It can be tricky deciding how to approach a proof; this problem, whose answer requires going in a very different direction than you might expect, provides some interesting insights into the nature of proof. The proof itself, in fact, is far less interesting than the process of getting there!
Prove \(\sqrt{4+2\sqrt{3}}-\sqrt{3}\) is a rational number
The question came from Mike at the end of August:
I am trying to prove that sqrt(4+2*sqrt(3)) – sqrt(3) is rational directly from the definition of a rational number.
First, I note that by definition and a number A is rational iff there exist integers n, m such that A = m/n and n is not zero.
The iff part is critical for my proof, since I will be proving the “only if” side of this definition.
Proof
Let sqrt(4+2*sqrt(3)) – sqrt(3) = m/n –> at this point I am not requiring m, n be integers but we will want to show that there are integer m, n that satisfy this and n ≠ 0
Carrying out some algebra, I am able to get to n^2 – m^2 + (2n^2 – 2nm)*sqrt(3) = 0
I realize that this equation is satisfied as long as m = n ≠ 0 which implies that the expression is rational and equals 1 (being the equivalence class where the numerators and denominators are the same and non-zero).
Therefore I have shown this irrational-looking expression is indeed rational. And, in fact, this expression does equal 1 if you just calculate it or ask WolframAlpha.
Question: I posted this to MathStackExchange as a proof verification and got a lot of pushback. Many are saying “you only proved that IF the expression is rational then it equals 1” but I don’t see why that is the case. If the expression were not rational, there would be no integer solutions to the equation I derived. I emphasized that the definition of rational is an iff statement, not if, so it is a valid move to show there exist integers that can serve as numerator and denominator. Nowhere in my proof did I ASSUME that – it could have been that only m, n irrational would do, in which case I could conclude it is NOT rational, again due to the iff.
Where am I going wrong?
Filling in the details, he assumes that $$\sqrt{4+2\sqrt{3}}-\sqrt{3}=\frac{m}{n}$$ for some numbers m and n, and presumably did something like this:
$$\sqrt{4+2\sqrt{3}}=\frac{m}{n}+\sqrt{3}\\n\sqrt{4+2\sqrt{3}}=m+n\sqrt{3}\\ n^2(4+2\sqrt{3})=m^2+2mn\sqrt{3}+3n^2\\ 4n^2+2n^2\sqrt{3}-m^2-2mn\sqrt{3}-3n^2=0\\ n^2-m^2+(2n^2-2mn)\sqrt{3}=0$$
And this is true if \(m=n\), as you can see by substituting. But is that enough?
Justifying the proof
Doctor Fenton was the first to answer:
Hi Mike,
I don’t see the definition of rational as an “if and only if” statement. To me, it is just an “if” statement: x ε R if there exist integers m and n, n ≠ 0, such that x = m/n. I don’t see any content to the statement the x is rational only if there exist such integers.
As we’ll see below, it isn’t wrong to use “iff” (“if and only if”) in the definition, but the “if” part (not the “only if”!) is the relevant part for proving a number rational.
You manipulated the equation
\(\sqrt{4+2\sqrt{3}}-\sqrt{3}=\frac{m}{n}\)
to obtain
\(n^2-m^2+(2n^2-2mn)\sqrt{3}=0\),
which can be rewritten as
\((n-m)(n+m)+2n(n-m)\sqrt{3}=0\)
\((n-m)[m+n(1+2\sqrt{3})]=0\)
so either \(n-m=0\) or \(m+n(1+2\sqrt{3})=0\) .
If the first factor is 0, then any non-zero value of n ( = m) solves the equation, and in particular, any non-zero integral value of n works. Therefore the expression is rational, because such solutions exist, and its value is 1.
Since he has found a solution, and m and n are both integers, he appears to have proved the claim.
But there’s a detail that has been missed:
I might point out that to obtain the modified equation required squaring a rewritten original equation, which can introduce extraneous solutions. If m – n ≠ 0, then
\(m+n(1+2\sqrt{3}=0\) ,
so \(m=-n(1+2\sqrt{3})\), or \(\frac{m}{n}=-1-2\sqrt{3}\) .
Note that \(\sqrt{4+2\sqrt{3}}=\sqrt{(1+\sqrt{3})^2}\) which has two square roots, 1 + √3 and -1 – √3 . If we use the negative square root, we have
\(\sqrt{4+2\sqrt{3}}-\sqrt{3}=-1-\sqrt{3}-\sqrt{3}=-1-2\sqrt{3}\)
which is the extraneous solution to the modified equation.
So there are actually two solutions to the equation, namely \(\frac{m}{n}=1\) and \(\frac{m}{n}=-1-2\sqrt{3}\); the former is what you get taking the positive root of \(4+2\sqrt{3}\), while the latter is what you get from the negative root. The former is the only one that matters (since the radical indicates only the positive root).
