# “Why do I always get 1 when I keep hitting the square root button on my calculator?”

I asked myself this question when I was a young boy playing around with the calculator.

Today, I think I know the answer, but I'm not sure whether I'd be able to explain it to a child or layman playing around with a calculator.

Hence I'm interested in answers suitable for a person that, say, knows what the square root of a square number is, but doesn't know about sequences, functions, convergence and the like.

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If you think this question should be CW, vote up this comment and I'll change it. – Rasmus Aug 25 '10 at 16:02
The technical way of putting it is that 1 is a "fixed point" of the square root function. – J. M. Aug 25 '10 at 16:08
@J. Mangaldan: it is not enough to say that 1 is a fixed point; 1 also needs to be attractive rather than repellent. – Qiaochu Yuan Aug 25 '10 at 16:14
(For example, 0 is also a fixed point, but on [0, infty) it is repellent.) – Qiaochu Yuan Aug 25 '10 at 16:15
You need to add that it's the unique (attractive) fixed point on (0, Infinity) and that (0, Infinity) lies within (actually is) its domain of attraction. – whuber Aug 25 '10 at 19:23

Here is how I had justified it to myself when I was a kid (I was convinced :-))

If $x > 1$ then $\sqrt x > 1$ and $x > \sqrt x$.

So we keep reducing the number while still being $> 1$.

Also, we cannot end up at a number $>1$ as then taking the square root would reduce it.

If we do end up at $1$, we stay there. Since the calculator has a limited precision, we end up at $1$, and pretty quickly.

Of course, the assumption here is that we do end up somewhere :-) (Which seemed justified by the fact that the first non-zero digit after the decimal point always seemed to reduce)

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So, how did you manage to convince yourself that taking square roots of fractions repeatedly has the same behavior? (Just kidding, you replace > with < of course). – J. M. Aug 25 '10 at 16:24
@J. Mangaldan: By considering $\frac{1}{x}$. – Aryabhata Aug 25 '10 at 16:30
You were a pretty smart kid! Up to the statement that any decreasing bounded below sequence has a limit, that is a perfectly good argument. And that statement more or less comes down to how you define real numbers in the first place... – David Speyer Aug 25 '10 at 18:18
@David: Thanks! Unfortunately, there has only been a decline :-( :-) You are right, it can be made rigorous. I suppose it comes from the axiom of existence of a supremum for subsets of R which are bounded above (or infimum). – Aryabhata Aug 25 '10 at 18:38

Make a picture of the usual spiral converging to the fixed point, in the style of

Where does this little obsession aparent on this site with 'layman' explanations? :P

Later. To make the picture it is better to use the cosine (so you do not even have to make the picture because that is what's in the wikipedia page) mainly because the iteration is seen more clearly: for the square root, the sequence converges boringly and fast to be interesting.

PS. You can draw this kind of pictures, assuming you have acces to Mathematica, with the following code:

f[t_] := Cos[t];
g = Plot[{f[t], t}, {t, -\[Pi]/2, \[Pi]/2},
AspectRatio -> 1, PlotStyle -> {Thick}, Epilog ->
Module[{pts = NestList[f, .1, 10]},
{
Thin, PointSize[0.015],
Line@ Flatten[Map[{#, {#[[2]], #[[2]]}} &, Partition[pts, 2, 1], 1], 1],
Red, Point[Partition[pts, 2, 1]]
}
]
]

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There's a picture... therefore, +1. – J. M. Aug 25 '10 at 16:15
I was even thinking about creating a layman tag. :) – Rasmus Aug 25 '10 at 16:19
Admittedly, the question about turning your clothes right-side-out was way wittier. – Rasmus Aug 25 '10 at 16:32
Notice that this picture is not of the correct function. The question was about square root, not cosine! – David Speyer Aug 25 '10 at 18:19
@David, sure (and hence the "in the style of"), but the one for the square root is too boring, and the cosine also has a key in most calculators! – Mariano Suárez-Alvarez Aug 25 '10 at 18:21

If they know about logarithms, try this:

