Recently, a teacher in Bangladesh asked us how to answer students’ common questions about his version of the proof that the square root of 2 is an irrational number. Some are questions about apparent gaps in the proof itself, others about how one would decide what to do in writing the proof. We’ll then compare that to several past answers, seeking to fill the gaps.
Answering students’ objections
The question came from a teacher, Manish. It’s a long one, so I’ll break it into parts and discuss them separately. As he is stating his proof step by step, together with questions from students, I’m going to put the proof in red, and the questions in bold, followed by his suggested answers.
Prove that √2 is irrational.
Proof:
Let √2 be rational.
Then √2 = p/q, p and q are integers, q is non zero, and they are coprime, or HCF(p,q)=1 for more clarity of concept for students.
Then squaring both sides, we get
2 = p^2/q^2
Now most important question which some good students ask is,
Sir, why you have done squaring, how we will get an idea that we need to do squaring?
My past reply to convince them:
It is easier to deal with 2 as compared to √2, like it is to perform mathematical operations on 2, it’s easier to add, subtract, multiply, divide by 2 as compared to √2 which is difficult to handle while using mathematical operations.
I don’t how accurate my reply is actually, but it calms down students many times.
As we’ll see below, I’d express some details a little differently; for example, I would not say that we “let √2 be rational” (as if we had the power to make it so), but that we suppose for the sake of argument that √2 is rational. And the facts about p/q that follow mean that any rational number can be written as a fraction in lowest terms (that is, the terms are relatively prime), so we assume we have done so. But we’ll focus on Manish’s questions, and perfect the details later.
So, the first question is, why do we choose to square? This is not a gap in the proof, but perhaps in the teaching of how to create a proof, as we’ll see more later.
I answered:
What I would say is that the concept of a square root is defined in terms of squaring: We say that √x = y when y2 = x (and y ≥ 0). So the most natural thing to do to prove something about a square root is to use the latter equation to study the former.
This is related to what you say.
To put it another way, we want to express \(\sqrt{2}=\frac{p}{q}\) without using a square root, so we can use familiar techniques; so we apply the definition, which is that this number’s square must be 2.
Also, as we’ll see again, in reading a proof, it can be helpful to first read through it (several times, usually) just following the logic, and only once we understand that, then think about how we might have thought to do that.
Why say that 2 divides \(p^2\)?
Back to Manish, continuing his proof by multiplying the last equation by \(q^2\):
Then
2 x q^2 = p^2 ——- eq(1)
From above equation we get
2 divides p^2
2 divides p also
Queries of students:::
Sir, how did you write 2 divides p^2?
My reply:
For example let’s take
2 x 3 = 6, it means
6 comes on table of 2
6 comes on table of 3
2 and 3 are factors of 6
We can choose any factor we like
We will select the factor which fits the situation
So if we observe our question it is √2, so our choice is 2
2 divides 6
Clearly “comes on table of 2” in his dialect means “is part of the multiplication table for 2”, and therefore means “is a multiple of 2”. His answer seems to be that 2 is the most relevant factor to talk about.
I answered:
I’m not sure what this question means. You seem to answer, WHY you say that; but the question as I see it seems to be, HOW do you know that is true? But the basic answer to either question is that the preceding statement, that 2q2 = p2, again represents a definition: We say that a divides b when b = ak for some integer k (that is, when b is a multiple of a). So on one hand, it is true that 2 divides p2, because it is equal to 2 times the integer q2; and on the other hand, we take note of that because divisibility is relevant to the question.
But mostly we do it because we are following in the footsteps of Euclid, who either was very smart, or had tried for a long time to prove this and finally saw that this would help, or was following in the footsteps of others who had first proved it! That is, we are not inventing these ideas for the first time, and each step in a proof will not in general be obvious to me the first time I see it!
So the teacher does this, in reality, because he has seen others prove it this way before, so he knows to do that. As we’ve pointed out before, a proof is a record of a path from given facts to a desired fact that we have found to work, and quite often it does not show how we would have invented these things to do! But if we were inventing this for the first time, we might just take note of the fact that the statement \(2q^2=p^2\) says, among other things, that since the LHS (left-hand side) of the equation is a multiple of 2, therefore the RHS (right-hand side) must also be; and another way to say this is that 2 divides \(p^2\). And both 2 (as Manish says) and divisibility (as I said) are relevant. We’ll see how relevant as we go on.
Why does 2 divide p?
Manish continued:
Next Query: Sir, how 2 divides p also you wrote?