Mike responded:
Thank you so much! I agree that the only substantive direction of the iff is the requirement that there exist integers whose ratio is that number. I guess I see definitions as establishing an equivalence between a term and a concept, hence my use of iff. You showed I actually don’t need it to be an iff anyway, contrary to what I said 🙂
So the bottom line is that my proof is valid, correct? The idea was simply to use the definition of a rational to derive an equation such that if there are any integer solutions then the original expression is rational and vice versa. However, I didn’t require that integers need be the only solutions, much like Diophantine equation have irrational solutions sometimes.
Doctor Fenton agreed:
Yes, your argument was correct. A real number is rational if it is possible to write it as a quotient of two integers. That doesn’t mean it cannot be written as a quotient of non-integers. For example, 1/2 = √2 / √8 .
And Mike was satisfied:
Thank you very much! So much for MathStackExchange — was halfway convinced I was missing something fundamental!
I’m going to continue to use this site instead to verify any questions I have.
Thanks for continuing to offer this great service 🙂
Catching some subtleties
But the commenter was actually correct, in a subtle way. I joined in to emphasize this point:
Hi, Mike.
I’d like to add a couple points.
First, I agree with your statement that any definition is in essence an “iff”: Something is a ___ if and only if it satisfies some description. Any rational number will fit the description, and anything that fits the description is a rational number.
But to prove that it is a ___, you are actually using the “if“, not the “only if“: It is a ___ because such and such is true.
In particular, a number is rational IF it can be written as a ratio of integers, so you are showing that, in fact, it can be written as a ratio of integers.
The “iff” nature of a definition comes from the fact that the description must be true of all rational numbers, and only rational numbers. If a number can be written as a fraction, then it is rational (that’s the “if”); and only numbers that can be written as a fraction can be called rational (that’s the “only if”).
It is common to misinterpret the “only if” part of “if and only iff”, as discussed here:
Necessary and Sufficient Conditions: If, or Only If?
When we say “A if and only if B”, that means, first, “A if B”, so that B implies A, and, second, “A only if B”, so that the only way A can be true is when B is true, so that A implies B: “B if A”.
Mike had claimed to be using the “only if” side; that would mean using the fact that a number is rational only if it can be written as a fraction, but that means that if a number is rational, then it can be written as a fraction, which is not what he did. Doctor Fenton had quietly mentioned that they were using the “if” (if a number can be written as a fraction, then it is rational), without pointing out the error (which is an error only in words, not in substance).
How about the proof?
What you are doing here is writing an equation in m and n, and showing that there is a solution consisting of integers. Ultimately, however, the proof consists in the final check, where you will show that your m and n actually work; the work to find them only works as a proof in itself if every step is reversible. You haven’t shown us the details of your work; that may be necessary in order to be completely sure whether your proof works.
Presumably your work shows that IF sqrt(4+2*sqrt(3)) – sqrt(3) = m/n, THEN n^2 – m^2 + (2n^2-2nm)*sqrt(3) = 0. I’ve done the same work; note that at one point you have to square both sides, which does not result in an equivalent equation; recall how you can get extraneous solutions by squaring.
We discuss this in Extraneous Solutions: Causes and Cures.
Doctor Fenton had mentioned how squaring can result in extraneous solutions; so it is entirely possible that 1 could have been the extraneous solution, rather than the correct one.
In fact, if our expression had been \(\sqrt{4+2\sqrt{3}}+\sqrt{3}\), that would be true. We’d get the equation $$n^2-m^2+(2n^2+2mn)\sqrt{3}=0$$ which factors as $$(n+m)\left[n(1+2\sqrt{3})-m\right]=0$$ giving solutions $$\frac{m}{n}=-1\text{ and }\frac{m}{n}=1+2\sqrt{3}$$
And in fact, \(\sqrt{4+2\sqrt{3}}+\sqrt{3}\approx4.464\), and \(1+2\sqrt{3}\approx4.464\). So this expression is irrational, though it looked for a moment as if it might be \(-1\). (Except that the square roots can’t be negative!)
Now you show, somehow, that m = n yields a solution. But note that it is not the finding of this solution that constitutes a proof, but the checking. And then, it is not the checking in the derived equation that proves rationality, but the checking in the original equation! And I suspect you didn’t do that, except by showing that a calculator produces that result, which is not really a proof.
So, depending on exactly what you wrote, your proof may be faulty, and the objections may be valid.
Finally, with regard to the question at the end:
I looked for what you might have written, and see that this problem has been covered several times on StackExchange, 8 years ago or more, but not your contribution. But it is true that if you showed only that the derived equation has integer solutions, that does not prove the original claim. It’s not that you assumed that the integers exist, but that you (apparently) assumed that the two equations are equivalent.
Again, can you show us the details, so we can be sure whether your proof is valid, and point out specific details if it is not? It doesn’t sound like the objections are quite right, but I suspect there is some truth in them.
