5

What is wrong in this proof. It seems correct to me but still doesn't make proper sense.

$$\sqrt{\cdots\sqrt{\sqrt{\sqrt{5}}}}=5^{1/\infty}=5^0=1$$

EDIT

So does this mean that

$5^{1/\infty} = 1$

$(5^{1/\infty})^\infty = 1^\infty$

But according to me, $1^\infty$ is indeterminate.

user3932000
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Arulx Z
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    What those periods in the first square root mean?? – Timbuc Apr 18 '15 at 13:59
  • It means square rooting infinitely – Arulx Z Apr 18 '15 at 13:59
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    $1/\infty$ is most definitely not something you can use in anything resembling a mathematical proof... – theage Apr 18 '15 at 14:00
  • Er...square root of what ? – Timbuc Apr 18 '15 at 14:00
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    Before asking for a correct proof, you may need to ask for a correct formalisation of the problem. –  Apr 18 '15 at 14:12
  • Let $a=\sqrt{...\sqrt{\sqrt{\sqrt{5}}}}$, then $a(a-1)=0$, so $a=0$ or $a=1$, thus this "quantity" does not exist?? What? am I hallucinating? What prevents $a$ from being zero? – Math137 Apr 18 '15 at 14:25
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    (Apart from the fact that infinite expressions are IMO delirious in general unless of a special kind for which meaning is given to them in special way, this one is particularly insidious:) Since there is both an innermost and an outermost square root sign (both visible), there is a serious question about what the dots mean if they suggest infinite repetition (and if it is finite we clearly need to know how many). Are they indexed, from the inside outwards by $\omega+1$ (so after taking the limit take one more root), or some other cardinal, or maybe they are not well-ordered in the first place? – Marc van Leeuwen Apr 18 '15 at 14:47
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    The fundamental error is in using $\infty$ as a number. – Thomas Andrews Apr 18 '15 at 21:02
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    You don't have any error except a lack of rigor; you're right that as you take more roots of 5 the result approaches 1. – Elliot Gorokhovsky Apr 18 '15 at 22:38
  • Treating $\frac{1}{\infty}$ as a number, or taking the $\infty$th power of both sides, is not something that math approves of. Or allows, even. – user3932000 Apr 19 '15 at 03:40
  • As for your last observation, note that $(5^{\frac{1}{\infty}})^\infty$ is indeterminate. – N. S. Apr 19 '15 at 04:10

8 Answers8

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What seems incorrect is because the language seems informal. Formally, one would write $$ x_0=5, \quad x_n = \sqrt{x_{n-1}} $$ to describe the iterations of the square roots. One computes easily that $$ x_n = x_0^{1/{2^n}} $$ and since $x_0^{1/n} \to 1$ for all $x_0 > 0$, so does $x_n$. (You can prove this by showing that the logarithm of the sequence goes to $0$ ; this is trivial.)

