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In particular, my question is about abstract mathematics such as group theory, analysis, topology, etc. where most textbooks are filled with exercises which require proof, and how to go about effectively "internalizing" proofs and solutions to exercises/theorems when you read them.

Often I will have the experience of trying out an exercise for a while--maybe a few hours, or days--but eventually giving up and ultimately consulting the textbook or Math.SE, etc. for a solution. However, more often than not while I do get the "aha" feeling of seeing the method or "trick" needed to prove the statement, I don't really feel as if I gained something I could apply to a brand new problem.

Recent example: Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.

I tried to solve this for a couple of hours, but I didn't make much progress. The solution in the book looks like this:

\begin{align*} \sqrt 2 + \sqrt 3 + \sqrt 5 &= r \in \mathbb Q\\ \\ \sqrt 2 + \sqrt 3 &= r - \sqrt 5\\ \\ (\sqrt 2 + \sqrt 3)^2 &= (r - \sqrt 5)^2\\ \\ 5+2\sqrt 6 &= r^2 + 5 -2r\sqrt 5\\ \\ 2\sqrt 6 + 2r\sqrt 5 &= r^2,\\ \end{align*}

so $2\sqrt 6 + 2r\sqrt 5$ is rational. Squaring this expression again will show that $\sqrt {30}$ is rational, which can be easily disproved.

But my question is, how does one read a solution to a problem in such a way where you can begin to comprehend what the thought processes were behind creating the solution? In the above example, it all makes sense, but why did they move $\sqrt 5$ to the RHS in the first place? Trial and error? Intuition?

Or in other words, are there any concrete studying techniques that can take you from "Yeah, that makes sense now that you mention it" to "Yeah, I can understand why they thought of this."

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    $\begingroup$ "why did they move $\sqrt5$ to the RHS in the first place? Trial & error? Intuition?" $\quad$ I'd say a combination of experience, intuition and trial & error (mostly the first). $\quad$ The expression $\sqrt 2 + \sqrt 3 + \sqrt 5$ looks amenable to squaring; $(\sqrt 2 + \sqrt 3 + \sqrt 5)^2$ feels like an unwieldy 2-step iterative operation, so one steps back then observes that $r - \sqrt 5$ as easy to square as the lone $r$ on the RHS. $\quad$ But this isn't a good example of a 'magic trick', since pushing $\sqrt5$ to the RHS (just makes the work feel easier and) isn't a crucial step. $\endgroup$
    – ryang
    Aug 12 at 4:45
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    $\begingroup$ This does not address your question, but I think it's still worthwile to mention it: "The solution in the book looks like this" Does the solution in the book really look like this? I'm asking because this is not even a mathematical argument; it's just a list of equations, without any comment on how these equations are related. $\endgroup$ Aug 12 at 10:37
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    $\begingroup$ I agree with @ryang that moving $\sqrt{5}$ to the right-hand side doesn't look crucial. You can also arrive at a contradiction starting from $\sqrt 2 + \sqrt 3 + \sqrt 5 = r$ just by squaring repeatedly, moving the rational terms to the right-hand side after each squaring. Moving $\sqrt 5$ to the right-hand side first looks like it's just cutting out a few steps by making both sides have two terms each, rather than having one side with three terms and one side with one term. $\endgroup$
    – Stef
    Aug 12 at 11:28
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    $\begingroup$ I find with problems like these that it helps to build intuition by solving a simpler but related problem. Start by proving that $\sqrt 2 + \sqrt 3$ is irrational first and see how that goes, then try the harder problem. $\endgroup$ Aug 12 at 16:17
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    $\begingroup$ How is this on-topic? "Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics." (my emphasis) $\endgroup$ Aug 13 at 10:59

3 Answers 3

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Attempt to solve a similar exercise with different numbers, see when it works and when it fails. Using the technique by yourself will help in internalizing it. That is, ask questions like:

  • does this work for different numbers in the square roots?
  • what about a different amount of square root expressions; is this phenomenon or argument specific to three?
  • what if the square root expressions are multiplied by integers or rational numbers?
  • what about other roots or other exponents?

You do not need to think about or solve all of these questions, but if you feel this particular trick is important to internalize, then you may want to look at some of them.

After having done it yourself, the techniques might accessible for you to use in solving other problems, or at least in solving more complicated problems where this is an intermediate stage.

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The main intuition behind the proof is that you are trying to obtain an equality of the form $\sqrt{x} = r'$, with $r' \in \mathbb{Q}$, because we known how to disprove that one.

