Prove that the square root of 5 is irrational.

Step 1: Assume $\sqrt{5}$ is rational

We begin by assuming, for the sake of contradiction, that $\sqrt{5}$ is a rational number. By definition, a rational number can 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., the fraction is in its simplest form).

Step 2: Square both sides

To eliminate the square root, we square both sides of the equation. This gives us $5 = \frac{a^2}{b^2}$. We can then rearrange this equation to $5b^2 = a^2$.

Step 3: Analyze divisibility

The equation $a^2 = 5b^2$ implies that $a^2$ is a multiple of 5, and thus $a^2$ is divisible by 5. A fundamental property of numbers states that if the square of an integer is divisible by a prime number, then the integer itself must also be divisible by that prime number. Since 5 is a prime number, it follows that $a$ must be divisible by 5.

Step 4: Express $a$ in terms of $5k$

Since $a$ is divisible by 5, we can express $a$ as $5k$ for some integer $k$. This substitution will allow us to explore the implications for $b$.

Step 5: Substitute $a$ back into the equation

Now, we substitute $a = 5k$ back into our equation $5b^2 = a^2$. This yields $5b^2 = (5k)^2$. We will simplify this expression to find information about $b$.

Step 6: Simplify and analyze divisibility of $b$

Simplifying $(5k)^2$ gives $25k^2$, so the equation becomes $5b^2 = 25k^2$. Dividing both sides by 5, we get $b^2 = 5k^2$. This means $b^2$ is divisible by 5. By the same property used earlier, if $b^2$ is divisible by 5, then $b$ must also be divisible by 5.

Step 7: Reach a contradiction

From our analysis, we found that both $a$ and $b$ are divisible by 5. This means that $a$ and $b$ share a common factor of 5. However, our initial assumption was that the fraction $\frac{a}{b}$ was in its simplest form, meaning $a$ and $b$ have no common factors other than 1. This creates a contradiction.

Step 8: Conclude $\sqrt{5}$ is irrational

Since our initial assumption (that $\sqrt{5}$ is rational) led to a contradiction, this assumption must be false. Therefore, $\sqrt{5}$ cannot be expressed as a fraction of two integers and is thus an irrational number.

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