Blogs, Mathematical Reasoning, Proofs

That’s irrational

One of my favorite proofs is that √2 is irrational.  The reason I like this proof so much, is that at first glance, it appears to be a very difficult thing to prove.  However, the final proof is actually very simple.  While the same technique we use to show √2 is irrational can be modified to work to show √3 is irrational, we will actually go through this proof as Fermat did so that we can see a different technique being used.

In the post today, I will focus on a general overview of the proof, therefore, if you would like to see the details of anything we talk about along the way, I will refer you to the proof write up.  In the proof write, I provide the details of the proofs, but not the background information, so it would be helpful to have both up if you are trying to get a complete picture.  Note that any theorem numbers refer to the numbers in the proof write up.  Furthermore, if you’d like to know more about these topics, I would refer you to Book of Proof by Richard Hammock.


Before we get into the proof itself I wanted to look at some background information.  First, what is a irrational number?  Well, an irrational number is a number that is not rational.  This answer doesn’t seem particularly helpful because it leads to the questions, then what is a rational number?  At least here we can be a little more specific.  We say a number r is rational if there exists integers p and q with q not 0 such that r=p/q.  Therefore, an irrational number is one that cannot be written as p/q for any integers p and q.

Now, if we want to directly show that √2 is irrational, this means that we would have to show that every choice of p and q would result in (p/q)2 not being 2. We would, therefore, have to check infinitely many possibilities for p and q, showing each does as we state above. Doing this one at a time is impossible, since we do not have an infinite amount of time, and doing this in a general way would still be extremely difficult. Therefore, we should just give up now, right?  Of course not, we just have to find another way to do things.

We will therefore work with the following logical process.  If we can show that something can’t be false, it must therefore be true.  In order to model this, we would assume that our theorem is false and show that this leads to a logical inconsistency.  This then implies the original statement had to be true.  For example suppose we want to show



  • Suppose √2=0.
  • Then 0=02 =(√2)2=2.
  • Since 0 is not equal to 2, we have a contradiction, so our theorem holds.

We will use this same technique to show that both √2 and √3 are not rational.

√2 is irrational


  • Suppose that √2 is rational.
  • Then √2=p/q where p,q are nonzero integers.
  • We may assume that the only (positive) common divisor of p and q is 1 .
    • If this were not the case, we could simplify p/q until it was.
  • Then 2=p2/q2, so 2q2=p2.
  • Therefore, p2 is even.
  • Now, p is even, since the product of two odds is odd.
  • Hence 2q2=(2k)2.
  • Now q2=2k2 so  q2 is even.
  • Hence q is even.
  • This contradicts that fact that the only common divisor of p and q is 1.
  • Hence √2 is irrational.

The great thing about this proof is instead of working with infinitely many numbers, we can focus on one number by assuming the √2 is rational.  This allows us to manipulate things algebraically so that we can get that both the bottom and top must be even, despite the fact that we assumed we had already simplified the fraction.

√3 is irrational

For this proof, we will again use a contradiction.  While we could do as we did above and show that both top and bottom are divisible by 3, I instead wanted to look at the proof that Fermat gave (see A History of Mathematics).  The contradiction he comes up with is interesting in its own right.


  • Suppose that √3 is rational.
  • Then √3=p/q where p and q are integers.
  • Then note that


  • Now, since 3/2 < √3 < 2, we get 3d-c< c and c-d< d.
  • Since we can continue to do this, we will create infinitely many positive integers less than c.
  • Since there are at most c-1 integers less than c, we have a contradiction.

The reason why I like this proof is that it shows a contradiction by creating an infinite number of elements in a set that we already know to be finite.  However, the proof really does rely on properties of 3 and approximations for its square root.  This makes it more difficult to generalize than the proof for √2.  Because of this, I would view the first proof as the prettier technique, but it is still fun to see both.


As I look over the proofs as they written up, there really is a swelling of joy within me to see these logical arguments come forward and prove that these things must actually be true.  While I don’t know if anyone else has this reaction to these proofs, I wish I could impart this feeling into everyone.  In particular, I find just trying to explain this feeling to my students helps them see a side of math that they may not have noticed before.  The simple elegance of a well made logical argument really is just wonderful.

If you enjoyed this post let me know by liking below.  If you didn’t enjoy it, comment on why.  Do you not like proofs?  Do you not see any beauty in logical arguments?  Our there other proofs you think are more interesting?  I’d love to hear what you think.  As long as there is interest, there are always more proofs to look at.  A few I plan on talking about in the future is the proof that the set of primes is infinite, that the set of irreducible and prime natural numbers are actually the same, a cube’s volume cannot be doubled using only straight edge and compass and a few more. 


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