Question 3

Suppose $\dfrac{1}{\sqrt{7}}$ is rational. Then there exist coprime integers $a,b$ with $b\neq0$ such that

$$\dfrac{1}{\sqrt{7}}=\dfrac{a}{b}.$$

Taking reciprocals (both sides are non‑zero) gives

$$\sqrt{7}=\dfrac{b}{a}.$$

Squaring yields

$$7=\dfrac{b^{2}}{a^{2}}\quad\Longrightarrow\quad 7a^{2}=b^{2}.$$

From $7a^{2}=b^{2}$ we see that $7$ divides $b^{2}$, hence $7$ divides $b$. Write $b=7c$ for some integer $c$. Substituting gives

$$7a^{2}=49c^{2}\quad\Longrightarrow\quad a^{2}=7c^{2}.$$

Now $7$ divides $a^{2}$, so $7$ divides $a$. Thus both $a$ and $b$ are multiples of $7$, contradicting the assumption that $a$ and $b$ are coprime. Hence the original assumption is false.

Suppose $7\sqrt{5}$ is rational. Then there exist coprime integers $p,q$ ($q\neq0$) such that

$$7\sqrt{5}=\dfrac{p}{q}.$$

Divide both sides by $7$:

$$\sqrt{5}=\dfrac{p}{7q}.$$

Squaring gives

$$5=\dfrac{p^{2}}{49q^{2}}\quad\Longrightarrow\quad 5\cdot49q^{2}=p^{2}.$$

Thus $p^{2}$ is divisible by $5$, so $5$ divides $p$. Write $p=5r$. Substituting:

$$5\cdot49q^{2}=25r^{2}\quad\Longrightarrow\quad 49q^{2}=5r^{2}.$$

Now $5$ divides the right‑hand side, so $5$ divides $49q^{2}$, and because $5$ does not divide $49$, it must divide $q^{2}$, hence $5$ divides $q$. Both $p$ and $q$ are multiples of $5$, contradicting coprimeness. Therefore $7\sqrt{5}$ cannot be rational.

Suppose $6+2\sqrt{5}$ is rational. Then there exist integers $m,n$ with $n\neq0$ such that

$$6+2\sqrt{5}=\dfrac{m}{n}.$$

Subtract $6$ from both sides:

$$2\sqrt{5}=\dfrac{m}{n}-6=\dfrac{m-6n}{n}.$$

Hence

$$\sqrt{5}=\dfrac{m-6n}{2n}.$$

Squaring yields

$$5=\dfrac{(m-6n)^{2}}{4n^{2}}\quad\Longrightarrow\quad 20n^{2}=(m-6n)^{2}.$$

The right‑hand side is a perfect square, so $5$ must divide $(m-6n)$. Write $m-6n=5k$. Substituting gives

$$20n^{2}=25k^{2}\quad\Longrightarrow\quad 4n^{2}=5k^{2}.$$

From $4n^{2}=5k^{2}$ we see $5$ divides the left side, so $5$ divides $n$. Then $5$ also divides $k$, and consequently $5$ divides $m-6n$ and $m$. Both numerator and denominator of the original fraction are multiples of $5$, contradicting the assumption that the fraction was in lowest terms. Hence $6+2\sqrt{5}$ is irrational.