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)="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.
