After reading and studying the proof of the irrationality of the square root of 2 by Tom Apostol, I began wondering if there were any other proofs. Obviously, there should be many proofs that show that the square root of 2 is an irrational number, right? This number has astounded mathematicians throughout the ages! Heck, an article I recently read on the subject says that there is a Babylonian tablet YBC 7289 which shows a “four-sexagesimal-place approximation to the square root of 2.” (Fowler and Robson, 1998) Four sexagesimal digits is roughly equivalent to six digits in base ten! This Babylonian approximation is:
1 + 24/60 + 51/(602) + 10/(603) =1.41421296…
So yeah, this number has been studied for a long time. One would hope that there are a plethora of different types of proofs that show that the square root of 2 is irrational. As it turns out, there are several proofs that do just that. I would like to take a moment and outline a few of my favorites.
It seems to me, though, that proving the irrationality of the square root of 2 usually involves prime factorization or the usage of parity. Also, most proofs that I’ve seen are proofs by contradiction. That is, they show that is irrational by showing the inconsistencies that would arise from the square root of 2 being rational.
The first and perhaps the simplest proof I will present is one such proof. This is perhaps one of the earliest proofs of the irrationality of the square root of 2. There is evidence to suggest that even Aristotle was aware of this proof. Such evidence can be found in Aristotle’s Prior Analytics (or Analytica Priora in Latin). Aristotle aside, this proof was first clearly and fully published in the well known Elements written by Euclid. It appears as Proposition 117 of Book X. Euclid may or may not have discovered this proof himself. It’s quite possible that it was discovered by another mathematician and the proof we see in Elements is merely referencing it.
Anyway, the proof by contradiction goes like this:
- First we assume that the square root of 2 is actually rational. If this is the case, then by the definition of a rational number, it can be written as a fraction a/b.
- If a and b share a common factor, then the fraction a/b can be reduced. Here, we assume that is an irreducible fraction and a and b have no common factors.
- If the square root of 2 = a/b then 2= (a2)/(b2) and (a2) = 2(b2).
- If this is the case, then (a2 has to be an even number, because 2(b2) is going to be some multiple of 2
- If (a2) is even, then a has to be even because an odd number times an odd number is always another odd number.
- If a is even, there must be some k such that a = 2k.
- If we substitute a = 2k into the equation found in step 3, we get (2k)2 = 4(k2) = 2(b2). This, of course, reduces to 2(k2) = (b2).
- Using the same logic as before, we can deduce that (b2) must be even and therefore b is also even.
- However, there is a contradiction. If both a and b are even, then the fraction is not irreducible because a and b have at least the common factor of 2. Therefore, the assumption that the square root of 2 can be written as is false and must be irrational.
Interesting proof, right? There is a video of this proof found on khanacademy.org and I would recommend you check it out. The site as a whole is also recommended, because it contains many other interesting mathematical proofs on the subject of rational and irrational numbers.
I will still present a few more proofs in this blog post, but due to size limitations, of course I cannot write all of the proofs that I have found here. The reader is encouraged also to check out http://www.cut-the-knot.org/proofs/sq_root.shtml, which is a website that contains a collection of 27 proofs that the square root of 2 is irrational.
Another great and simple proof, which is found on the above website, proves the irrationally of the square root of 2 by unique factorization. It goes as follows:
- Again, we assume that is the square root of 2 actually rational. It can therefore be written as a fraction a/b.
- We again assume that is an irreducible fraction and a and b are coprime.
- Observe that b cannot be 1, because there is no integer a that equals the square root of 2.
- By the fundamental theorem of arithmetic, all numbers greater than 1 are either prime numbers or can be written as the product of prime numbers.
- Therefore, there must be a prime number p that divides b but does not divide a. If p did divide a, the fraction wouldn’t be irreducible.
- If we square both sides of the equation found in step 1, we obtain 2 = (a^2)/(b^2).
- a2 can be factored as the product of all its primes squared
- The prime p that divides b will also divide and it cannot divide . and will retain the same prime numbers as a and b.
- We can then conclude that if a/b is an irreducible fraction, then (a2)/(b2) cannot be reduced to an integer and therefore 2 does not equal (a2/b2) and the square root of 2 must be irrational.
The careful reader will observe that the two proofs I have provided thus far are remarkably similar. However, the fundamental concepts used in each proof are different. The first one relies on the usage of parity to show a contradiction in the square root of 2’s fractional representation, though it does to some degree rely on prime factorization. The second one shows through prime factorization a contradiction in that if the square root of 2 was written as a fraction a/b, then (a2)/(b2) could not be reduced to a single integer like 2. It does this without the assumption of any number’s parity.
Finally, I’d like to present a geometric proof that the square root of 2 is irrational. I have attached a picture for this proof to help with visualization. This proof actually uses the Pythagorean Theorem to prove the square root of 2 is irrational.
- Once again we will do a proof by contradiction and suppose that the square root of 2 is rational.
- We can make an isosceles right triangle where the hypotenuse and legs do not share any common factors and where the legs are some positive integer.
- Of course, since this triangle is a 45, 45, 90 triangle, the hypotenuse will be equal to one of the legs times the square root of 2.
- The legs of the triangle must be equal and their squares also must be equal. It follows from the Pythagorean Theorem that one half of the hypotenuse squared will be equal to one of the legs squared.
- It then follows from the Pythagorean Theorem that since (hypotenuse2) = (leg2) + (leg2) = 2(leg2), then the square of the hypotenuse has to be an even number.
- If the square of the hypotenuse is an even number then the hypotenuse must be an even number.
- Because (hypotenuse2) = 2(leg2), we know that the square of the hypotenuse can be evenly divided two times and still have a positive integer area. (It can be divided by 2 and leg). Dividing the square of the hypotenuse only once will result in an even number
- This means that the square of the leg must be even and therefore the leg itself must be even.
- If hypotenuse and leg are both even then they share a common factor and our assumption is false and the square root of 2 must be irrational.
It should be well accepted by now that the square root of 2 is without a doubt irrational. The simplest rational approximation is perhaps 17/12 and the most accurate rational approximation we have so far is 5 million digits long. Perhaps the most efficient approximation to is the fraction 99/70, which is estimated to only be off by about .01%. Even still, these approximations are assuredly not the actual value of the square root of 2. Despite the fact that we’ve successfully calculated the first 5 million digits of the square root of 2, what the Babylonians approximated thousands of years ago still continues to be irrational to this day.
–Fowler, David; Robson, Eleanor (1998), “Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context”, Historia Mathematica 25 (4): 366–378 Retrieved from http://www.sciencedirect.com/science/article/pii/S0315086098922091
-Square root of 2. (n.d.) Retrieved January 25th, 2015 from Wikipedia http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality
-Alexander Bogomolny (n.d.) Square Root of 2 is Irrational Retrieved from
-Salman Khan. Proof that square root of 2 is irrational. Retrieved from https://www.khanacademy.org/math/algebra/ratio-proportion-topic/alg-rational-irrational-numbers/v/proof-that-square-root-of-2-is-irrational