Zero Exponent Rule

[mathdad]

Let a = any integer except for 0. It is very common for math textbooks to say a^0 = 1 without providing a reason.

Questions

1. Why does any integer (except for 0) raised to the zero power equal one?

2. Does this rule apply to expressions?

For example:

(x + 2)^0 = 1

You say?

 

[blamocur]

Let a = any integer except for 0. It is very common for math textbooks to say a^0 = 1 without providing a reason.

Because $a^m \cdot a^n = a^{m+n}$

 

[mathdad]

Because $a^m \cdot a^n = a^{m+n}$

Can you further explain?

 

[blamocur]

Can you further explain?

$a^m \cdot a^0 = a^m$

 

[mathdad]

$a^m \cdot a^0 = a^m$

How about a^m • a^0 = a^(m + 0) = a^m?

How about a^m • a^0 = a^m • 1 = a^m?

Yes?

 

[khansaheb]

How about a^m • a^0 = a^m • 1 = a^m?

Yes?

In the above calculation, you are assuming

a^0 = 1

But in your op you wanted the REASON for THAT assumption !!!

 

[blamocur]

How about a^m • a^0 = a^(m + 0) = a^m?

How about a^m • a^0 = a^m • 1 = a^m?

Yes?

Just divide both sides by $a^m$.
Another way to reason: $a^0 = a^m \cdot a^{-m}$.

 

[mathdad]

In the above calculation, you are assuming

a^0 = 1

But in your op you wanted the REASON for THAT assumption !!!

I accepted the textbook's definition as true.

 

[mathdad]

Just divide both sides by $a^m$.
Another way to reason: $a^0 = a^m \cdot a^{-m}$.

Very good.

 

[fresh_42]

My favorite argument is as follows: Taking a number to some power is an abbreviation for a multiplication: $$a^n=\underbrace{a\cdot a\cdot a\ldots a\cdot a}_{n-times}$$
If $n=0$ then this is an empty multiplication, i.e. something neutral. And the neutral element of multiplication is $1.$ So doing nothing equals $1.$ The same holds for additions. Empty additions are the neutral element,in that case $0.$ E.g.
$$\sum_{\substack{k=1\\ 1 \nmid k}}^n k =0.$$
The term $0^0 =1$ is usually also set being $1$ because it makes the most often occurring formulas easier to handle. $0^0=0$ is an extremely rare convention.

$$\begin{array}{lll} (a+b)^n &=\displaystyle{\sum_{k=0}^n \dfrac{n!}{k!(n-k)!}a^kb^{n-k}}\\[18pt] 0^0&=(-1+1)^0\\[18pt] &=\displaystyle{\sum_{k=0}^0 \dfrac{0!}{k!(0-k)!} (-1)^k 1^{0-k}}\\[18pt] &= \dfrac{0!}{0!(0-0)!} (-1)^0 1^{0-0}\\[18pt] &=\dfrac{1}{1\cdot 1} 1\cdot 1 \\[18pt] &=1 \end{array}$$with the convention that $0!=1$ by the empty product argument.

 

[mathdad]

My favorite argument is as follows: Taking a number to some power is an abbreviation for a multiplication: $$a^n=\underbrace{a\cdot a\cdot a\ldots a\cdot a}_{n-times}$$
If $n=0$ then this is an empty multiplication, i.e. something neutral. And the neutral element of multiplication is $1.$ So doing nothing equals $1.$ The same holds for additions. Empty additions are the neutral element,in that case $0.$ E.g.
$$\sum_{\substack{k=1\\ 1 \nmid k}}^n k =0.$$
The term $0^0 =1$ is usually also set being $1$ because it makes the most often occurring formulas easier to handle. $0^0=0$ is an extremely rare convention.

$$\begin{array}{lll} (a+b)^n &=\displaystyle{\sum_{k=0}^n \dfrac{n!}{k!(n-k)!}a^kb^{n-k}}\\[18pt] 0^0&=(-1+1)^0\\[18pt] &=\displaystyle{\sum_{k=0}^0 \dfrac{0!}{k!(0-k)!} (-1)^k 1^{0-k}}\\[18pt] &= \dfrac{0!}{0!(0-0)!} (-1)^0 1^{0-0}\\[18pt] &=\dfrac{1}{1\cdot 1} 1\cdot 1 \\[18pt] &=1 \end{array}$$with the convention that $0!=1$ by the empty product argument.

I must admit that most of what you said here is over my head. However, your reply makes great future study notes. This is what I'm looking for. No argument, no nonsense, no waste of precious time. Go straight to the math and stay on the subject at hand. Well-done fresh_42. By the way, what does your username mean?

 

[fresh_42]

By the way, what does your username mean?

Nothing, except for the 42, of course. IIRC then I invented it decades ago on a MS-platform with chat rooms that doesn't exist anymore since years. It has the advantage that it doesn't use names of any scientific or popular celebrities and that it sounds young. Well, that once was the case. Those registration pages always surprise me by "choose a username" although I know it comes. Then I'm sitting in front of the PC like a cow in front of a tv and don't know what to type and remember it.

The 42 is of course a reminiscence to Douglas Adams.

 

[mathdad]

Nothing, except for the 42, of course. IIRC then I invented it decades ago on a MS-platform with chat rooms that doesn't exist anymore since years. It has the advantage that it doesn't use names of any scientific or popular celebrities and that it sounds young. Well, that once was the case. Those registration pages always surprise me by "choose a username" although I know it comes. Then I'm sitting in front of the PC like a cow in front of a tv and don't know what to type and remember it.

The 42 is of course a reminiscence to Douglas Adams.

Thanks for sharing the story of your username and for not getting upset with my textbook questions.

 

[fresh_42]

Thanks for sharing the story of your username and for not getting upset with my textbook questions.

De nada. Your questions are grateful. They force me to get to the point and think about what really counts rather than drowning in technical babble. Math can be very complicated, but I think the basics shouldn't be. I once managed to teach a six-year-old how to deal with negative numbers. Not that I want to say your questions were on that level. It's just an anecdote. She proudly presented that she could count backwards and all I had to do was ask why she stopped at zero and if counting backwards was stepping backwards there would be no reason to stop at a certain point. It worked. I always hated it when teachers said "impossible" or "too difficult" and those impossible and difficult subjects came up only a year later.

 

[mathdad]

De nada. Your questions are grateful. They force me to get to the point and think about what really counts rather than drowning in technical babble. Math can be very complicated, but I think the basics shouldn't be. I once managed to teach a six-year-old how to deal with negative numbers. Not that I want to say your questions were on that level. It's just an anecdote. She proudly presented that she could count backwards and all I had to do was ask why she stopped at zero and if counting backwards was stepping backwards there would be no reason to stop at a certain point. It worked. I always hated it when teachers said "impossible" or "too difficult" and those impossible and difficult subjects came up only a year later.

De nada? Are you Hispanic?

 

[fresh_42]

De nada? Are you Hispanic?

No, but there is no proper English phrase for it, and "de nada" is usually understood. I would have said "da nich für".

 

[mathdad]

No, but there is no proper English phrase for it, and "de nada" is usually understood. I would have said "da nich für".

Thank you.