# Is sqrt21 real number, rational number, whole number, Integer, Irrational number?

Jul 26, 2015

It is an irrational number and therefore real.

#### Explanation:

Let us first prove that $\sqrt{21}$ is a real number, in fact, the square root of all positive real numbers is real. If $x$ is a real number, then we define for the positive numbers $\sqrt{x} = \text{sup} \left\{y \in \mathbb{R} : {y}^{2} \le x\right\}$. This means that we look at all real numbers $y$ such that ${y}^{2} \le x$ and take the smallest real number that is bigger than all of these $y$'s, the so called supremum. For negative numbers, these $y$'s don't exist, since for all real numbers, taking the square of this number results in a positive number, and all positive numbers are bigger than negative numbers.

For all positive numbers, there is always some $y$ that fits the condition ${y}^{2} \le x$, namely $0$. Furthermore, there is an upper bound to these numbers, namely $x + 1$, since if $0 \le y < 1$, then $x + 1 > y$, if $y \ge 1$, then $y \le {y}^{2} \le x$, so $x + 1 > y$. We can show that for each bounded non empty set of real numbers, there is always a unique real number that acts as a supremum, due to the so called completeness of $\mathbb{R}$. So for all positive real numbers $x$ there is a real $\sqrt{x}$. We can also show that in this case ${\sqrt{x}}^{2} = x$, but unless you want me to, I will not prove this here. Lastly we note that $\sqrt{x} \ge 0$, since $0$ is a number that fits the condition, as stated before.

Now for the irrationality of $\sqrt{21}$. If it were not irrational (so rational), we could write it as $\sqrt{21} = \frac{a}{b}$ with $a$ and $b$ whole numbers and $\frac{a}{b}$ simplified as much as possible, meaning that $a$ and $b$ have no common divisor, except for $1$. Now this means that $21 = {a}^{2} / {b}^{2}$.

Now we use something called the prime factorization of the natural numbers. This means we can write down each positive whole number as a unique product of prime numbers. For $21$ this is $3 \cdot 7$ and for $a$ and $b$ this is some arbitrary product of primes $a = {a}_{1} \cdot \ldots \cdot {a}_{n}$ and $b = {b}_{1} \cdot \ldots \cdot {b}_{m}$. The fact that the only common divisor of $a$ and $b$ is $1$ is equivalent to the fact that $a$ and $b$ share no primes in their factorization, so there are ${a}_{i}$ and ${b}_{j}$ such that ${a}_{i} = {b}_{j}$. This means that ${a}^{2}$ and ${b}^{2}$ also don't share any primes, since ${a}^{2} = {a}_{1} \cdot {a}_{1} \cdot \ldots \cdot {a}_{n} \cdot {a}_{n}$ and ${b}^{2} = {b}_{1} \cdot {b}_{1} \cdot \ldots {b}_{m} \cdot {b}_{m}$., therefore the only common divisor of ${a}^{2}$ and ${b}^{2}$ is $1$. Since ${a}^{2} = 21 {b}^{2}$, this means ${b}^{2} = 1$, so $b = 1$. Therefore $\sqrt{21} = a$. Note that this only holds under the assumption that $\sqrt{21}$ is rational.

Now we could of course run through all whole positive numbers smaller than $21$ and check if squaring them gives $21$, but this is a boring method. To do it in a more interesting way, we turn again to our primes. We know that ${a}^{2} = {a}_{1} \cdot {a}_{1} \cdot \ldots \cdot {a}_{n} \cdot {a}_{n}$ and $21 = 3 \cdot 7$, so $3 \cdot 7 = {a}_{1} \cdot {a}_{1} \cdot \ldots \cdot {a}_{n} \cdot {a}_{n}$. On the left side, every prime occurs only once, on the right hand, every prime occurs at least twice, and always an even amount of times (if ${a}_{1} = {a}_{n}$ it would for instace occur at least four times). But as we've stated, these prime factorizations are unique, so this can't be right. Therefore $21 \ne {a}^{2}$, so $a \ne \sqrt{21}$, meaning that our earlier assumption of $\sqrt{21}$ being rational turns out to be wrong, therefore $\sqrt{21}$ is irrational.

Note that the same argument holds for any positive whole number $x$ with a prime factorization where one of the primes apears an uneven number of times, since the square of a whole number always has all of its prime factors apearing an even amount of times. From this we conclude that if $x$ is a positive whole number ($x \in \mathbb{N}$) has a prime factor that occurs only an uneven amount of times, $\sqrt{x}$ will be irrational.

I'm aware that this proof may seem a bit long, but it uses important concepts form mathematics. Probably in any high school curriculum, these kind of reasonings are not included (I'm not 100% sure, I don't know the curriculum of each high school in the world), but for actual mathematicians, proving stuff is one of the most important activities they do. Therefore I wanted to show you what kind of mathematics is behind taking the square root of things. What you need to take away from this, is that indeed $\sqrt{21}$ is an irrational number.