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Which side is larger? $$ \sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2} \stackrel{?}{\lessgtr} 5 $$

Without using a calculator, computer, or estimating square roots, please determine which side has the larger value.

Rules of the game: In this puzzle, you have to manipulate inequalities between two sums of radicals (one on the left hand side, and another one on the right hand side). You start with the two expressions $L:=\sqrt{15}-\sqrt{7}+\sqrt{5}+\sqrt{2}$ and $R:=5$ as given above. You are only allowed to perform the following three operations:

  • Add the same value $\Delta$ to $L$ and to $R$, which yields a new left hand side $L+\Delta$ and a new right hand side $R+\Delta$.
  • Multiply both $L$ and $R$ by the same non-negative real number $c$, which yields a new left hand side $c\cdot L$ and a new right hand side $c\cdot R$.
  • Square $L$ and $R$ (given that $L$ and $R$ are non-negative), which yields a new left hand side $L^2$ and a new right hand side $R^2$.

The goal is to reach an inequality with integers on both sides.

There is a "nifty way" of doing this that moves a quantity (a radical) over first. After squaring both sides, and combining integers on respective sides, the integers can be subtracted away. Then you'll be left with the combination of three unlike radicals on one side versus a single fourth unlike radical on the opposite side. But, you should keep going until you have one integer on one side, versus one integer on the other side.

Gamow
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Olive Stemforn
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  • @hmmn You can get $5 = 2^2 + 1^2$ and $2 = 1^2 + 1^2$ but no straight-forward right-angle triangles for $15$ or $7$. – Paul Evans Mar 18 '16 at 01:22
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    This question was posted here, because the following similar question, which was meant also meant as a puzzle question was not welcomed here: http://math.stackexchange.com/questions/1695391/which-is-larger-sqrt2-sqrt3-sqrt5-sqrt7-or – Olive Stemforn Mar 18 '16 at 02:57
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    @Olive: You mean you're posting off-topic questions deliberately because other questions were off-topic? – Deusovi Mar 18 '16 at 05:42
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    @2012rcampion -- You must have gone awry somewhere. Your right-hand side should be: $-5\sqrt{2} \ + \ \sqrt{35} \ + \ \sqrt{105} \ \ $ instead. And, even with that, you would not be done, because you need to move another radical around and square at least one time more. – Olive Stemforn Mar 18 '16 at 05:53
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    @2012rcampion - When I posted a similar problem in the Mathematics section, intending for it to be a challenge, they stated it looked as if I posted it as if it were a math problem I was posting without showing any work/attempts of my own. But it isn't that. It's not a homework problem. – Olive Stemforn Mar 18 '16 at 06:21
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    I'm finding this to be like a maze with some very convenient cancellations, but haven't yet found the cleanest route to the cheese. I hope this gets reopened so that a solution might even be presented in maze format. – humn Mar 18 '16 at 06:32
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    @hmmn Something like http://i.stack.imgur.com/nB9HK.png ? – 2012rcampion Mar 18 '16 at 06:50
  • So far I haven't found a way to justify the choice of rearrangement at each step other than "it works," and it looks like you can do the same thing with any sum of radicals. – 2012rcampion Mar 18 '16 at 06:54
  • @hmm It should be, I had the computer brute-force all arrangements at each step. Basically each node is an expression; the edges are labeled by two subsets of each expression which, when squared and subtracted, produce the next expression. The recursion stops when all terms are positive (or negative, but I take out all common factors including -1). – 2012rcampion Mar 18 '16 at 07:09
  • I'm starting to get it, @2012rcampion. It looks more than complete and should be seen by the doubters at math.se. OliveStemforn, have you seen it laid out like this before? Wow! – humn Mar 18 '16 at 07:13
  • @ hmmm - No, I have not seen a computer brute-force out all arrangements, if that is what you are asking. It sounds as if I were to get lost looking at it (figuratively speaking). – Olive Stemforn Mar 18 '16 at 07:21
  • – Olive Stemforn Mar 18 '16 at 07:42
  • Another note in favor of this puzzle. The specific numbers seem to have been selected deliberately to allow manual solution to be like a detective story with manageable clues that simplify the mystery as they fit together. – humn Mar 18 '16 at 08:05
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    @hmmn Try $\sqrt{2} + \sqrt{3} + \sqrt{5} \lessgtr \sqrt{6} + \sqrt{7}$, or $\sqrt{2} + \sqrt{3} + \sqrt{5} \lessgtr \sqrt{7} + \sqrt{11}$. I think the properties of this problem are shared by pretty much all sums of radicals. – 2012rcampion Mar 18 '16 at 08:52
  • @hmmn or 2012rcampion: can either of you point me to an explanation of what's going on with your process? I can't follow that picture at all. If it's not too much trouble, that is. – question_asker Mar 18 '16 at 12:03
  • Not sure where to point, @question_asker, other than the revised question. Pretty sure that 2012rcampion figured it out from a comment exchange with OliveStemforn. – humn Mar 18 '16 at 12:19
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    This puzzle is far beyond text book math. There is no general theory for this area. I would expect that the solution will heavily depend on the particular choice of the numbers 15, 7, 5, 2. – Gamow Mar 18 '16 at 12:52
  • Surely multiplication of both sides by any number would also be allowed or not? – Ivo Mar 18 '16 at 13:54
  • @ Ivo Beckers - Well, at one point division by 2 on both sides was allowed. That's multiplication on both sides by 1/2. – Olive Stemforn Mar 18 '16 at 20:19
  • Well, @2012rcampion, you now have more than a margin in which to post the roadmap. While you weren't looking, hexomino(+Gamow) found my favorite route on the map but, having seen hexomino work through another puzzle, there's no doubt the solution here isn't plagiarism. If it means anything to you, though, these comments record your getting to the summit first. (Welcome back to awake mode, OliveStemforn, thanks again for adding this summit to the landscape.) – humn Mar 18 '16 at 21:50
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    If your approach differs at all from what you've seen here, OliveStemforn, why not post it as an answer? The two approaches we've seen so far, even if fundamentally similar, look different enough to be extra interesting in each other's company. OPs' intended solutions in their original forms seem very welcome here. You could even mention how you chose the particular numbers that work out so well. I learn a lot from seeing how a surprising puzzle came together. – humn Mar 19 '16 at 06:47