I did later find what appears to be Mike’s question and the pushback he got (in which he subsequently added a correction based on our comments).
Mike understood the idea:
You are correct that I assumed that if my derived equation had integer solution then the original was rational. Sounds like that was the issue people were getting at since my derived equation also has a solution that implies m/n is irrational
I thought thar the existence of extraneous solutions wouldn’t be an issue as long as one of them are rational.
However, I think that is wrong now. I’d need to show, as a check, that sqrt(4+2*sqrt(3)) – sqrt(3) – 1 = 0 and indeed it does (simple two steps of algebra).
I answered:
Yes; in searching for what you might have said on StackExchange, I saw several ways to do this, but once you know (e.g. by the work you did, or by a calculator) to “guess” that the expression equals 1, you can show that it is in several ways. One, perhaps yours, is to note that 4+2√3 = 1+2√3+3 = (1+√3)2 (as it must if this equation is true), so that √(4+2√(3)) – √(3) = √((1+√3)2) – √(3) = 1+√3 – √3 = 1.
Completing the proof
Mike now showed his finished proof:
Correct, I just did the following:
√(4+2√(3)) – √(3) = 1 (check my “guess” from my prior solution) –>
√(4+2√(3)) = 1+ √(3) –>
4+2√(3) = 1 + 2√(3) + 3 = 4+2√(3) QED (?)
I told Mike this was good; but looking at it now, I see that he started with a form of the claimed solution and just squared both sides; the resulting equation, as we’ve seen, could be true even if the original was false (that is, if the signs were wrong), so this still isn’t quite a proof. To correct that, we might also show that both sides of \(\sqrt{4+2\sqrt{3}}=1+\sqrt{3}\) are positive, so that the last line does imply the previous line, since the square root is the positive number whose square is the given number.
In his StackExchange correction, he wrote
So I conferred with a mathematics professor, and he confirmed my reasoning is valid, but incomplete. In my above proof, I square the expressions in one of my steps, which introduces extraneous solutions (i.e., my steps are not reversible).
Therefore, all I have shown is that 1 may be the value of \(\sqrt{4+2\sqrt{3}}-\sqrt{3}\). The proof now requires me to show that, indeed, \(\sqrt{4+2\sqrt{3}}-\sqrt{3}=1\)
Thankfully, this is an easy final check: $$\sqrt{4+2\sqrt{3}}-\sqrt{3}=1 \Rightarrow\sqrt{4+2\sqrt{3}}=1+\sqrt{3} \Rightarrow 4+2\sqrt{3}=1+2\sqrt{3}+3=4+2\sqrt{3}$$
The last line is just what he wrote here.
I would prefer to start not with what we want to prove, but with what I said earlier, together with an explicit mention of positivity:
Note that \(4+2\sqrt{3}=1+2\sqrt{3}+3=(1+\sqrt{3})^2\), and that \(1+\sqrt{3}>0\), so that \(\sqrt{4+2\sqrt{3}}=1+\sqrt{3}\).
Therefore \(\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{(1+\sqrt{3})^2}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\). Since this is rational, we have completed the proof.
Incidentally, since the ultimate answer is to show that the value is equal to the value we determined from the original equation, it wasn’t necessary to use \(\frac{m}{n}\) in that equation; we could have just set the expression equal to x and solved for that, then checked the answer as I did here. The main work to find the value would then look like this:
Suppose that \(\sqrt{4+2\sqrt{3}}-\sqrt{3}=x\). Then $$\sqrt{4+2\sqrt{3}}=x+\sqrt{3}\\ 4+2\sqrt{3}=x^2+2x\sqrt{3}+3\\4+2\sqrt{3}-x^2-2x\sqrt{3}-3=0\\1-x^2+(2-2x)\sqrt{3}=0\\(1-x)(1+x)+2(1-x)\sqrt{3}=0\\(1-x)(1+x+2\sqrt{3})=0$$ so either \(1-x=0\) or \(x+(1+2\sqrt{3})=0\).
If \(x=1\), then we have a rational number! To show that it is, …
In addition to simply recognizing the perfect square, or solving as we just did, there are a couple standard ways to “de-nest” a radical (when possible), which would provide our answer. But, indeed, all the work of discovering that the value of the expression is (that it, appears to be) 1 is not technically part of the proof itself; one could just use the calculator, as Mike did, in order to see the apparent answer, and then the entire work of proving is to show that it actually is.
Of course, that minimal proof is rather unsatisfying when presented without a way to find the answer, isn’t it? His original attempt at a proof is what we have called the “exploratory phase“, finding a hypothesis that we could then prove. That phase is an essential part of our work (whether it involves solving an equation, or just looking at the problem in the right way), but can be omitted in the final proof – unless, like us, you want to show students how proofs are done.