math problem

[Towlie]

There's a class of numbers called the square numbers, which are integers like 1, 4, 9, 16, 25, etc.

I found it stated on a few different web pages that only the square numbers have rational square roots. I didn't find a proof but I'm sure there's one out there somewhere.

Therefore, if and only if the quantity 1 + 4X is a square number, the square root of that quantity is rational, and the expression I gave in my post #4

[latex]({1 + \sqrt{1 + 4X})/2[/latex]

is rational.

 

[Thabiguy]

So for the function

f(x) = sqrt(x+sqrt(x+sqrt(x+sqrt(x+...

The problem with this is that this is not a function. The right-hand side doesn't give a way to uniquely determine the result. Consider for example x=0. The right-hand side is then sqrt(sqrt(sqrt(sqrt(... - Under the interpretation that we seem to have chosen here, there are two equally plausible "values" for such right-hand side, but a function can't have two results.

Let me elaborate on that. The ellipsis can sometimes be used as a shorthand for a certain rule. For example, you can ask: what is the value of 1+1/2+1/4+1/8+... - What you mean by that is that you are asking for the limit of a sequence whose first element is 1, second element is 1+1/2, third element is 1+1/2+1/4, and so on. You're wondering whether, if you keep doing this, the sequence will converge to a certain value. In this case, it does, and the value is 2.

That is all well, as long as the sequence is well-defined. For example, it is sufficient to know the first element of the sequence and a rule to determine subsequent elements (recursive definition).

In this case, the situation is not so clear. Here we have a certain operation,

g(x) = sqrt(a+x)

and the question is: what is the value of g(g(g(g(...))))? Or, what number do you get when you perform that operation infinitely many times?

Well, the obvious question should be: on what? We have a rule to determine subsequent elements of the sequence, but we were never given the first element. So how can we know what the result is?

We can get around that by interpreting the ellipsis as a shorthand for a different kind of problem. The implicit assumption here is that g(g(g(...))) does correspond to some number, and that the ellipsis stands for the same thing as the whole expression. But then the ellipsis, no matter where we put it, should also correspond to the same number. So the question boils down to this: for what number does g(x)=x? Or, more formally, what is the fixed point of g(x)?

For the given operation, this gives the equation x=sqrt(a+x), and the solution to that equation are any non-negative roots of the quadratic equation x²-x-a=0.

For a<-1/4, there are no solutions, for a=-1/4 there is a single solution (1/2), for -1/4<a<=0 there are two solutions (for 0, these are 0 and 1), and for every a>0 there is a single solution.

(ETA: Another way to wave off the problem of not having the first element of the sequence might be to note that for a>0, the sequence of repeated application of g(x) will converge to the same value for any first element >= -a (the fixed point is attractive). That approach, however, would break down for a<=0.)

b) why those values? That summation is just 2*(1+2+3+...+n)

Because partial sums of even numbers (2+4+6+...) are of the form x²-x, which are the numbers you're looking for (x²-x=a). Similarly, partial sums of odd numbers (1+3+5+7+...) are square numbers.

 

[Thabiguy]

I found it stated on a few different web pages that only the square numbers have rational square roots. I didn't find a proof but I'm sure there's one out there somewhere.

I really doubt there's a proof out there, because it's not true. 1/4 is not a square number, but it does have a rational square root.

But it is true that square numbers are the only integers with rational square roots (i.e. that any rational square root of an integer must itself be integer). The proof is not difficult and I'll leave it as an exercise for inquisitive readers.

 

[Towlie]

I really doubt there's a proof out there, because it's not true. 1/4 is not a square number, but it does have a rational square root.

I accidentally left out "natural numbers", meaning "integers." The question that I found is actually "Which natural numbers have rational square roots?" and the answer is "Only the square roots of the square numbers."

It's a grammatical slip-up, implying that the square roots of the square numbers have rational square roots, but the intended meaning is correct. I just misstated it.