Question

classify $(sqrt{5}+\frac{1}{5})$ as rational or irrational, and explain your reasoning.
a suppose an irrational + rational = rational, then irrational = rational - rational. this means irrational = rational. thus the irrational number $sqrt{5}$ plus the rational number $\frac{1}{5}$ equals $(sqrt{5}+\frac{1}{5})$, which is both rational and irrational.
b suppose an irrational + rational = rational, then irrational = rational - rational. this means irrational = rational, which is a contradiction. thus the irrational number $sqrt{5}$ plus the rational number $\frac{1}{5}$ equals the rational number $(sqrt{5}+\frac{1}{5})$.
c suppose an irrational + rational = rational, then irrational = rational - rational. this means irrational = rational, which is a contradiction. thus the irrational number $sqrt{5}$ plus the rational number $\frac{1}{5}$ equals the irrational number $(sqrt{5}+\frac{1}{5})$.
d suppose an irrational + rational = rational, then irrational = rational - rational. this means irrational = rational, which is a contradiction. thus the irrational number $sqrt{5}$ plus the rational number $\frac{1}{5}$ equals $(sqrt{5}+\frac{1}{5})$, which is neither rational nor irrational

Explanation:

We know that the set of rational numbers is closed under subtraction. If we assume that \(\sqrt{5}+\frac{1}{5}\) is rational, then \(\sqrt{5}=(\sqrt{5}+\frac{1}{5})-\frac{1}{5}\). Since \(\frac{1}{5}\) is rational, if \(\sqrt{5}+\frac{1}{5}\) were rational, then \(\sqrt{5}\) (an irrational number) would be equal to the difference of two rational numbers (which is rational). This is a contradiction. So \(\sqrt{5}+\frac{1}{5}\) must be irrational.

Answer:

C. Suppose an irrational + rational = rational, then irrational = rational - rational. This means irrational = rational, which is a contradiction. Thus the irrational number \(\sqrt{5}\) plus the rational number \(\frac{1}{5}\) equals the irrational number \((\sqrt{5}+\frac{1}{5})\).