Beginning Analysis, need help with proof.?

Question

I need help proving if this is true or false:
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For each ε>0, there exists positive integers m and n such that:

0 < n^(1/2) - m^(1/2) - π < ε
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I've messed around with it and I think it's true, but I'm stuck on how to prove it.

Answer 1

Pick an integer n such that 1/(2n)= pi.

Now sqrt(n^2+k)-sqrt(n^2+k-1) = 1/[sqrt(n^2+k)+sqrt(n^2+k-1)]

which is <= 1/[sqrt(n^2+0)+sqrt(n^2+0)] = 1/(2n) which is

less than epsilon/2. Since sqrt(n^2+k)-sqrt(n^2+k-1) =

[sqrt(n^2+k)-n] - [sqrt(n^2+k-1)-n], and the first quantity in brackets is >= pi and the second one is < pi (since k is least) and finally each is < epsilon/2, it follows that

pi<=[sqrt(n^2+k)-n]
(We are taking two steps of size < epsilon/2 and the start is < pi and after the first step we are past pi.)

Of course the <= must be < since pi is not algebraic.

So the first integer is n^2 +k and the second one is n^2.

Answer 2

For any positive m, the real number:
x = (sqrt(m) + pi)^2
satisfies
sqrt(x) - sqrt(m) = pi

Let n be the integer ceil(x), the least integer greater than or equal to x. So:
m >= x
sqrt(m) >= sqrt(x) = sqrt(n) + pi
sqrt(m) - sqrt(n) - pi >= 0

Now show that the above value is less than sqrt(m) - sqrt(m-1). Finally show that lim[m->inf](sqrt(m+1)-sqrt(m)) is zero and you're done.

For the last part, note that sqrt(m+1)-sqrt(m) is the integral from m to m+1 of dt/(2*sqrt(t)). ["The integral of the derivative of sqrt(t)"] You know the maximum value of the integrand, so you can give an upper bound on the integral which will vanish as m goes to infinity.

Thanks for the problem!