When is an Integer a Rational Number, and are All Ratios Rational, Even $\frac{\sqrt{7}}{2}$?

Any number for which it is possible to express as the ratio or quotient of integers is a rational number. So yes, $5$ is rational, because it is possible to express this as $\frac 51, \frac {10}{2}...$.

$5$ is also an integer. Every integer is a rational number, but not all rational numbers, e.g. $\frac 12$, is an integer. We know $\frac 12$ is rational, because it is the quotient of two integers.

However, $\dfrac {\sqrt 7}{2}$ is not a ratio of integers. It is a ratio of a non-integer, namely $\sqrt 7$, over an integer. So $\dfrac{\sqrt 7}{2}$ is not rational.

$\sqrt{7}$ is not a rational number.

You can prove it by the absurd.

Suppose that you can write it as $\sqrt{7}=\frac{a}{b}$ with $a$ and $b$ natural numbers.

You can suppose without loss of generality that $a$ and $b$ are coprime. Decompose $a$ and $b$ in prime factors: $a=a_1^{n_1}\dots a_k^{n_k}$ and $b=b_1^{m_1}\dots b_l^{m_l}$. All the $a_i$ are primes and different of all the $b_j$ (also primes) because $a$ $b$ are coprime.

$\sqrt{7}=\frac{a}{b}$ give you $7=\frac{a^2}{b^2}$ hence $a^2=7\cdot b^2$ hence $a_1^{2n_1}\dots a_k^{2n_k}=7b_1^{2m_1}\dots b_l^{2m_l}$ hence one of the $a_i$ is $7$ (lets say $a_1=7$). We have: $7^{2n_1-1}\dots a_k^{2n_k}=b_1^{2m_1}\dots b_l^{2m_l}$ hence $7$ divide $b_1^{2m_1}\dots b_l^{2m_l}$ hence one of the $b_j$ is $7$ contradiction with $a$ $b$ coprimes.