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