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

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Raising any non-zero number to the power of zero results in one due to the properties of exponents, specifically that \( a^m / a^m = a^{m-m} = a^0 \), which simplifies to one when \( a \neq 0 \). This holds true regardless of whether the base is positive or negative, as demonstrated by the example of \( (-2)^0 = 1 \) because the negative is enclosed in parentheses. Conversely, \( -1^0 \) equals negative one because the negative sign is outside the exponentiation operation, leading to \( -1 \cdot 1^0 = -1 \). The discussion also notes that \( 0^0 \) is undefined, emphasizing the importance of the base being non-zero for the exponent rule to apply. Understanding these principles clarifies why any base raised to the power of zero equals one.
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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 ( ) ?
 
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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
 
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ineedhelpnow said:
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.
 
OMGMathPLS said:
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

$\chi$ $\sigma$
 
OMGMathPLS said:
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
 
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,$$\left|Y\right|^0=\left|Y\right|^{\left|\emptyset\right|}=\left|Y^{\emptyset}\right|=1,\quad \forall \left|Y\right|=0,1,2,\ldots$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

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