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I wondered if using square roots to prove that $N^0=1$ is valid (where $N$ is any real number). The way I propose to do this is as follows:

We consider when $x > 0$. If we do an iterative method:

$x_2=\sqrt{x_1}$

$x_3=\sqrt{x_2}$

and so on... we get that this would tend towards 1, so $x_n=1$ as $n \rightarrow \infty $.

This is equivalent to writing $$(x)^{\frac{1}{2}*\frac{1}{2}*...*\frac{1}{2}}$$ which tends towards $(x)^0$, and we know that this tends towards 1.

Would this be valid? And how would one prove this for $x \leq 0$, and perhaps formulate it better than I have managed to?

  • What is $N$ ? A natural number? – Improve Jan 14 '17 at 10:42
  • What is your definition of $N^0$, first ? – Watson Jan 14 '17 at 10:42
  • I think this is a circular, invalid argument: in both cases , how do you know that $;x_n\xrightarrow[n\to\infty]{}1;$ ?? – DonAntonio Jan 14 '17 at 10:42
  • @Watson edited! –  Jan 14 '17 at 10:43
  • @DonAntonio by using the behaviour of square roots. –  Jan 14 '17 at 10:43
  • What's the point considering two cases and using the same argument ? –  Jan 14 '17 at 10:45
  • @Watson I'm not sure I understand what you mean. I just mean it is any real number to the power of 0. –  Jan 14 '17 at 10:45
  • @YvesDaoust I'm not fully sure how to explain my thinking, I was hoping I could have help formulating my proof more precisely. –  Jan 14 '17 at 10:47
  • Notice that if $N=0$, this claim is invalid. – projectilemotion Jan 14 '17 at 10:53
  • @Leonhard Would you mind going a little deeper into that "behaviour of square roots" thing? I think this could be the base of the probably mistake (either yours or mine, certainly) – DonAntonio Jan 14 '17 at 10:53
  • @DonAntonio This is the aspect I'm not sure how to mathematically explain. I mean something like this http://math.stackexchange.com/questions/3283/why-do-i-always-get-1-when-i-keep-hitting-the-square-root-button-on-my-calculat –  Jan 14 '17 at 10:56

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Assuming that you are using one of the inequalities

$$n\le\sqrt n\le1\text{ or }1\le\sqrt n\le n$$

to squeeze, you are indeed showing that

$$\lim_{x\to0}n^x=1,$$ if the limit exists.

But

  • this is not sufficient to prove that the limit exists (as you just use the particular exponents $x=2^{-k}$),

  • this does not "prove" $n^0=1$, which is a pure matter of convention, but proves that the function $n^x$ is continuous at $0$ when you admit that $n^0:=1$.