# How do you find the set in which the real number sqrt93 belongs?

Nov 29, 2016

$\sqrt{93}$ is an algebraic irrational number.

#### Explanation:

The prime factorisation of $93$ is:

$93 = 3 \cdot 31$

Since not all of the prime factors occur an even number of times (in fact neither of them do), $\sqrt{93}$ is irrational.

One way of proving that might go as follows:

Suppose $\sqrt{93} = \frac{p}{q}$ for a pair of positive integers $p , q$ with $p > q$.

Without loss of generality we can assume that this is the smallest such pair of integers.

Then:

${\left(\frac{p}{q}\right)}^{2} = 93$

So:

${p}^{2} = 93 {q}^{2}$

Hence ${p}^{2}$ is divisible by both $3$ and $31$.

Since $3$ and $31$ are prime, $p$ must also be divisible by $3$ and $31$ (since factorisation into a product of primes is unique up to a unit).

Hence:

$p = 93 k$ for some integer $k$ and we find:

$93 {q}^{2} = {p}^{2} = {\left(93 k\right)}^{2} = {93}^{2} {k}^{2}$

Then dividing both ends by $93$ we find:

${q}^{2} = 93 {k}^{2}$

Hence:

${\left(\frac{q}{k}\right)}^{2} = 93$

So $\sqrt{93} = \frac{q}{k}$

Now $p > q > k$, so $q , k$ are a smaller pair of integers whose quotient is $\sqrt{93}$, contradicting our assertion that $p , q$ was the smallest such pair.

Hence there is no such pair of integers and $\sqrt{93}$ is therefore irrational.

$\sqrt{93}$ is also an algebraic number. That is, it is the root of a polynomial equation with rational coefficients, namely:

${x}^{2} - 93 = 0$

Some irrational numbers are algebraic, but most are transcendental. Numbers like $e$ and $\pi$ are transcendental: They satisfy no polynomial equation with rational coefficients.