Home
DFS
SB
OD
SDN
Let's see why. Let's try out some variables for a using the formula a ^ (a ^ (b - 1)) as it becomes 0 ^ (0 ^ (b - 1)).
| 0 ✱ 4 = 0 ^ (0 ^ (4 - 1)) = 0 ^ (0 ^ 3) = 0 ^ 0 = undefined |
| 0 ✱ 2 = 0 ^ (0 ^ (2 - 1)) = 0 ^ (0 ^ 1) = 0 ^ 0 = undefined |
| 0 ✱ 1 = 0 ^ (0 ^ (1 - 1)) = 0 ^ (0 ^ 0) = 0 ^ (undefined) = undefined |
| 0 ✱ 0 = 0 ^ (0 ^ (0 - 1)) = 0 ^ (0 ^ −1) = 0 ^ (undefined) = undefined |
So there you have it. You can't use the Superpowerition expression 0 ✱ b for any variable b at all since this results in the formula 0 ^ (0 ^ (b - 1)), and you can't take zero to the power of zero or a negative number.
There is a debate, however, whether the expression 0⁰ is undefined or the number 1.
According to the rules of exponentation for some numbers, n⁰ = 1. Examples: 1⁰ = 1, -1⁰ = 1, 6⁰ = 1, 3,616⁰ = 1, even π⁰ = 1.
And for values of n (where n > 0) in the expression 0ⁿ: 0⁹ = 0, ⁴ = 0, 01/2 = 0, 0¹ = 0, and yes, 0π = 0
Let's let n⁰ and 0ⁿ as well as n¹1 and 1ⁿ, battle it out with each other, going from 9 to -2:
| Values for n | n⁰ | 0ⁿ | n¹ | 1ⁿ |
|---|---|---|---|---|
| 9 | 1 | 0 | 9 | 1 |
| 8 | 1 | 0 | 8 | 1 |
| 7 | 1 | 0 | 7 | 1 |
| 6 | 1 | 0 | 6 | 1 |
| 5 | 1 | 0 | 5 | 1 |
| 4 | 1 | 0 | 4 | 1 |
| 3 | 1 | 0 | 3 | 1 |
| 2 | 1 | 0 | 2 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | ? | 0 | 1 |
| -1 | 1 | ∞ | -1 | 1 |
| -2 | 1 | ∞ | -2 | 1 |
As you can see, n⁰ and 1ⁿ are always 1. n¹1 is always n. 0ⁿ is 0 as long as n is greater than 0, but once it hits 0, should there be a meaning? Also, if you attempt to take 0 to the exponent of a negative number, you get an error. You could for example try to convert 0-2 to another expression using this rule for negative exponent conversion:
a-b = 1 ÷ ab = (1 ÷ -b) √ a
And apply it to this:
0-2 = 1 ÷ 02 = (1 ÷ -2) √ 0
And you can see that you get a division by zero error, so therefore, you can't raise 0 to a negative exponent.
So if 0 raised to a positive number is 0, and 0 raised to a negative number is ∞ according to some, then what is 0 raised to the power of 0? If 0⁰ is 1 in the above calculations for 0 ✱ n, here is what we could get:
| 0 ✱ 4 = 0 ^ (0 ^ (4 - 1)) = 0 ^ (0 ^ 3) = 0 ^ 0 = [1?] |
| 0 ✱ 2 = 0 ^ (0 ^ (2 - 1)) = 0 ^ (0 ^ 1) = 0 ^ 0 = [1?] |
| 0 ✱ 1 = 0 ^ (0 ^ (1 - 1)) = 0 ^ (0 ^ 0) = 0 ^ [1?] = [0?] |
| 0 ✱ 0 = 0 ^ (0 ^ (0 - 1)) = 0 ^ (0 ^ −1) = 0 ^ [∞] = [1?] |
As you can see from 0 ✱ 4 and 0 ✱ 2, the answer could work out to be 1. For 0 ✱ 1, the answer could be 0, but for 0 ✱ 0, since infinity could be one number lower than the lowest negative number and one number higher than the highest positive number, based on the way zero to positive powers are 0 and zero to negative powers are infinity, and zero to the power of zero could be 1, zero to the power of infinity might be 1 as well.
Continue on with your debates on raising zero to negative, zero and infinity exponents.
© 1995-2023. davesfunstuff.com. All Rights Reserved. Reproduction of any part of this website without expressed written consent is prohibited.