My reply:
Let us take any prime number, like suppose 2, and
If 2 divides square of 6
Then 2 will divide 6 also.
Many students are not convinced even by this explanation so I go further deep.
I tell them when we say 2 divides 6^2, it 6 is multiplied twice and 6 contains prime factor 2 in its prime factorization at least once, so basically 2 cancels 2 itself, not exactly 6 as it appears while cancelling.
It calms them down a bit.
I explain the theorem also along with but that out too much load on student mind and exhausts them badly.
This is a key step in the proof, and deserves both explicit mention, and a careful explanation. “The theorem” presumably refers to the general fact I’ll be explaining here.
I answered:
An example doesn’t really prove anything; and I’m not sure what “canceling” means here.
I might just say this:
The square of an even number is always even:
(2n)2 = 4n2 = 2(2n2), which is even.
The square of an odd number is always odd:
(2n+1)2 = 4n2 + 4n + 1= 2(2n2+2n)+1, which is odd.
So if a square is even, the number must be even! If it weren’t, then the square would have been even!
Here I am both using words that should make sense to a student, and using algebra to demonstrate it conclusively. We’ll see more on this theorem later; the important thing is that we at least state the theorem, making it clear that it is a general fact that can be proved.
Why say c is an integer?
Back to Manish:
Then I say
As we know 2 divides p also
So let’s write division format
p/2 = c (integer)
Query from student:
Sir, why you wrote integer in brackets?
My reply:
I take back to same example:
2 × 3 = 6, 2 divides 6
6/2 = 3 ( and is integer)
We got fixed number here, because dividend is defined fixed, that is 6.
Same when we say 2 divides p.
It means 2 is a factor of p and on remainder will come 0 without going into decimals, so hence always an integer.
But since value of p is not fixed so answer also will be in variable, so let it be c.
We say p = 2c.
I answered:
I would simply say that, again, the definition of “a divides b” is “b = 2k for some integer k”, so if 2 divides p, then by definition p = 2 times an integer, and so p/2 is an integer. So the important fact is that your c is an integer.
(Really, though, I don’t think the division statement is needed at all, but just “\(p=2c\)” for some integer c.)
Reaching the conclusion
Now Manish finished the proof:
Then I tell them put this value in eq(1)
2 × q^2=4 × c^2
q^2=2 × c^ 2
Then I tell them to repeat the same pattern above
2 divides q^2
2 divides q also.
Then 2 divides both p and q.
But p, q are coprime.
So this is in contradiction.
It is due to wrong assumption.
So √2 is irrational.
Explanations I provide along with this are
If contradiction word you find difficult to write, you can replace by word wrong.
Which assumption is wrong:
Assuming rational
That means if something is not rational than it must be _________?
Students reply irrational.
We’ve finished a proof, though we can refine it to make it a little clearer to the students. But there’s one more question:
Can’t you prove it more quickly?
Manish again:
Important remark: Once an intelligent student asked me a query which I couldn’t answer, I still don’t have answer to it even now.
It is as follows:
When we do squaring both sides
We get
p^2 = 2 × q^2
He said, Sir, p and q are integers, so their squares would be perfect squares and in their prime factorization all prime factors would be in pairs with no unpaired prime remaining behind.
But here LHS is perfect square.
But RHS is not perfect square as 2 is left unpaired, so contradiction had arisen in this particular step then why to do further so lengthy?
I couldn’t answer it
Even today after 5-6 years I don’t have answer to this query.
The student is not wrong! I answered:
There is often more than one way to prove a theorem! This idea could be the basis of a different proof, one that is based on unique factorization, which is not the central idea in the “standard” proof. But stating that proof carefully would require some precise explanations of what “unpaired” means.
We’ll see a version of this approach later.
Are there more complete proofs?
Now, I looked at several answers to questions about this proof on the Ask Dr. Math site, and most of them skip over the detail of proving that if a^2 is even, then a must be even! (They just say it is easily proved.) But one did cover that issue, so I’ll quote it here.
We’ll look at this below. I pointed out that part of the answer parallels the student’s thought, and referred to another answer we’ll also see below, which is specifically about the needed theorem.
Now, when I ask Google for “proof that square root of 2 is irrational”, at this point the AI summary says merely, “According to number theory, if the square of an integer is even, the integer itself must be even. Therefore, a is even.” So it does what we usually do: it quotes the theorem without proof!
It also gave me this page, which gives a very gentle explanation:
Euclid’s Proof that √2 is Irrational (Math is Fun)
This also explains the step of interest. You may find the whole page useful.
