $a^2$ is a multiple of 3, which means $a$ itself must be a multiple of 3.

Step 3: Substitute 'a' as a multiple of 3

Since $a$ is a multiple of 3, we can write $a = 3c$ for some integer $c$ . Substituting this into the equation $3b^2 = a^2$ , we get $3b^2 = (3c)^2$ , which simplifies to $3b^2 = 9c^2$ . Dividing both sides by 3, we find $b^2 = 3c^2$ . This implies that $b^2$ is a multiple of 3, and therefore $b$ must also be a multiple of 3.

Step 4: Reach a contradiction

From the previous steps, we concluded that both $a$ and $b$ are multiples of 3. This means that $a$ and $b$ have a common factor of 3. However, our initial assumption was that $a$ and $b$ are coprime, meaning they have no common factors other than 1. This creates a contradiction.

Step 5: Conclude irrationality

Since our assumption that $\sqrt{3}$ is rational leads to a contradiction (that $a$ and $b$ have a common factor of 3, despite being assumed coprime), our initial assumption must be false. Therefore, $\sqrt{3}$ cannot be rational, which means it must be irrational.

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