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I have always known that $a^n=a*a*a*.....$(n times)

Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?

I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$.

But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?

Any help is highly appreciated.

Empty
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Soham
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  • This is only a comment: indeed, this kind of definition for exponential only applies when $n$ is a positive integer. So the idea will fail for $e^x$ such that $x \notin \mathbb{Z}_+$. And the solution you see is the most nature one: it is the only way to preserve the properties of exponential function unchanged while extending its definition. – Asydot Sep 24 '15 at 16:35

5 Answers5

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That's the intuition; but the powers are properly defined by recursion from $$ a^0=1,\quad a^{n+1}=a\cdot a^n $$ so $a^0=1$ by definition. It is a sound definition, because it agrees with the property $a^{m+n}=a^m\cdot a^n$ for any natural $m$ and $n$.

Think to what you do when you have a heap of candies to count. You start from zero and take one candy at a time, uttering the corresponding number: one, two, three, and so on.

Similarly, if you have to know how many candies are in a bunch of heaps, you can count each heap and write down the number. Then you start from zero, add the first number, then the second and so on (at this stage you already know how to perform symbolic sums).

For multiplication it's the same, but you start from one! So $a^0=1$, then $a^1=a\cdot 1$, $a^2=a\cdot a^1$, and so on, each time multiplying by $a$ until you arrive at $n$ and you have your $a^n$.

egreg
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  • I have two things to say-1.What was going in the mind of the man who gave the definition?2.I want to know the exact meaning of $a^0$. – Soham Sep 24 '15 at 16:21
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    @tatan Historically $a^n$ was defined, for $n>1$, as the product of $n$ factors equal to $a$; but then the definition got extended to $a^1=a$ and $a^0=1$, because they agree with the property $a^{m+n}=a^m\cdot a^n$. Nobody woke up one day with the recursive definition: this came when people realized that the “naïve” one needed to be made rigorous. Mathematics needs conventions; since $a^n$ reveals useful, why not extending the definition in a way that's consistent with the already known properties? – egreg Sep 24 '15 at 16:25
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It's defined that way (except, usually, for $a = 0$) because it's most consistent to do so. The empty product is defined to be $1$ because $1$ is the multiplicative identity, much as the empty sum is defined to be $0$ because $0$ is the additive identity. Both of these definitions allow for the pattern that arises from successive multiplication to be "extended backward" to a product (or sum) with zero terms.

The extension of exponents to zero, to negative numbers, to the rationals, to the reals, and to complex numbers, in each case continues a pattern identified in the previous, "smaller" domain. Those patterns are useful; that is why they are defined that way. Nothing stops you from defining them differently—nothing, that is, except that they generally are less useful that way.

Brian Tung
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You can see that in this way:

$$a^0 = a^{m - m}$$

for every value of $m$. Using the properties of powers we have:

$$a^{m-m} = \frac{a^m}{a^m} = 1$$

Because the two terms are identical so they are canceled. So

$$a^0 = 1$$

Enrico M.
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  • $a^{m-n}=\frac{a^m}{a^n}$ holds only if $m\not=n$. – Empty Sep 24 '15 at 16:17
  • @S.Panja-1729-Really?Can you give a link where this is stated? – Soham Sep 24 '15 at 16:18
  • @S.Panja-1729 Why?? There is no meaning for that.. It holds also for $m = n$. – Enrico M. Sep 24 '15 at 16:19
  • Check any H.S. standard book.... – Empty Sep 24 '15 at 16:22
  • $a^0=1$ is a definition...See the answer of ' egreg '. – Empty Sep 24 '15 at 16:24
  • @S.Panja-1729: If your point is that the argument is circular because the demonstration depends on a property that is a result of the thing Henry's trying to demonstrate, you might have said so! :-) At any rate, OP might clarify whether what is desired is a rigorous demonstration, or an intuitive justification. – Brian Tung Sep 24 '15 at 16:28
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The definition you give only applies when n is an integer greater than zero. It does not apply to n = 0 or negative n because it doesn't make sense to talk about multiplying together zero a's or a negative number of a's.

However, using the principle that you mention, we can extend this definition to zero and negative numbers. This allows us to give a meaning to a to the power of zero or negative exponents. That meaning is no longer "a certain number of a's multiplied together" but something that works consistently with the definition we already have for positive n.

Paul G
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You are correct in that $a$ is repeated zero times.

$a^n = a*a*a*... = 1 * a*a*a*...$

And so $a^0 = 1$ when $a$ is repeated zero times

Russ H
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