4 Answers4

26

The answer is

The right hand side is bigger

I imagine the line of reasoning the author wants is as follows:

$2352 ~>~ 1$
$\Rightarrow~ 28\sqrt{3} ~>~ 1$
$\Rightarrow~ 3+196+28\sqrt{3} ~>~ 200$
$\Rightarrow~ \sqrt{3}+14 ~>~ 10\sqrt{2}$
$\Rightarrow~ 3+14\sqrt{3} ~>~ 10\sqrt{6}$
$\Rightarrow~ 21+7+14\sqrt{3} ~>~ 10+15+10\sqrt{6}$
$\Rightarrow~ \sqrt{21} + \sqrt{7} ~>~ \sqrt{10} +\sqrt{15}$
$\Rightarrow~ \sqrt{21} -\sqrt{15} + \sqrt{7} ~>~ \sqrt{10}$
$\Rightarrow~ \sqrt{105} -\sqrt{75} + \sqrt{35} ~>~ \sqrt{50}$
$\Rightarrow~ -\sqrt{50} ~>~ -(\sqrt{105} -\sqrt{75} + \sqrt{35})$
$\Rightarrow~ - 2(\sqrt{105} -\sqrt{75} + \sqrt{35}) ~<~- 2\sqrt{50}$
$\Rightarrow~ 27 - 2(\sqrt{105}-\sqrt{75} + \sqrt{35}) ~<~ 27- 2\sqrt{50}$
$\Rightarrow~ 15+7+5 -2(\sqrt{105}-\sqrt{75}+\sqrt{35})~<~25+2-2\sqrt{50}$
$\Rightarrow~ (\sqrt{15} - \sqrt{7} + \sqrt{5})^2 ~<~ (5 - \sqrt{2})^2$
$\Rightarrow~ \sqrt{15} - \sqrt{7} + \sqrt{5} ~<~ 5 - \sqrt{2}$
$\Rightarrow~ \sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2} ~<~ 5$

