Can you use induction from $n$ to $n^2$ to prove a statement is true for all non-negative real numbers? Suppose we have a statement, and we want to prove is true for all $\mathbb{R}^{+}$. Assume the base case is "the statement can be proven true for any sufficiently small values". So now can we prove the statement by having the induction step be from $n$ to $n^2$?
Thank you!

Edit:
The original statement is one of my homework questions, so please do not write out the whole proof directly. It looks like they are trying to make the induction step from $n$ to $n^2$, but I am wondering about the validity of the proof.

The recursion $T(n)$ is: $T(n) = \sqrt{n}T(\sqrt{n}) + n$. Assuming that $T(n)$ is constant for sufficiently small $n$, show by induction that $T(n) = \theta(nlog_{2}log_{2}n)$

 A: The homework question you quote uses $n$, and so I wouldn't interpret it as being abut real numbers -- especially in computational complexity $n$ is usually taken to implicitly range over the natural numbers.
The induction principle you suggest is certainly doesn't work for the reals, or even for the positive reals. If it did, you could use it to prove that every positive real is less than $1$ (which is known not to be the case), for example.
You could change the base case to be "the property is true  for $[1-\varepsilon,1+\varepsilon]$ for some $\varepsilon>0$", and then you would conclude the property is true for all strictly positive reals.

In the actual homework question, I think you need to do something like interpret $T(\sqrt n)$ as $T(\lfloor \sqrt n\rfloor)$. Most probably the problem author imagined that "sufficiently small $n$" includes large enough values that the rounding doesn't upset things anyway.
You can then make do with long induction on the naturals.
A: Well, induction can be used for the set of natural numbers and more generally any set which is well-founded.
But the set of non-negative rational numbers or reals under the standard ordering is not well-founded (as e.g. the subset of positive rationals or reals has no minimum).
So the answer is no.