How Ask Dr. Math has explained this
Let’s first look at the sort of answer I had first seen, that doesn’t directly deal with the theorem about even squares.
A brief proof
Here is a typical question and answer, from 1996:
Proof that Sqrt(2) is Irrational Dr. Math - For years I've tried to recall the proof that the square root of 2 is irrational. The proof I recall begins with the assumption that the square root of 2 is rational and therefore = a/b. The proof then shows this is impossible. I remember the proof being clever and elegant.
Doctor Tom answered with a very brief version:
Yup. Assume that it is rational, of the form a/b, where a/b is reduced to lowest terms. So (a/b)*(a/b) = 2, or a^2 = 2b^2. Since a^2 is even, a must be even, say a = 2c so (2c)^2 = 2b^2, or 4c^2 = 2b^2 or 2c^2 = b^2, so b must also be even. So in a/b, both a and b are even, but we assumed we'd reduced the fraction to lowest terms. We've got a contradiction, so sqrt(2) must be irrational.
This is essentially Manish’s proof, reduced to its core. Many of the students’ questions would be raised about this, but it is complete as far as it goes.
A fuller proof
In 2005 (included in the same page), reader Mitch expressed doubts similar to those of Manish’s students:
Dear Dr. Math, I saw the proof on your website for the square root of 2 being irrational. I can follow all the steps, but the proof doesn't seem valid to me. I looked around the web and saw very similar descriptions for this proof and they too all seem invalid to me. All the proofs start with the idea that a rational number can be written as the ratio of two integers, say a/b, and that for any ratio there exists exactly one fully reduced fraction (where no integer greater than 1 exists that can be evenly divided into both the numerator and denominator.) What I see is there may exist a fraction that is not fully reduced. Even if I assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction. I would want to see that no fully reduced fraction of integers exists anywhere to see the conclusion. Thanks in advance for any help.
The issue here lies at the center of the argument: Is it sufficient to show there is no fraction in lowest terms? It is, in fact; but that requires some theorems from number theory that have not been explicitly stated.
Doctor Rick answered:
Hi, Mitch. The version shown above is very brief, more an outline of the proof, and shouldn't be judged as if it were a formal proof. Here's a little more formal version that addresses your concerns about the ratio a/b being in lowest terms.
What he’ll do here is to back up in the assumption, and not initially assume the fraction is in lowest terms.
To prove: The square root of 2 is irrational. In other words, there is no rational number whose square is 2. Proof by contradiction: Begin by assuming that the thesis is false, that is, that there does exist a rational number whose square is 2. By definition of a rational number, that number can be expressed in the form c/d, where c and d are integers, and d is not zero. Moreover, those integers, c and d, have a greatest common divisor, and by dividing each by that GCD, we obtain an equivalent fraction a/b that is in lowest terms: a and b are integers, b is not zero, and a and b are relatively prime (their GCD is 1). Now we have [1] (a/b)^2 = 2 Multiplying both sides by b^2, we have [2] a^2 = 2b^2 The right side is even (2 times an integer), therefore a^2 is even. But in order for the square of a number to be even, the number itself must be even. Therefore we can write [3] a = 2f Using this to replace a in [2], we obtain [4] (2f)^2 = 2b^2 [5] 4f^2 = 2b^2 [6] 2f^2 = b^2 The left side is even, therefore b^2 must be even, and by the same reasoning as before, b must be even. But now we have found that both a and b are even, contradicting the assumption that a and b are relatively prime. Therefore the assumption is incorrect, and there must NOT be a rational number whose square is 2.
This is essentially what I would have recommended as the main proof to teach, with a reasonable level of detail.
That's my proof. Admittedly it still leaves out some steps, for instance a proof that if the square of a number is even then the number itself must be even. But I filled in the part that had you confused.
There is nothing wrong with stating a theorem without proof; but it’s a good idea to be prepared with one, in case a wise student asks!
Proving an even square must be the square of an even number
Now let’s look at the proof of that fact that I referred Manish to, in a question from 2002:
If n^2 is Even, n is Even I have to show that if n^2 is even then n is also even. I don't know where to start. I would love your help. Thanks.
Doctor Paul answered with a variation on the proof by contradiction we’ve been looking at:
Try a proof by contraposition (are you familiar with this method of proof? a statement and its contrapositive are logically equivalent so proving the contrapositive of the original statement actually proves the original statement).
