Next, we square both sides of the equation √3 = (a)/(b) to eliminate the square root. This gives us 3 = (a²)/(b²). Multiplying both sides by b² results in the equation 3b² = a². This equation implies that a² is a multiple of 3.

Next, we square both sides of the equation √3 = (a)/(b) to eliminate the square root. This gives us 3 = (a²)/(b²). Multiplying both sides by b² results in the equation 3b² = a². This equation implies that a² is a multiple of 3.

According to a fundamental property of numbers, if a prime number (like 3) divides the square of an integer (a²), then it must also divide the integer itself (a). Therefore, a must be a multiple of 3. We can express this by writing a = 3c, where c is some integer.

According to a fundamental property of numbers, if a prime number (like 3) divides the square of an integer (a²), then it must also divide the integer itself (a). Therefore, a must be a multiple of 3. We can express this by writing a = 3c, where c is some integer.

Now, we substitute a = 3c back into our equation 3b² = a². This gives us 3b² = (3c)², which simplifies to 3b² = 9c². Dividing both sides by 3, we get b² = 3c². This new equation implies that b² is also a multiple of 3.

Now, we substitute a = 3c back into our equation 3b² = a². This gives us 3b² = (3c)², which simplifies to 3b² = 9c². Dividing both sides by 3, we get b² = 3c². This new equation implies that b² is also a multiple of 3.

Similar to the reasoning for a, since 3 divides b², it must also divide b. Therefore, b is a multiple of 3.

Since our initial assumption led to a contradiction, the assumption must be false. Therefore, √3 cannot be a rational number, which means it must be an irrational number.

Prove that the square root of 3 is irrational.

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

Step-by-step solution

Step 1: Assume $\sqrt{3}$ is rational

To prove that $\sqrt{3}$ is irrational, we will use the method of proof by 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 and rearrange

Next, we square both sides of the equation $\sqrt{3} = \frac{a}{b}$ to eliminate the square root. This gives us $3 = \frac{a^2}{b^2}$ . Multiplying both sides by $b^2$ results in the equation $3b^2 = a^2$ . This equation implies that $a^2$ is a multiple of 3.

Step 3: Deduce $a$ is a multiple of 3

According to a fundamental property of numbers, if a prime number (like 3) divides the square of an integer ( $a^2$ ), then it must also divide the integer itself ( $a$ ). Therefore, $a$ must be a multiple of 3. We can express this by writing $a = 3c$ , where $c$ is some integer.

Step 4: Substitute $a=3c$ and simplify

Now, we substitute $a = 3c$ back into our equation $3b^2 = a^2$ . This gives us $3b^2 = (3c)^2$ , which simplifies to $3b^2 = 9c^2$ . Dividing both sides by 3, we get $b^2 = 3c^2$ . This new equation implies that $b^2$ is also a multiple of 3.

Step 5: Deduce $b$ is a multiple of 3

Similar to the reasoning for $a$ , since 3 divides $b^2$ , it must also divide $b$ . Therefore, $b$ is a multiple of 3.

Step 6: Reach a contradiction

From our deductions, we found 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 7: Conclusion

Since our initial assumption led to a contradiction, the assumption must be false. Therefore, $\sqrt{3}$ cannot be a rational number, which means it must be an irrational number.

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