The intuition behind moving $\sqrt 5$ to the right-hand side is that squaring a sum of two square roots gets rid of one root, whereas squaring a sum of three square roots doesn't get rid of a root directly. Indeed: $$(\sqrt a + \sqrt b)^2 = a + 2 \sqrt {ab} + b$$ goes from two roots down to one root; whereas $$(\sqrt a + \sqrt b + \sqrt c)^2 = a + b + c + 2 (\sqrt {ab} + \sqrt {ac} + \sqrt {bc})$$ goes from three roots to three roots, which doesn't immediately look like progress.

That being said, I agree with @ryang that moving $\sqrt{5}$ to the right-hand side doesn't look crucial. You can also arrive at a contradiction starting from $\sqrt 2 + \sqrt 3 + \sqrt 5 = r$ just by squaring repeatedly, moving the rational terms to the right-hand side after each squaring.

Here is a proof without moving $\sqrt 5$ to the right-hand side.


Assume $\sqrt 2 + \sqrt 3 + \sqrt 5 = r \in \mathbb{Q}$.

We are going to square sums of three terms more than once, so let's do it with variables first so we have a formula to apply: \begin{align*} (a + b + c)² & = (a + b)² + 2((a+b)c) + c² \\ (a + b + c)² & = a^2 + 2ab + b² + 2ac + 2bc + c² \\ (a + b + c)² & = a^2 + b² + c² + 2(ab + ac + bc) \\ \end{align*}

Now we can square our rational expression: \begin{align*} \sqrt 2 + \sqrt 3 + \sqrt 5 & = r \in \mathbb{Q} \\ (\sqrt 2 + \sqrt 3 + \sqrt 5)² & = r^2 \\ 2 + 3 + 5 + 2 (\sqrt 6 + \sqrt {10} + \sqrt {15}) & = r^2 \\ \sqrt 6 + \sqrt {10} + \sqrt {15} & = r_1 \in \mathbb{Q} \\ (\sqrt 6 + \sqrt {10} + \sqrt {15})^2 & = {r_1}^2 \\ 6+10+15+2(\sqrt {60} + \sqrt{90} + \sqrt{150}) & = {r_1}^2 \\ \sqrt {60} + \sqrt{90} + \sqrt{150} & = r_2 \in \mathbb{Q} \\ 2 \sqrt {15} + 3 \sqrt{10} + 5 \sqrt{6} & = r_2 \\ \end{align*} Now we recognise that $r_1$ and $r_2$ are both expressed as a combination of the same three square roots, $\sqrt {15}$, $\sqrt{10}$ and $\sqrt{6}$. So we can eliminate one of the three roots with a subtraction:

\begin{align*} (2-2) \sqrt {15} + (3-2) \sqrt{10} + (5-2) \sqrt{6} & = r_2 - 2 r_1 \\ \sqrt{10} + 3 \sqrt{6} & = r_3 \in \mathbb{Q} \\ \end{align*}

Finally we have only two roots, so now squaring again will leave us with only one root. \begin{align*} \sqrt{10} + 3 \sqrt{6} & = r_3 \\ (\sqrt{10} + 3 \sqrt{6})^2 & = {r_3}^2 \\ 10 + 6 \sqrt{60} + 54 & = {r_3}^2 \\ \sqrt {15} & = r_4 \in \mathbb{Q} \\ \end{align*}

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are there any concrete studying techniques that can take you from "Yeah, that makes sense now that you mention it" to "Yeah, I can understand why they thought of this."

Technically, yes, but not at the beginner's level. Try to read Polya's "How to solve it?". The main advice there (if you forget about various tricks and details) is just "relate it to something familiar". It applies to the analysis of proofs just as well as to the proof design. But to follow it you have to have some database of "familiar things" and an effective algorithm for extracting them and evaluating the degree of similarity. Both come with practice. In the beginning you just look for similar words (This problem is about, say, a simplex in $R^n$. What other problems about a simplex in $R^n$ have I seen? Can I use the same technique here?). Then you slowly acquire a feeling for a general setup (so the simplex problem can be related to a problem about contractive mappings, say). And so on. It all takes time and patience.

As to your particular problem, the answer is another ubiquitous general principle (also discussed in Polya, I believe): Simplify whenever possible, making sure that you have some tangible gain at each step!

Here it works like that. When you see an equation with square roots, your first instinct (from high school, say) should be to square something to get rid of those nasty radicals. But if you square the LHS as it is, you start with 3 radicals and you end up with 3 radicals after the squaring, so not much is gained. The reason is that the product of two different radicals is a radical again more often than not, so you have three radicals and three pairwise products. You want to reduce the number of square roots. So you want to have only 2 or fewer of them on one side. Hence, to the other side one of them goes (which one does not matter in this problem though sometimes it does). Now you have a really tangible gain (down to 2 from 3). Then you go from 2 to 1 and, bingo, you can relate it to the famous ancient proof of the irrationality of $\sqrt 2$ only, perhaps, with a different prime now.

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