Prove that \sqrt{5} is irrational and find whether 3/13 is a terminating or non-terminating repeating decimal.

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Step-by-Step Solution

Step 1: Proof by Contradiction for $\sqrt{5}$

To prove that $\sqrt{5}$ is irrational, we use the method of proof by contradiction. We start by assuming the opposite, that $\sqrt{5}$ is a rational number. If $\sqrt{5}$ is rational, it can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are coprime integers (meaning they have no common factors other than 1) and $b \neq 0$.

Step 2: Square both sides and rearrange

Squaring both sides of the equation $\sqrt{5} = \frac{a}{b}$ gives $5 = \frac{a^2}{b^2}$. Multiplying both sides by $b^2$ yields $5b^2 = a^2$. This equation implies that $a^2$ is a multiple of 5, which means $a$ must also be a multiple of 5.

Step 3: Substitute $a = 5c$ and simplify

Since $a$ is a multiple of 5, we can write $a = 5c$ for some integer $c$. Substituting this into the equation $5b^2 = a^2$, we get $5b^2 = (5c)^2$, which simplifies to $5b^2 = 25c^2$. Dividing both sides by 5 gives $b^2 = 5c^2$. This implies that $b^2$ is a multiple of 5, and therefore $b$ must also be a multiple of 5.

Step 4: Conclusion for $\sqrt{5}$

We have shown that both $a$ and $b$ are multiples of 5. This means that 5 is a common factor of $a$ and $b$. However, our initial assumption was that $a$ and $b$ are coprime integers, meaning they have no common factors other than 1. This is a contradiction. Therefore, our initial assumption that $\sqrt{5}$ is rational must be false. Hence, $\sqrt{5}$ is irrational.

Step 5: Analyze the denominator for $3/13$

To determine if a rational number $\frac{p}{q}$ (in simplest form) has a terminating or non-terminating repeating decimal expansion, we examine the prime factors of its denominator $q$. If the prime factorization of $q$ contains only 2s and/or 5s, the decimal expansion is terminating. Otherwise, it is non-terminating repeating.

Step 6: Conclusion for $3/13$

The prime factorization of the denominator, 13, is simply 13. Since 13 is not of the form $2^m \times 5^n$ (where $m$ and $n$ are non-negative integers), the decimal expansion of $\frac{3}{13}$ will be non-terminating and repeating.