• taking repeated square roots of a positive real number is the same as repeatedly dividing its logarithm by 2,
• repeatedly dividing something by 2 gets you to 0, and
• the exponential of 0 is 1.
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... and wave hands about continuity :) – Mariano Suárez-Alvarez Aug 25 '10 at 16:30
Sure. But it is easy to convince someone with a calculator that the exponential of something close to 0 is close to 1. – Qiaochu Yuan Aug 25 '10 at 16:40
I'd have said raising something to 0 is the same as dividing a number by itself... but that works too. – J. M. Aug 25 '10 at 16:42
@Mariano: but that's certainly far less hand-waving than your "proof by picture". All that hand-waving - esp combined with the spiraling - is RSI exacerbating merely thinking about it! – Bill Dubuque Aug 25 '10 at 21:41

Dualizing reduces it to: repeatedly squaring $1+\epsilon$ goes to $\infty$. But that's clear by the Binomial theorem.

Namely: given any positive $\;\epsilon, \;$ we can force $\;\;\; 2^n\epsilon > x \;\;$ by choosing $\: n \:$ large enough.

Therefore we have that $\;\;\; (1+\epsilon)^{2^n} \ge 1+2^n \epsilon > x \ge 1\quad$ by the Binomial Theorem.

So, taking $\;2^n\:$'th roots: $\;\;\; 1+\epsilon \;\ge\; x^{1/2^n} \;\ge\; 1. \;\;$ Hence $\; x^{1/2^n}\to 1\;$ as $\;n\to\infty$

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Nice! You don't even need the Binomial Theorem; you only need to demonstrate that (1+e)^2 > 1 + 2e, which is obvious. – whuber Aug 25 '10 at 20:26
Indeed, one could trade off the Binomial theorem for iterating that inequality. But probably anyone capable of understanding this already knows the Binomial Theorem. – Bill Dubuque Aug 25 '10 at 20:43

A nice series of solutions has already been given. It's hard to improve on any of them. But because this is an elementary question, it deserves an elementary answer. (Although "elementary" is subjective, being able to punch buttons on a calculator requires little mathematical sophistication, so we should attempt to minimize the formality and maximize the intuition in the explanation.)

When you begin with a value x on the calculator with x > 1, note that its square root lies between 1 and the midpoint of the interval [1, x].

(Proofs: The midpoint is (x+1)/2. Because ((x+1)/2)^2 - x = ((x-1)/2)^2 >= 0, the midpoint is no greater than the root of x. Non-algebraic version: draw graphs of the relations x = y^2 and y = (1+x)/2 on the same plot and note that the latter never falls below the former.)

It follows immediately (by induction if you want to be formal) that the iterated root, although never falling below 1, decreases faster than the iterated average with 1, which obviously (on geometric or arithmetic grounds) converges to 1.

The case of 0 < x < 1 is reduced to the case 1/x > 1 because Sqrt(1/x) = 1/Sqrt(x).

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I think a better question is to analyze how the roots of a number X approach 1, depending on X. I.e. xa ~ 1 + a(lnX). Not only does this introduce the log, it gives a natural derivation of e which is easy to remember. I never understood what was "natural" about the natural log until I was an undergraduate, and I think the real value of this sequence is the way it leads so memorably and naturally to the derivation of e.

It's easy to see why you arrive at one from repeatedly taking square roots (as moron answered), since the sequence is strictly decreasing and bounded below by 1. I think a motivated person of almost any age can understand that, as long as it's explained slowly and clearly.

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if(x == 0): sqrt(x) == 0
if(0 < x < 1): x < sqrt(x) < 1
if(x == 1): sqrt(x) == 1
if(1 < x ): 1 < sqrt(x) < x


Except for the cases x == 0 and x == 1, x != sqrt(x) so it must move, and it must move towards 1, so it will approach 1 to any accuracy you choose.

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There's a logical gap here. It's consistent with all your assertions, for example, for iterated square roots of x > 2 to converge down to 2 and for iterated roots of x, 1 < x <= 2, to converge down to 1. At a minimum you need to invoke continuity. – whuber Aug 26 '10 at 1:01
If you add something about how far towards 1 it moves that would solve the problem, but I'm to lazy to figure out how to add that. – BCS Aug 26 '10 at 1:09