EKons
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hexomino
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    A great solution, by the way. The central step seems to be that the value 27 cancels out in lines 4 and 5 from the bottom. – Gamow Mar 18 '16 at 14:19
  • Okay, I think I've fixed it but please do check again. It's a bit more complicated than I first thought. – hexomino Mar 18 '16 at 14:25
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    Ugh.. somehow I had it in my mind that a negative number squared was also negative otherwise I probably also found it. Nice job! Personally I would write the steps in the other direction though because now you have to read from bottom to top to see what steps you did – Ivo Mar 18 '16 at 14:26
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    My insight from this is that in order to solve any future equations of the kind $\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}\pm\sqrt{d} \stackrel{?}{\lessgtr} e$ you just have to calculate $\frac{a+b+c+d-e^2}{2}$ and that will be the square root you have to move to the other side to eliminate a square root – Ivo Mar 18 '16 at 14:38
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    @hexomino: I have slightly revised your solution (divided once by sqrt(3) and once by sqrt(5)). It is the same argument, just the numbers are a little bit smaller (and easier to parse). – Gamow Mar 18 '16 at 14:40
  • @Gamow That's much nicer, thank you. I had actually misread your first comment as referring to the glaring mistake in my original "solution" – hexomino Mar 18 '16 at 14:42
  • @Ivo: So what would you do in case a=9999, b=1001, c=101, d=11, and e=145? (And assume left hand side $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}$.) – Gamow Mar 18 '16 at 14:59
  • @Gamow yeah then it wouldn't work. It only works if $\frac{a+b+c+d-e^2}{2}$ is equal to $a$,$b$,$c$ or $d$. And I think that when that isn't the case it won't be that easy to come with a solution to the inequality – Ivo Mar 18 '16 at 15:04
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    Cheers, hexomino! Cheers, @Gamow! 2012rcampion's treasure map shows that 1 < 2352 is indeed the lowest possible residue. Congratulations as well on keeping the signs straight. – humn Mar 18 '16 at 22:06
  • Well done, but halfway you multiply by $-2$ and that is against the rules (you may only multiply by positive numbers) – Willemien Mar 31 '16 at 18:31
  • @Willemien, fair point. But multiplication by negative numbers can be achieved by several permissible steps. First, subtract the total of both sides from both sides of the inequality, then multiply by a positive constant. I've added an extra line for clarity. – hexomino Mar 31 '16 at 21:45
  • If we were allowed to estimate the value of the square roots, then is there a way to realise that the right side is larger, without making very precise calculations ? – Hemant Agarwal Jun 17 '21 at 21:29
  • @HemantAgarwal I should think so, two decimal places would probably be enough. – hexomino Jun 17 '21 at 21:39
  • Thanks. How did you realise though that finding square roots to 2 decimal places ( and not 1 or 3 decimal places) would give the answer ? I tried with 1 decimal place and in that case, I got the wrong answer. – Hemant Agarwal Jun 17 '21 at 21:48
  • @HemantAgarwal Maybe one of the two other solutions given (by Rosie F and Toby Mak)? They seem to be more concise, at least – hexomino Jun 17 '21 at 21:52
  • @HemantAgarwal 2 decimal places was mainly a guess. I didn't think 1 would be enough, 3 will also work. – hexomino Jun 17 '21 at 21:55
  • Ok. Secondly, let's say that this was the only condition : "Without using a calculator, computer, or estimating square roots, please determine which side has the larger value." There is no other condition other than this. Is there an easier way than the one that you have given, to solve this question ? – Hemant Agarwal Jun 17 '21 at 21:56
  • @HemantAgarwal Have you looked at the other two answers? Maybe they would be considered easier? – hexomino Jun 17 '21 at 21:58
  • I saw all the answers . When I look at your answer from the bottom to the top, then I find it to be the most intuitive of all the answers. I would have also taken a similar approach, moved $ \sqrt{15} , \sqrt{5} $ or $ \sqrt{2} $ to the right, squared both the sides and then proceed further . How did you realise that it should be $ \sqrt{2} $ that should be moved to the right side and not $ \sqrt{15} $ or $ \sqrt{5} $? Was it just trial and error or was there some intuition behind moving $ \sqrt{2} $ ? – Hemant Agarwal Jun 18 '21 at 05:15
11

