Technique in simplifying this (involves square roots)

[yik-boh]

\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}


What technique/method could I use to simplify that one fast? It was asked in a math contest here in our country and the question is only for 20 seconds. The answer is \sqrt{2}. You know any technique for that one guys?

 

[Gerenuk]

I don't know of a general technique, but here I'd probably see that squaring the expression makes sense.

The change in sign infront of the 2 looks suspiciously like (a+b)(a-b)=a^2-b^2 and even this special case indicates simplification (\sqrt{5}+2)(\sqrt{5}-2)=5-2^2=1!

So I would decide to bring these terms together which is conveniently done by squaring the fraction. On the top you use the binomial theorem (a+b)^2=a^2+2ab+b^2. I would see that the a^2 and b^2 terms would cancel the 2. And I would see that the cross term simplifies 2ab=2\sqrt{5-2^2}=2. You also lose a square root on the bottom. This all I would have done in my head being lead but the coincidences just mentioned.

The rest is down to luck that incidently
\frac{2\sqrt{5}+2}{\sqrt{5}+1}
cancels.

 

[Mark44]

I don't know of a general technique, but here I'd probably see that squaring the expression makes sense.

No it doesn't. If you square the expression you will get a new expression with a different value. The goal here is to rewrite the original expression in a simpler form that is equal to the original expression.

The change in sign infront of the 2 looks suspiciously like (a+b)(a-b)=a^2-b^2 and even this special case indicates simplification (\sqrt{5}+2)(\sqrt{5}-2)=5-2^2=1!

So I would decide to bring these terms together which is conveniently done by squaring the fraction. On the top you use the binomial theorem (a+b)^2=a^2+2ab+b^2. I would see that the a^2 and b^2 terms would cancel the 2. And I would see that the cross term simplifies 2ab=2\sqrt{5-2^2}=2. You also lose a square root on the bottom. This all I would have done in my head being lead but the coincidences just mentioned.

The rest is down to luck that incidently
\frac{2\sqrt{5}+2}{\sqrt{5}+1}
cancels.

 

[trambolin]

So? You can take the square root of the result in the end...

 

[statdad]

If the question is to evaluate the original expression then the work done above is correct: it's been shown that a^2 = 2, so it follows a = \sqrt a.
Philosphically, I wouldn't call this a "demonstration", or "simplification", since I think of those as referring to steps that start with the original expression and then, through application of algebraic steps that leave the basic value unchanged, reach a specified result.

That said, the idea of squaring this thing and seeing where that leads was pretty darn clever.

 

[trambolin]

Your philosophy is not correct. This result is.

 

[Mark44]

You can run into problems when you square things and then take square roots; namely introducing extraneous solutions.

Simple example
x = -2
x^2 = (-2)^2 = 4
x = 2 or x = -2

 

[statdad]

Your philosophy is not correct. This result is.

That's a little rash, isn't it? The comment of mine (to which you object) follows
along the point of Mark44's response. (Please do not take this to mean that I am so bold
as to speak for him: I'm pointing out my comment's origin.) I think I could make a very good argument for the work not being, in strict terms, a derivation.

But, as I noted (and you seem to have ignored) I stated that the observation and attack proposed by Gerenuk was clever, and never stated it didn't work.

 

[statdad]

Mark44: true; clearly it works in this case because the original expression is positive, but your concern should have been noted.

yik-boh, if I may ask, what was the exact wording of the question (if you have it).

 

[Gerenuk]

No it doesn't. If you square the expression you will get a new expression with a different value. The goal here is to rewrite the original expression in a simpler form that is equal to the original expression.

I'm sorry if you don't see it, but I do. That is the point of mathematical competitions that people are equipped with many different tricks that give them an advantage.

 

[Mark44]

Yes, I see what you did, but my point is that you have to be careful when applying operations that aren't reversible when simplifying an expression. Squaring an expression is not generally a reversible operation. As statdad pointed out, in this case the original expression was positive, so squaring and then taking the square root is valid in this case.

