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I am trying to understand why $1^{x}=1$ for any $x\in\mathbb{R}$

Is it OK to write $1^{x}$? As the base 1 should not equal 1 for $1^{x}$ to be an exponential function?

Is $1^{x}=1$ just because it is defined to be so?

If possible please refer me to a book or article that discusses this topic.

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You don't have to specifically define $1^x$ to be something. It follows from the definition of exponentiation in general.

The story is roughly as follows: first there was exponentiation with exponents being natural numbers. That is just repeated multiplication of the base with itself. Then people noticed that this exponentiation obeys the rule \begin{eqnarray} a^{n+m} = a^na^m. \end{eqnarray} This suggests that if we want to define $a^0$, then we want it to satisfy $a^m = a^{m+0} = a^ma^0$, so that it is natural to define $a^0 = 1$. Now, we can use the same rule to extend this operation to all integral exponents, i.e. positive and negative. Since we want $a^na^{-n} = a^{n-n} = a^0 = 1$, we are forced to define $a^{-n} = \frac{1}{a^n}$.

So, this way you have defined exponentiation with arbitrary integral exponents. Now, going back to the original exponentiation with natural numbers, you also notice that \[(a^n)^m = a^{nm} \] and $a^1 = a$. Applying the same considerations as above, you are forced to define $a^{\frac{1}{n}} = \sqrt[n]{a}$. This way, you have now extended exponentiation to all rational exponents. Note that all these rules imply that $1^x = 1$ for $x\in \mathbb{Q}$. Now, you extend all this to all real exponents using continuity, so you still have $1^x = 1\;\forall x\in \mathbb{R}$.

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Nicely explained. But it might be best to stop at real exponentiation of positive bases. To fully explain complex exponents, you need to choose a branch of log or view complex exponentiation as multiple-valued; it's a different story from real exponentiation. See – Jonas Kibelbek Dec 10 '10 at 10:38
@Jonas I reread what I wrote and the complex part really reads ambiguously (as though one could just extend exponentiation as a single-valued continuous function). I decided to follow your advice and delete that part altogether. – Alex B. Dec 10 '10 at 13:50
It's all good. I mostly wanted to link to the other thread, for the rest of the story. I found it very instructive and interesting to read the different perspectives on how to think about 1^i. – Jonas Kibelbek Dec 10 '10 at 14:28

Once you move past elementary topics, definitions become much more fundamental in mathematics. So, in a formal sense, you're right that the reason $1^x = 1$ for all $x \in \mathbb{R}$ is that the definition of $1^x$ makes this so. The emphasis on definitions comes from the use of mathematical proofs; the only way to make a rigorous proof about exponentiation is to start with a rigorous definition. So, in a sense, all formal mathematical propositions are true because the definitions have been chosen to make them true.

However, we have a clear motivation behind exponentiation, and if the definition of real number exponentiation did not make $1^x = 1$ for all real $x$, then the definition would have been changed. We don't make up mathematical definitions at random - they are motivated by our informal ideas about the objects we are studying.

The role of this motivation can be seen more clearly by considering complex exponentiation. Unlike natural number exponentiation, complex number exponentiation is not based on repeated multiplication; it's based on logarithms and the function $\exp(z)$. So the definition of complex exponentiation does not imply that $1^i = 1$, and mathematicians are OK with that. There are several possible values for $1^i$, only one of which is $1$. I explained this in this answer.

One thing that is often confusing at first is that there are really several different exponentiation functions, with different domains, all of which are denoted with the notation $x^y$.

Finally, you asked whether $f(x) = 1^x$, as a function from $\mathbb{R}$ to itself, is an exponential function. Many calculus books seem to include a special clause in their definitions that makes this not be an exponential function. However, things would work just as well if you did call it an exponential function. It's just a question of terminology. The only downside to calling $1^x$ an exponential function is that, when stating some results, you might have to add an exception to get rid of $1^x$. Instead of saying "Every exponential function" you would say "Every exponential function except $1^x$". Of course, students in a class need to adopt the conventions of the class so that everyone can understand them. But if you were writing a math book alone on a desert island you could adopt whatever terminology you wanted.

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I think you will at least agree that $1^x$=$1$ if $x$ is any natural number (since this is just 1 times itself x times). We can extend this to all integer values of $x$ by using the facts $a^0$=$1$ for all non-zero $a$ and $x^{-c}$=$1\over{x^c}$. Then we have $1^x$=1 for all rational numbers $x$ by the fact $1^{b\over{c}}$ is the cth root of $1^b$ which is 1 by previous results. Finally we exdend this to all real values of $x$ by using the following definition of the exponential function from calculus.

"If $t$ is irational, then $a^t$ is defined to be the limit as n approaches infinity of $a^{t_n}$ where {$t_n$} is any sequence of rational numbers converging to $t$ (assuming this limit exists)."

In the case of $1^t$ we already have that $1^{t_n}=1$ for all rational $t_n$ so any such sequence as described above is the constant sequence, 1,1,1,..... which converges to 1.

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Note, that, in a sense, it is as non-conceptual to call $a^0=1$ or $x^{-c}=\frac{1}{x^c}$ a fact, as it is to say that $1^x=1$ is a fact. See also my answer below. – Alex B. Dec 10 '10 at 8:50

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