Here's another approach, which starts from smaller integer residuals than hexomino's solution.

$\qquad 14 < 15$
$\Rightarrow \sqrt{14} < \sqrt{15} $
$\Rightarrow 4\sqrt{105} < 30\sqrt2 $
$\Rightarrow 28 + 4\sqrt{105} + 15 < 25 + 30\sqrt2 + 18 $
$\Rightarrow 2\sqrt7 + \sqrt{15} < 5 + 3\sqrt2 \qquad\qquad \text{___ [1]} $

and

$\qquad 35 < 36 $
$\Rightarrow \sqrt{35} < 6 $
$\Rightarrow 6\sqrt{35} < 36 $
$\Rightarrow 32 < 63 - 6\sqrt{35} + 5 $
$\Rightarrow 4\sqrt2 < 3\sqrt7 - \sqrt5 $
$\Rightarrow 3\sqrt2 - \sqrt7 + \sqrt5 < 2\sqrt7 - \sqrt2 \quad \text{___ [2]} $

Finally, bringing these lines of argument togegther,

$2\sqrt7 + \sqrt{15} + 3\sqrt2 - \sqrt7 + \sqrt5 < 2\sqrt7 + \sqrt{15} + 2\sqrt7 - \sqrt2 \quad \text{by [2]} $ $ \qquad\qquad < 5 + 3\sqrt2 + 2\sqrt7 - \sqrt2 \quad \text{by [1]} $
$\Rightarrow \sqrt{15} - \sqrt7 + \sqrt5 < 5 - \sqrt2 $
$\Rightarrow \sqrt{15} - \sqrt7 + \sqrt5 + \sqrt2 < 5 $

I admit it isn't a single chain of inferences.

Rosie F
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Yet another method:

$$105 > 49 \Rightarrow \sqrt{105} > 7 \Rightarrow 22 +2\sqrt{105} > 22+14$$ $$\Rightarrow (\sqrt{15})^2 + (\sqrt{7})^2 + 2 \sqrt{15} \sqrt{7} > 36$$ $$\Rightarrow \sqrt{15} + \sqrt{7} > 6 \tag{1}$$

And then:

$$9^2 \times 10 < 29^2 \Rightarrow \sqrt{10} < \frac{29}{9} \Rightarrow 7 + 2 \sqrt{10} < \frac{29 \cdot 2 + 7 \cdot 9}{9}$$ $$\Rightarrow (\sqrt{5})^2 + (\sqrt{2})^2 + 2\sqrt{10} < \frac{121}{9}$$ $$\Rightarrow \sqrt{5} + \sqrt{2} < \frac{11}{3} \Rightarrow \frac{8}{6} +\sqrt{5} + \sqrt{2} < 5$$ $$\Rightarrow \frac{15-7}{\sqrt{15} + \sqrt{7}} +\sqrt{5} + \sqrt{2} < 5 \tag{*}$$ $$\Rightarrow \sqrt{15} - \sqrt{7} +\sqrt{5} + \sqrt{2} < 5$$

(*): From (1), increasing the denominator decreases the fraction; strictly smaller than LHS of previous step

Toby Mak
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-14

The right side is larger because

You square the roots and do the sum, which equals 15 and you suqare the right side which gives you 25.

Trixy
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