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2006-08-24 12:10:10 · 11 answers · asked by Anonymous in Science & Mathematics Mathematics

11 answers

Because raising something to the power of zero is the same as dividing a number by itself.
This site explains it well:
http://mathforum.org/dr.math/faq/faq.number.to.0power.html

2006-08-24 12:13:39 · answer #1 · answered by Anonymous · 2 0

It comes about because of the way powers are defined.

We know squared and cubed, those are intuitively area and volume. It almost makes sense that the power "1" is just the number itself. When we allow x^1.5, it's not so clear what it means. We have square root which is the 1/2 power. Then cube roots is (1/3) and on down to "nth" roots.

But take a number, say 2, and look at:
2^(1/2) = 1.4142...
2^(1/3) = 1.2599...
2^(1/4) = 1.1892...

and so on down.

It works the same with any number
999^(1/2) = 31.6069...
999^(1/3) = 9.99666...
999^(1/4) = 5.62200...

999^(1/100) = 1.07150...

even starting with numbers less than 1

0.5^(1/2) = 0.70710...
0.5^(1/3) = 0.79370...
0.5^(1/4) = 0.84089...

0.5^(1/100) = 0.99309...
0.5^(1/1000) = 0.99930...

What this is saying is that to get 0.5 by multiplying a number times itself 1000 times you have to use a number that is close to 1; in fact about 0.99930709305.

So as the the exponent gets smaller (the n in "nth" root gets larger), the nth root gets closer and closer to 1.

We say the limit as n => infinity of x^(1/n) is 1.
And the limit as n => infinity of 1/n is 0,
So in the limit, x^0 = 1.

2006-08-24 12:52:09 · answer #2 · answered by bubsir 4 · 0 0

Note: I use the caret symbol " ^ " to indicate exponents (or powers!). I also use / for division and fractions.

Think about dividing powers of 3, for example.

(3 ^ 6) / (3 ^ 2) = (3x3x3x3x3x3) / (3x3)
cancel two threes from top and bottom...
= (3x3x3x3) / 1 = 3 ^ 4.

Notice that 3 ^ 6 divided by 3^2 is 3^4, where 6-2=4. Subtract the exponents to get the answer.

Now think about this one....

(3^4) / (3^4) = ???

Think of the two ways to express this. One is 3 ^ 0, by subtracting the exponents. The other way is to say that since the top and bottom are the same, that they divide to get 1.

So 3^0 and 1 are the same! You can do this exercise with other numbers to see that any number x^0 = 1.

2006-08-24 12:42:30 · answer #3 · answered by Polymath 5 · 0 0

since a ^ 3 / a ^ 2 = a ^ (3 - 2), so that in other words you subtract the two values of the powers, a ^ 0 is the same as

a ^ 3 / a ^ 3 = a (3 - 3) = a ^ 0

but any number divided by itself is also 1 to begin with

2006-08-24 14:24:13 · answer #4 · answered by fmg134s 2 · 0 0

everything is to the 0 power is 1 meaning that the 0 power really is another way of saying the same or one.

2006-08-24 12:16:42 · answer #5 · answered by Anonymous · 0 0

Originally, powers x^n are only defined for exponents n which are natural numbers. It turns out that one of the properties of powers is that

x^(n + m) = x^n * x^m

that is, if you add exponents you can also multiply two powers with the same base.

Defining powers with other exponents, such as zero, negative numbers, or fractions, is only useful if this fundamental rule is still true. In particular, we want

x^(0 + n) = x^0 * x^n

Because 0 + n = n, this is the same as

x^n = x^0 * x^n

and therefore x^0 must be equal to 1.

2006-08-24 12:17:48 · answer #6 · answered by dutch_prof 4 · 1 1

Numbers to the 0 power are the same as multiplying a number by 1. 0 power means multiplying a number by itself 0 times. not multiplying the number by 0

2006-08-24 12:15:30 · answer #7 · answered by Anonymous · 0 2

x^0 = 1 so that the rules involving exponents have no exceptions.

2006-08-24 12:31:12 · answer #8 · answered by davidosterberg1 6 · 0 0

(x^n)/(x^n) = x^(n - n) = x^0

as you can see you have identical values on top and bottom, and as we know, when we have the same exact value in the denominator and numerator and we divide them, we get 1, so

x^0 = 1

2006-08-24 13:11:29 · answer #9 · answered by Sherman81 6 · 0 0

consider this

z^y x z^-y = z^0

bear in mind that z^-y = 1/z^y

which implies that z^y x z^-y = z^y/z^y

z^y/ z^y = 1

where z and y are any integer

2006-08-24 12:30:59 · answer #10 · answered by Aslan 6 · 0 0

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