Why does raising a number to the power of 0 always result in 1?

[OMGMathPLS]

Why is base to power 0 always 1, even if it's .2^0 it =1?

Is it just counting the 1 time the .2 is existing? Is that why?And - base ^ 0 is negative whatever the number is so why is that not a negative one?

Because (-2)^2 is 4 but no because a negative times a negative is a positive.

And with parentheses -2^2 is -4, because the number is done first and then the - is kept afterwards.

So why is (-2)^0 positive 1 and -1^0 a negative?

Is it because we're doing the same thing by adding the - after when unenclosed with a ( ) ?

 

[ineedhelpnow]

my explanation for it:

take $\frac{x^2}{x^2}$ when x is not equal to 0. obviously it's equal to 1, right? now by the law of subtraction for exponents $x^{2-2}=1$ which is $x^0=1$ this holds for all numbers

 

[Prove It]

my explanation for it:

take $\frac{x^2}{x^2}$ when x is not equal to 0. obviously it's equal to 1, right? now by the law of subtraction for exponents $x^{2-2}=1$ which is $x^0=1$ this holds for all numbers

More generally: Since $\displaystyle \begin{align*} \frac{a^m}{a^n} = a^{m-n} \end{align*}$, if the powers are the same then $\displaystyle \begin{align*} \frac{a^m}{a^m} = a^0 \end{align*}$. But the top and bottom are equal, and so they cancel leaving 1. So $\displaystyle \begin{align*} a^0 = 1 \end{align*}$.

Just note, $\displaystyle \begin{align*} 0^0 \end{align*}$ is undefined.

 

[chisigma]

Why is base to power 0 always 1, even if it's .2^0 it =1?

Is it just counting the 1 time the .2 is existing? Is that why?And - base ^ 0 is negative whatever the number is so why is that not a negative one?

Because (-2)^2 is 4 but no because a negative times a negative is a positive.

And with parentheses -2^2 is -4, because the number is done first and then the - is kept afterwards.

So why is (-2)^0 positive 1 and -1^0 a negative?

Is it because we're doing the same thing by adding the - after when unenclosed with a ( ) ?

In the field of discrete mathematics a precise description of why it is $\displaystyle a^{0}=1$ no matter which is a [even a=0...] il given in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2479

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

So why is (-2)^0 positive 1 and -1^0 a negative?

Is it because we're doing the same thing by adding the - after when unenclosed with a ( ) ?

To answer your last question, yes, we are doing the same thing. $(-2)^0$ is simply $1$ because $(-2)^0=(-2)^{1-1}=\frac{(-2)}{(-2)}=1$. The negatives cancel out because they are enclosed in parenthesis (). However, $-1^0=(-1) \cdot 1^0$ has its negative outside the parenthesis so that operation is done last: $-1^0=(-1) \cdot 1^0=(-1) \cdot 1=-1$ :D

 

[Fernando Revilla]

Another interpretation using the concept of empty function:

If we want $\left|Y^X\right|=\left|Y\right|^{\left|X\right|}$ to hold for $X=\emptyset,$ then,$$\left|Y\right|^0=\left|Y\right|^{\left|\emptyset\right|}=\left|Y^{\emptyset}\right|=1,\quad \forall \left|Y\right|=0,1,2,\ldots$$