Hope that helps,

Patrick Da Silva
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  • Apparently the OP meant infinite square roots at the beginning, otherwise his use of infinity is even more odd. – Timbuc Apr 18 '15 at 14:03
  • What do you mean "at the beginning", that makes no sense to me – Patrick Da Silva Apr 18 '15 at 14:04
  • @Pa Neither to me, yet how else can one explain the infinity in the OP's post? Even more interesting, that'd be infinite square roots... of what ? – Timbuc Apr 18 '15 at 14:05
  • @Timbuc I appreciate your discussion. I wonder if it would have made any difference if the dots were placed more to the right or "at the end". We may start a discussion about use of dots in math and when it is proper. But I think it was said enough already, thank you. – Mirko Apr 18 '15 at 14:37
  • @Mirko If the OP has used $;n;$ anywhere, somewhere , maybe I, and perhaps others, could have deduced easily that he meant the iteration of the square roots $;n;$ times ...but he didn't. In fact, from first sight, I even though that he would be working with some $;p$-adic ring or field. – Timbuc Apr 18 '15 at 14:39
  • I would write more formally what the OP, apparently, meant: $$a_1=\sqrt5;,;;a_2=\sqrt{\sqrt5},,\ldots,a_n=\overbrace{\sqrt{\sqrt{\ldots \sqrt5}}}^{n;\text{square roots}}$$ and then also use somehow $;n;$ in my expression for limit. That way one could hardly get misled by the expression. – Timbuc Apr 18 '15 at 14:42
  • @Timbuc Now I get it, you have seen too much, that is why it looked ambiguous :) I am blissfully ignorant about the possible (mis)use of dots in the context of $p$-adics. BTW in your previous comment you guessed the intended meaning of the question. Did you do this on your own (and if so, how)? Or did you just copied what the others did? – Mirko Apr 18 '15 at 14:45
  • @Mirko Jeje...don't worry about that: it will come in time (or not...). It is just a rather useful, in certain aspects, way of writing a $;p$-adic number in a way ressembling the infinite expression of real numbers with decimal expansion...bit in the $;p$-adic case it end to the right, with the periods on the left, unlike the real, and usual, case. – Timbuc Apr 18 '15 at 14:48
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    @Timbuc : What else could he have meant? I don't see any other plausible option. – Patrick Da Silva Apr 18 '15 at 16:27
  • @PatrickDaSilva I already wrote elsewhere: infinite square roots, which means god knows what, but which seemed to be so because of the OP's use of that infinite sign. "Plausible" is the key word here: it really didn't look like a plausible/logical/makes-sense question, and this is reflected by several comments/answers here. Read in particular Mark's comment below the question. – Timbuc Apr 18 '15 at 20:02
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    @Timbuc : No offense, but in my opinion it is just arrogant to tell the OP is not being clear when he probably doesn't know how to formulate his question with a 100% clear mathematical precision. It is more user-friendly to guess his answer, make oneself clear, and then if by any chance the OP did not mean what we guessed, discuss with him and figure out what the actual question was. I understand your mathematical worries, but we're not here to pass an exam and be 100% right all the time but rather to help people and learn all together! – Patrick Da Silva Apr 18 '15 at 22:35
  • @PatrickDaSilva I strongly disagree with you. Imo, mathematics students must be well aware of the fact that they must try to be as precise and clear as possible, since sometimes even little imprecisions can lead to huge misunderstandings/mistakes. In this case I honestly didn't know what the OP meant, and reading what he wrote ($;5^{1/\infty}$) made it even more confusing as I was sure he meant infinite square roots at the beginning ...and I'm not the only one who seems to have thougt that way. I think it truly is arrogant to believe students being corrected are going to be offended. – Timbuc Apr 18 '15 at 22:48
  • @Timbuc : I'm not going against that! I'm saying that in the helping process, instead of just saying "what are you saying?" using bold letters and stuff, it might be nicer to suggest what the OP might mean and start a discussion with him rather than simply tell the OP that he is confused. If he wasn't, he wouldn't have asked a question in the first place. – Patrick Da Silva Apr 18 '15 at 23:09
  • @PatrickDaSilva Perhaps I missed something, but I can't remember now when did I write the OP in bold letter and etc. "what are you saying?" BTW, the OP's first comment, below mine, explicitly says "it means square rooting infinitely"...I think both he and I, and perhaps others, already learned a good deal of all these discussions, and maybe next time he'll be more careful with the way he writes down things in mathematics...just as many, perhaps most, of us learned in the past. – Timbuc Apr 18 '15 at 23:12
  • @Timbuc : I meant the two separately, about the bold. To prevent looking arrogant I usually reserve the use of bold only to warn a reader that a detail of my comment is very important to read. If I just want to attract attention on some point, I usually use italic ; it looks less aggressive but still attracts attention, whereas bold tends to mean "danger".

    Of course many of us learned from this. Have fun on the website.

     – Patrick Da Silva Apr 19 '15 at 00:16
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I guess you mean this: $$\lim_{n\to\infty}5^{1/2^n}=1$$

ajotatxe
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  • Perhaps he did, yet that weird use of periods in that root and then the weirder use of infinity makes this doubtful. – Timbuc Apr 18 '15 at 14:02
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    @Timbuc both the dots and reciprocal of infinity make sense and may be properly formalized as several answers did instead of making fun for not expressing the obvious meaning in the way that is thought proper – Mirko Apr 18 '15 at 14:08
  • @Mirko What's the meaning of those periods at the beginning of the square root ? Can you guess? I can't. Also, I can't guess who's making fun of whom or what for who knows what. – Timbuc Apr 18 '15 at 14:10
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    @Timbuc they are not at the beginning they are at the end. Look from the inside to the outside. Read from right to left as necessary (and as it is usually done for function composition). How did all those people that posted answers interpreted the question in a similar way? – Mirko Apr 18 '15 at 14:15
  • @Mirko Why would anyone read from right to left? That is a mathematical expression and as such, and unless otherwise stated, it must be read from left to right, and thus it begins with periods in the square root. What other people understood is something I can't answer. – Timbuc Apr 18 '15 at 14:18
  • @Timbuc because everyone reads function composition from right to left, you know very well. But you are right. – Mirko Apr 18 '15 at 14:23
  • @Mirko I beg to difer greatly. What is closer to the truth, perhaps, is that in order to find out what the composition is many people usually read the comoposition from right to left. That could be true. – Timbuc Apr 18 '15 at 14:25
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Let $x=\sqrt{\sqrt{...\sqrt{5}}}$. $$x^2 = \sqrt{\sqrt{...\sqrt{5}}} = x$$ $$x = 0 \, or \, 1 $$

Verify:$$\sqrt{5}=2.24$$ $$\sqrt{\sqrt{5}}=1.50$$ $$...$$ From the pattern we can see $x$ tends to $1$, so $x = 0$ is rejected.