So what you want to do is assume that n is not even (i.e., it is odd) and show that n^2 is not even (i.e., it is odd).
n is odd means that you can write n = 2*k + 1 for some integer k. Then
n^2 = 4*k^2 + 4*k + 1
= 2 * (2*k^2 + 2*k) + 1
which is clearly odd.
This completes the proof.
This is essentially the same as the quick proof I gave earlier. We discuss proof by contraposition here.
An alternative proof, using unique factorization
Now let’s look at the 1999 page I first referred Manish to:
Sqrt2 Irrational I would like to know the answer to the question, How do you prove that the square root of 2 (or other roots of other real numbers) is irrational? I am not able to understand the argument.
Rather than “other real numbers”, he should have said “other prime numbers”, or perhaps “other integers that are not perfect squares.”
Doctor Floor answered with a proof that does not assume the fraction is in lowest terms:
Hello, G. Arjun, Thanks for your question! To prove that sqrt(2) [sqrt means square root] is an irrational number, we have to show that it cannot be written as a/b, where a and b are integers. Such a proof has to be done indirectly: 1. We assume that sqrt(2) could be written as a/b for integer a and b; 2. We show that this leads to a contradiction. So something impossible has to be derived from our assumption. The fact that we derive something impossible from the fact that sqrt(2) = a/b shows that sqrt(2) = a/b must be false, and we have proven the theorem.
This is the core idea of an indirect proof.
Let's assume that sqrt(2) could be written as a/b for integer a and b. We can derive:
sqrt(2) = a/b [sqare both sides]
2 = a^2/b^2 [multiply by b^2; ^2 means squared]
2*b^2 = a^2
a^2 = 2*b^2
This all looks familiar. But now he uses unique factorization, rather than reduced fractions:
Now we have two integers in the equation, 2*b^2 and a^2. We can factor these integers into primes, and find:
a^2 = 2^m * {product of odd primes}
2*b^2 = 2^n * {product of odd primes}
Because a^2 is a square, m has to be even.
Because b^2 is a square, and "2*" brings an additional factor 2, n has to be odd.
So m and n are not equal.
But if a^2 = 2*b^2, then the factorisation into primes of these two has to be equal. So, m and n should be equal.
We have found a contradiction. And our assumption, that sqrt(2) = a/b for integers a and b, cannot hold. This proves that sqrt(2) is irrational.
Notice that this uses the idea Manish’s student asked about, but stated a little more carefully.
Another alternative: Our proof
That same day, between that question and its answer, another student, Wayne, asked essentially the same question:
Can you prove to me how the square root of two is irrational?
Doctor Rob answered him:
There are a couple of ways to do that. Both are proofs by contradiction; that is, they assume that sqrt(2) is rational, and derive an impossible conclusion from that assumption. Here is one: Assume sqrt(2) = a/b, reduced to lowest terms (that is, a and b have no common factor besides 1). Then 2 = (a/b)^2 = a^2/b^2, so 2*b^2 = a^2. That means that a^2 is even, which implies that a is even. Then write a = 2*c, a^2 = 4*c^2 = 2*b^2, so 2*c^2 = b^2. That means that b^2 is even, which implies that b is even. This means that a and b have the common factor 2, which is a contradiction. Thus no such fraction a/b can exist, and sqrt(2) is irrational.
This is our basic proof, stated without identifying or proving the key fact about even squares.
And another: Lowest terms, literally
A third proof takes a difference perspective on “lowest terms”, emphasizing “lowest” rather than “no common factors”. Doctor Rob continues:
Here's another:
Assume sqrt(2) = a/b, with a and b positive integers, and least denominator among all such fractions. Then
2*b^2 = a^2,
2*b^2 - a*b = a^2 - a*b, 4*b^2 > 2*b^2 = a^2 > b^2,
b*(2*b-a) = a*(a-b), 2*b > a > b,
(2*b-a)/(a-b) = a/b = sqrt(2), b > a - b > 0 and 2*b - a > 0.
Now (2*b-a)/(a-b) is a fraction equal to sqrt(2) with positive integer numerator and denominator, but smaller denominator than a/b has. This is a contradiction. Thus no such fraction a/b can exist, and sqrt(2) is irrational.
This is written rather compactly. The left-hand column subtracts ab from both sides, then factors, then divides both sides by \(b(a-b)\), to obtain a new fraction for \(\sqrt{2}\); the right-hand column derives an inequality that shows that we have obtained a new fraction (of positive integers) with the same value, but a smaller denominator, resulting in the contradiction.
There are yet more ways to approach the proof, some not using indirect proof at all, but including them would make this too long. Maybe I’ll include them next time.