MTH 18101 Exam 3 Version 1 3 4. (16 pt) Assume $\sqrt{5}$ has been proven to be an irrational number. Using the definition of a rational of number and basic algebra, give a step-by-step proof by contradiction that $s=\frac{5}{4}-\frac{3 \sqrt{5}}{7}$ is irrational. No credit if proved by another method!

Step 1 ：Assume that $s$ is rational, i.e., $s = \frac{p}{q}$ for some integers $p$ and $q$ with $q \neq 0$.

Step 2 ：Then we have the equation $\frac{5}{4}-\frac{3 \sqrt{5}}{7} = \frac{p}{q}$.

Step 3 ：Rearrange this equation to solve for $\sqrt{5}$, we get $\sqrt{5} = \frac{7p - 5q}{12q}$.

Step 4 ：If $p$ and $q$ are integers, then $7p - 5q$ and $12q$ are also integers.

Step 5 ：Therefore, $\sqrt{5}$ can be expressed as a ratio of two integers, which contradicts the assumption that $\sqrt{5}$ is irrational.

Step 6 ：Hence, our initial assumption that $s$ is rational must be false.

Step 7 ：Final Answer: The number $s=\frac{5}{4}-\frac{3 \sqrt{5}}{7}$ is \boxed{\text{irrational}}.