Raising to the power of 0 or 1

[blueridge]

I know that any number (except for zero), any variable and any term raised to the zero power is 1. I don't know the reason why this is true. Can someone explain why?

Samples:

59^0 = 1

R^0 = 1

(a + b)^0 = 1

 

[blueridge]

I know that any number, any variable and any term raised to the first power equals itself. I don't know the reason why this is true. Can someone explain why?

Samples of what I mean:

45^1 = 45

x^1 = x

(y + 3)^1 = y + 3

 

[DrClaude]

Moderator's note: I have merge the two questions because they are closely related.

 

[blueridge]

Moderator's note: I have merge the two questions because they are closely related.

Ok.

 

[docnet]

For nonzero $a$, $a^0=1$, because
$$\frac{a^n}{a^n}=a^{n-n}=a^0.$$

For any $a$, $a^1=a$, because
$$a^1\cdot a^1=a^{1+1}=a^2.$$

 

[blueridge]

For nonzero $a$, $a^0=1$, because
$$\frac{a^n}{a^n}=a^{n-n}=a^0.$$

For any $a$, $a^1=a$, because
$$a^1\cdot a^1=a^{1+1}=a^2.$$

Can you explain in words why a^0 = 1 and a^1 = a? Your reply is a textbook definition. I can look it up myself. Know what I mean?

 

[FactChecker]

$a^n$ is 1 multiplied by $a$ $n$ times. When $n=0$, that is just 1. When $n=1$, that is $1 \cdot a = a$.

 

[pasmith]

Can you explain in words why a^0 = 1 and a^1 = a? Your reply is a textbook definition. I can look it up myself. Know what I mean?

If we don't define them that way (and also define a^{-1} = 1/a) the index law a^{n+m} = a^na^m does not hold for all integer n and m.[/i]

 

[blueridge]

$a^n$ is 1 multiplied by $a$ $n$ times. When $n=0$, that is just 1. When $n=1$, that is $1 \cdot a = a$.

Does this also apply to numbers and terms?

For example, 5^0 = 1 and 5^1 = 5.
Another example is (a + b)^0 = 1 and (a + b)^1 = a + b.

Yes?

 

[blueridge]

If we don't define them that way (and also define a^{-1} = 1/a) the index law a^{n+m} = a^na^m does not hold for all integer n and m.[/i]

What do you mean if we don't define them that way?

 

[Vanadium 50]

59^0 = 1

Does this also apply to numbers and terms?

For example, 5^0 = 1

Huh? You are asking us what you say you already know.

 

[blueridge]

Huh? You are asking us what you say you already know.

I am asking for a clear, concise, to the core explanation of why a^0 = 1 and a^1 = a. This is all I am asking. This is not an abstract algebra problem.

 

[pbuk]

What do you mean if we don't define them that way?

He means if we define them a different way.

I can look it up myself. Know what I mean?

Don't be rude.

Let's try again from the start.

You know that $a^3 = a(a^2)$ and $a^4 = a(a^3)$, yes? These are just examples of a general definition (an inductive definition) $a^n = a(a^{n-1})$: in the first example we have used $n = 3$ and in the second example $n = 4$.

Now what happens if we take $n = 2$?
$$ \begin{align}
\nonumber a^2 =&\ a(a^{2-1})\\
\nonumber a^2 =&\ a(a^1)\\
\nonumber \frac{a^2}a =&\ \frac{a(a^1)}a \\
\nonumber \left( \frac a a \right)a =&\ \left( \frac a a \right) a^1 \\
\nonumber a =&\ a^1 \\
\nonumber a^1 =&\ a
\end{align}
$$

Now you try this taking $n = 1$.

 

[blueridge]

He means if we define them a different way.


Don't be rude.

Let's try again from the start.

You know that $a^3 = a(a^2)$ and $a^4 = a(a^3)$, yes? These are just examples of a general definition (an inductive definition) $a^n = a(a^{n-1})$: in the first example we have used $n = 3$ and in the second example $n = 4$.

Now what happens if we take $n = 2$?
$$ \begin{align}
\nonumber a^2 =&\ a(a^{2-1})\\
\nonumber a^2 =&\ a(a^1)\\
\nonumber \frac{a^2}a =&\ \frac{a(a^1)}a \\
\nonumber \left( \frac a a \right)a =&\ \left( \frac a a \right) a^1 \\
\nonumber a =&\ a^1 \\
\nonumber a^1 =&\ a
\end{align}
$$

Now you try this taking $n = 1$.

Perfectly done.

 

[WWGD]

Or consider: $a^0 *a^0=a^{0+0}=a^0$. It follows $a^0=1$.

 

[blueridge]

Or consider: $a^0 *a^0=a^{0+0}=a^0$. It follows $a^0=1$.

Very good.

 

[fresh_42]

$a^n$ is an abbreviation for $a^n=\underbrace{a\cdot a\ldots a}_{n-times}$ so $a^1=a$ and $a^0$ is an empty product and equals the neutral element of multiplication, the $1.$

The same holds for empty sums that are equal to the neutral element of addition, the $0.$
$$
\sum_{\substack{k=1\\1\nmid k}}^n k=0
$$

 

[Mark44]

Thread closed as this new member has been banned.