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

  • Thread starter Thread starter OMGMathPLS
  • Start date Start date
  • Tags Tags
    Exponent
AI Thread Summary
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.
OMGMathPLS
Messages
64
Reaction score
0
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 ( ) ?
 
Mathematics news on Phys.org
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
 
Last edited:
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,$$\left|Y\right|^0=\left|Y\right|^{\left|\emptyset\right|}=\left|Y^{\emptyset}\right|=1,\quad \forall \left|Y\right|=0,1,2,\ldots$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
17
Views
2K
Replies
10
Views
2K
Replies
7
Views
2K
Replies
55
Views
5K
3
Replies
105
Views
6K
Replies
2
Views
1K
Replies
36
Views
6K
Back
Top