James Pak
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  • You have a good idea here. Could you explain it more clearly? – Rory Daulton Apr 18 '15 at 15:02
  • I understand that: I mean you should add your clarification to your answer. The other unclear part is your parenthetical comment. That should be expanded after you state "$x=0$ or $x=1$". The terseness seems confusing to me. --Why did you delete your comment? The material was good, I simply recommend that you put that comment into your main answer. – Rory Daulton Apr 18 '15 at 15:13
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    This is a good way to derive possible values for $x$, but it is not in itself a proof that any of the values actually work. (Try solving $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ [what? the same value for $x$ in each?].) – Kyle Miller Apr 19 '15 at 01:45
  • @KyleMiller Nice counterexample. Maybe using limit identities is enough to find x, just like other answers. – James Pak Apr 20 '15 at 14:43
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$\sqrt{...\sqrt{\sqrt{\sqrt{5}}}}=\lim x_n$, where $x_n=5^{(1/2^n)} (n\in\mathbb{N})$

Blind
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$$n=\sqrt{...\sqrt{\sqrt{\sqrt{5}}}}$$

Since there's an infinite cascade of roots, we can add 1 (or as many as we want):

$$n=\sqrt{...\sqrt{\sqrt{\sqrt{\sqrt{5}}}}}$$

Replacing by the $n$ from the first equation:

$$n=\sqrt{n}$$

or

$$n^2=n$$

Since square roots of a number greater than 1 never can become less than 1, we can discard the solution $n=0$, so we can divide both sides by n, and

$$n=1$$

stevenvh
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    That argumentation only works if you already know that the sequence converges. Otherwise you'll find that $1+2+4+8+\ldots=-1$ because it obviously fulfils the equation $2x+1=x$, and $-1$ is the only solution to that equation. – celtschk Apr 18 '15 at 16:35
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    @celtschk: If we agree that square roots are positive , the fact that extraction of a square root for a number bigger than 1 yields a smaller number, assures we are safer grounds: a monotonically decreasing sequence of positive numbers. – P Vanchinathan Apr 19 '15 at 04:27
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    @PVanchinathan: Yes, that's a simple proof that the sequence converges. But the answer is missing such a proof. – celtschk Apr 19 '15 at 08:42
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$$\lim_{n\to\infty}x^{1/2^n}=1$$ because the exponent approaches zero

Jimmy360
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  • The exponent approaching zero is not sufficient. Think of $0^0$. – stevenvh Apr 18 '15 at 15:23
  • @stevenvh The base is fixed! Your comment doesn't apply. – egreg Apr 18 '15 at 15:50
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    @egreg - I know it doesn't, but saying it does not apply doesn't count. For a rigorous proof he has to show it, otherwise his proof is incomplete, as you no doubt know. – stevenvh Apr 18 '15 at 16:22
  • @stevenvh Actually in a lot of contexts/mathematics, $0^0=1$ is perfectly fine :). (Yes, it's a point of contention) – Alan Apr 19 '15 at 04:10
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Note that $\lim_{n\to \infty}x_n^{y_n}=(\lim_{n\to \infty}x_n)^{\lim_{n\to \infty}y_n}$

Paul
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Using $1/\infty$ doesn't make sense, as already stated, because ∞ is not a number, it is a concept. I believe this is the closest thing to what you're really asking.

$$\lim_{n\to\infty}5^{1/n}=1$$

This is true. n never actually reaches infinity, but we find what happens as it gets closer and closer. And it does, indeed, approach 1.

This is because as n gets larger and larger, 1/n gets smaller and smaller. It is known as infinitesimal, essentially zero. It's a little more complicated than just zero (if you're interested, learn calculus, specifically limits and infinitesimals) but for the purposes of this, it's as good as zero. And because $5^{0}=1$, the limit shown above is equal to 1.

Edit: I'd also like to add that using infinity as a number is a bad idea.

$$1+\infty = \infty$$ $$2+\infty = \infty$$ $$1+\infty = 2+\infty$$ $$1 = 2$$

Daffy
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