# Is sqrt7 a rational, irrational, natural, whole, integer or real number?

Jul 10, 2015

$\sqrt{7}$ is an irrational number. That is, it cannot be expressed as $\frac{p}{q}$ for some integers $p$ and $q$ with $q \ne 0$

#### Explanation:

How do we know that $\sqrt{7}$ is irrational?

For a start, $7$ is a prime number, so its only positive integer factors are $1$ and $7$.

Next suppose $\sqrt{7} = \frac{p}{q}$ for some positive integers $p$ and $q$.

Suppose further that $\frac{p}{q}$ is in lowest terms - that is $p$ and $q$ have no common factor apart from $1$. If this were not so, then we could just divide $p$ and $q$ by that common factor.

Since we are given $\sqrt{7} = \frac{p}{q}$, it follows that:

${p}^{2} / {q}^{2} = {\sqrt{7}}^{2} = 7$

Multiply both ends by ${q}^{2}$ to get:

${p}^{2} = 7 {q}^{2}$

Since $q$ is a positive integer, $7 {q}^{2}$ is a positive integer divisible by $7$.

So ${p}^{2}$ is a positive integer divisible by $7$.

But if ${p}^{2}$ is divisible by $7$, then $p$ must be divisible by $7$ (since $7$ is prime).

So $p = 7 k$ for some positive integer $k$.

Then we have $7 {q}^{2} = {p}^{2} = {\left(7 k\right)}^{2} = 7 \cdot 7 {k}^{2}$

Divide both ends by $7$ to get:

${q}^{2} = 7 {k}^{2}$

Since $k$ is a positive integer, $7 {k}^{2}$ is a positive integer multiple of $7$. So ${q}^{2}$ is divisible by $7$.

Since ${q}^{2}$ is divisible by $7$, $q$ must be divisible by $7$ (since $7$ is prime).

So both $p$ and $q$ are divisible by $7$, contradicting the assumption that they were both in lowest terms.

So our original supposition must be wrong and there are no integers $p$, $q$ such that $\frac{p}{q} = \sqrt{7}$