To prove that √3 is irrational, we will use the method of proof by contradiction. We start by assuming the opposite, that √3 is a rational number. If √3 is rational, it can be expressed as a fraction a/b, where a and b are integers, b ≠ 0, and a and b have no common factors other than 1 (i.e., they are in their simplest form or coprime).

To prove that √3 is irrational, we will use the method of proof by contradiction. We start by assuming the opposite, that √3 is a rational number. If √3 is rational, it can be expressed as a fraction a/b, where a and b are integers, b ≠ 0, and a and b have no common factors other than 1 (i.e., they are in their simplest form or coprime).

Multiply both sides by b². This equation shows that a² is a multiple of 3. This implies that a² is divisible by 3.

According to a property of prime numbers, if a prime number p divides a², then p must also divide a. Here, 3 is a prime number, and since 3b² = a², it means 3 divides a², so 3 must divide a.

Divide both sides of 3b² = 9c² by 3 to get b² = 3c². This shows that b² is a multiple of 3, which implies that b² is divisible by 3. Therefore, similar to the case of a, 3 must also divide b.

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

Prove that $\sqrt{3}$ is irrational.

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

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. If $\sqrt{3}$ is rational, it can be expressed as a fraction $a/b$ , where $a$ and $b$ are integers, $b \neq 0$ , and $a$ and $b$ have no common factors other than 1 (i.e., they are in their simplest form or coprime).

Step 2: Square both sides

To eliminate the square root, we square both sides of the equation. This operation maintains the equality if both sides are positive. This gives us an expression relating $a^2$ and $b^2$ .

Step 3: Rearrange the equation

Multiply both sides by $b^2$ . This equation shows that $a^2$ is a multiple of 3. This implies that $a^2$ is divisible by 3.

Step 4: Conclude that 'a' is divisible by 3

According to a property of prime numbers, if a prime number $p$ divides $a^2$ , then $p$ must also divide $a$ . Here, 3 is a prime number, and since $3b^2 = a^2$ , it means 3 divides $a^2$ , so 3 must divide $a$ .

Step 5: Express 'a' as a multiple of 3

Since $a$ is divisible by 3, we can write $a$ as $3c$ , where $c$ is some integer. This substitution will allow us to find a similar relationship for $b$ .

Step 6: Substitute 'a' back into the equation

Substitute the expression for $a$ (i.e., $3c$ ) back into our rearranged equation $3b^2 = a^2$ . Squaring $3c$ gives $9c^2$ .

Step 7: Simplify and conclude that 'b' is divisible by 3

Divide both sides of $3b^2 = 9c^2$ by 3 to get $b^2 = 3c^2$ . This shows that $b^2$ is a multiple of 3, which implies that $b^2$ is divisible by 3. Therefore, similar to the case of $a$ , 3 must also divide $b$ .

Step 8: Reach a contradiction

We have shown that both $a$ and $b$ are divisible by 3. This means that 3 is a common factor of $a$ and $b$ . However, our initial assumption stated that $a$ and $b$ have no common factors other than 1. This contradiction arises from our initial assumption that $\sqrt{3}$ is rational.

Step 9: Conclude that $\sqrt{3}$ is irrational

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

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