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

Solution 1:

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)

Solution 2:

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

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

NB. The picture comes from the Wikipedia page on fixed points.

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]]
     }
   ]
]

Solution 3:

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.

Solution 4:

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 $\;n\to\infty$

Solution 5:

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).