Let us first prove that sqrt(21)√21 is a real number, in fact, the square root of all positive real numbers is real. If xx is a real number, then we define for the positive numbers sqrt(x)="sup"{yinRR:y^2<=x}. This means that we look at all real numbers y such that y^2<=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<=x, namely 0. Furthermore, there is an upper bound to these numbers, namely x+1, since if 0<=y<1, then x+1>y, if y>=1, then y<=y^2<=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 RR. 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)>=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)=a/b with a and b whole numbers and 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*7 and for a and b this is some arbitrary product of primes a=a_1*...*a_n and b=b_1*...*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*a_1*...*a_n*a_n and b^2=b_1*b_1*...b_m*b_m., therefore the only common divisor of a^2 and b^2 is 1. Since a^2=21b^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*a_1*...*a_n*a_n and 21=3*7, so 3*7=a_1*a_1*...*a_n*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 21nea^2, so anesqrt(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 inNN) 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.
