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It is intuitive for $2^n$, if $n$ is an integer, to exist.

How do we know that less intuitive values such as $2^\frac{1}{2}$, $2^\sqrt{2}$, $2^\pi$ etc exist?

I'd like to accept that $2^x$ is continuous, but how can we be sure of the existence of the number when $x$ is something obscure, like an irrational number?

Trogdor
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  • Because every value is smooth. There isn't a positive real that will make it jump suddenly. – Calculus Nov 09 '14 at 12:07
  • Your question and your question's title don't match: you don't want to know about continuity of $;2^x;$ but rather about how is that defined when $;x;$ is not "a nice" real number. – Timbuc Nov 09 '14 at 12:07
  • The title is matching my question.

    My question is regarding the continuity of $2^x$. But to be sure of the continuity, we must be sure that the curve doesn't have any 'holes' in it, hence why I am questioning the existence of a function value when $x$ is 'not nice'.

     – Trogdor Nov 09 '14 at 12:09

2 Answers2

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This is usually dealt with by defining $$2^x = e^{x \ln 2}$$ Such a definition works provided we ultimately have a way to define $e^x$ so we can be sure it's continuous.

One way this is done is by defining the exponential function as the the inverse of natural logarithm, which is defined as the definite integral $$\ln x = \int_1^x \frac{dt}{t}$$ This integral is known to exist and be continuous by theorems about integrals of continuous functions on closed intervals. As $\ln$ is also strictly monotonically increasing, the inverse exists and is also continuous

Simon S
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Read this answer. It's pretty good. We just can define these numbers in "most intuitive" way.

Community
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Jihad
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