Is $\sqrt{7}$ $sqrt7$ a rational, irrational, natural, whole, integer or real number?

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Is $\sqrt{7}$ $sqrt7$ a rational, irrational, natural, whole, integer or real number?

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Answer 1 · 2015-07-10T23:48:38

$\sqrt{7}$ $sqrt(7)$ is an irrational number. That is, it cannot be expressed as $\frac{p}{q}$ $p/q$ for some integers $p$ $p$ and $q$ $q$ with $q \neq 0$ $q != 0$

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide both ends by $7$ $7$ to get:

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

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

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

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

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

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Is $\sqrt{7}$ $sqrt7$ a rational, irrational, natural, whole, integer or real number?

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