 

[Gerenuk]

Oh I see, sorry. You are right.
I was assuming that within the 20sec one wouldn't consider all special cases of mathematics. As a physics I also don't care about issues like convergence (most of the time)

 

[statdad]

Gerenuk, your final comment reminds me of a saying from one of my old professors. It was the first "mathematical statistics" course, senior level, so pre-measure theory. Each time he would move a derivative, or limit, through an integral or infinite series, he would preface it by saying "Now we'll pretend we're physicists".
No slight intended, then or now, of course: his meaning was that, with the toolset we had, as long as the procedure gave the correct result, we should be happy (and leave the math justification for later).

 

[trambolin]

That's a little rash, isn't it? The comment of mine (to which you object) follows
along the point of Mark44's response. (Please do not take this to mean that I am so bold
as to speak for him: I'm pointing out my comment's origin.) I think I could make a very good argument for the work not being, in strict terms, a derivation.

But, as I noted (and you seem to have ignored) I stated that the observation and attack proposed by Gerenuk was clever, and never stated it didn't work.

Sorry, if it sounded as a mean argument but I don't understand why philosophy comes in over a shortcut. Everything is positive but it does not matter if they were not if you do the book keeping. Since square root is a well-defined 'function'. There is no reversing whatsoever. Besides the quantity is not a variable, it is a number, so you cannot just introduce extraneous solutions.

 

[Mark44]

Sorry, if it sounded as a mean argument but I don't understand why philosophy comes in over a shortcut. Everything is positive but it does not matter if they were not if you do the book keeping. Since square root is a well-defined 'function'. There is no reversing whatsoever. Besides the quantity is not a variable, it is a number, so you cannot just introduce extraneous solutions.

My point was not about the square root being a well-defined function, but about other operations being one-to-one functions; i.e., reversible functions. Also, it doesn't matter whether we're talking about variables or numbers. After all, variables are numbers that we just don't happen to know at the moment.

Here's a very simple example

1. -2
2. (-2)² - Square it
3. 4 - Result after squaring
4. 2 - Take the square root

Notice that we didn't end up with the same number we started with. This is because the squaring operation is not always reversible.

Here's a slightly more complicated example

1. 5pi/6 (in radians)
2. sin(5pi/6) - Take the sine of the number
3. .5 - The sine of the number
4. sin^-1(.5) - Take the inverse sine of the number
5. pi/6 - Result

Notice that again we didn't end up with what we started. In this case the sine function is not 1-to-1 and therefore not always reversible.

 

[trambolin]

I still don't understand why you are looking from this angle. If you can read a little bit more carefully, and as I mentioned before, these are numbers and we can do the book-keeping of the signs, of course if you blindly square it and take it back you get different numbers. But we are not computers we know what we started with. There is not a single issue of confusion, I think this is a waste of time to discuss this.

 

[Jame]

Both numerator and denominator are real positive numbers, i.e. so will the quotient be and therefore we know to take the positive square root.

 

[statdad]

My point was a little different. If you are asked to simplify something, numerical or not, that means you are to go through a series of steps, each time replacing the previous expression with a new one that represents the same quantity but is different in appearance (think of rationalizing the denominator as a simple case).

If you merely need to find the value of an expression (like the current problem) then
any method of attack is fair game (squaring the expression, as here).

 

[yik-boh]

I'm sorry guys if I started a little debate here just for a simple problem. That wasn't my intention.

So the right method was squaring the whole expression first then just be observant of the factoring methods like DTS, and the like so you would be able to simplify fast. Am I right? :)

 

[statdad]

yik-boh - you have nothing to feel sorry for - you did nothing more than ask an interesting question.

Concerning your question in your most recent post: given the information we have about the problem, I would say "Yes"

 

[yik-boh]

Thanks for the help guys. I really appreciate it.