4 ms·

How is nothing multiplied nothing times 1? I feel dumb asking this,but, why is it not 0 or NaN. 2^4 == 2222 == 16 0^0 == 0000 == 0 Why is it not that simple?

gizmo686 6y ago
Why should the answer be 0? We have an intuition that 0 is the identity element for addition, so it makes sense to treat it as the starting point for many things. However, when we look at multiplication, 0 looks very different. Instead of preserving the other value, as it does with addition, multiplying by 0 changes the other value in an information destroying way (e.g if you know that x * 0 = 0, you have no idea what x is). This property of 0 is so distinct from the other numbers, that when we consider the integers (or reals, or complex numbers) as a structure that only has multiplication, but not addition, we typically exclude 0 from consideration. Since (discrete) exponentiation is a multiplicative process, it makes sense to consider it in terms of the integers under multiplication; and so excluding 0. When we consider integers under multiplication, we notice that 1 has the same properties that 0 had under addition. Namely, 1 * x = x for all x. Given this, it makes sense to treat 1 as the "empty" element, the same way you want to do with 0 in additive contexts.
- badrabbit 6y ago
  Thank you for the explanation, the reasoning sort of makes sense but I still can't wrap my head around how on the left side of the equation there is nothing(0) on the right hand there is something. Something cannot come out of nothing when considering natural phenomena in reality.
Koshkin 6y ago
In the context of multiplication, raising a number to the zero-th power is a NOP (no operation, or nothing). So, for any x, x^0 = 1 times (x^0) = 1 times nothing = 1.
- gizmo686 6y ago
  Multiplying by a number raised to the zero-th power is a NOP. This comes up frequently in the context of polynomials, where the constant term is often expressed as the x^0 term. E.g in f(x) = 3x^2 + 2x^1 + 1x^0 , we would like to have f(0)=0. In many cases this is important because we express polynomials as a sum: f(x) = sum (n from 0 to 2) { (n+1)x^n }, so having x^0 behave reasonably at x=0 allows us to avoid needed to special case the n=0 term.
conradludgate 6y ago
Let's take 2^3 * 2^5, we know that we can just add the numbers in the exponent to get 2^8. We can extend this, for example what would 2^3 * 2^0 be? If we want our rule to be consistent, it would be 2^3, implying that 2^0 = 1. This works fine for all numbers except for 0. Like you pointed out, 0^5 = 0, 0^1 = 0, so it would make sense that 0^0 = 0? but if x^0 = 1, we have an issue.
- badrabbit 6y ago
  Why is x^0=1? How can something be generated out of any equation where the only number is 0 (nothing)
  - conradludgate 6y ago
    > We can extend this, for example what would 2^3 * 2^0 be? If we want our rule to be consistent, it would be 2^3, implying that 2^0 = 1 replace 2 with x. Another way to see it is 2^2 = 4 = 8 / 2 = 2^3 / 2 2^1 = 2 = 4 / 2 = 2^2 / 2 2^0 = ? how about 2^1 / 2 = 1 Again, replace 2 with x and the same follows. With this example though, when x = 0, it's obvious that it no longer works since 0^0 would be 0^1 / 0 = 0/0

gizmo686 6y ago

Why should the answer be 0? We have an intuition that 0 is the identity element for addition, so it makes sense to treat it as the starting point for many things. However, when we look at multiplication, 0 looks very different. Instead of preserving the other value, as it does with addition, multiplying by 0 changes the other value in an information destroying way (e.g if you know that x * 0 = 0, you have no idea what x is). This property of 0 is so distinct from the other numbers, that when we consider the integers (or reals, or complex numbers) as a structure that only has multiplication, but not addition, we typically exclude 0 from consideration. Since (discrete) exponentiation is a multiplicative process, it makes sense to consider it in terms of the integers under multiplication; and so excluding 0. When we consider integers under multiplication, we notice that 1 has the same properties that 0 had under addition. Namely, 1 * x = x for all x. Given this, it makes sense to treat 1 as the "empty" element, the same way you want to do with 0 in additive contexts.

badrabbit 6y ago
Thank you for the explanation, the reasoning sort of makes sense but I still can't wrap my head around how on the left side of the equation there is nothing(0) on the right hand there is something. Something cannot come out of nothing when considering natural phenomena in reality.

Koshkin 6y ago

In the context of multiplication, raising a number to the zero-th power is a NOP (no operation, or nothing). So, for any x, x^0 = 1 times (x^0) = 1 times nothing = 1.

gizmo686 6y ago
Multiplying by a number raised to the zero-th power is a NOP. This comes up frequently in the context of polynomials, where the constant term is often expressed as the x^0 term. E.g in f(x) = 3x^2 + 2x^1 + 1x^0 , we would like to have f(0)=0. In many cases this is important because we express polynomials as a sum: f(x) = sum (n from 0 to 2) { (n+1)x^n }, so having x^0 behave reasonably at x=0 allows us to avoid needed to special case the n=0 term.

conradludgate 6y ago

Let's take 2^3 * 2^5, we know that we can just add the numbers in the exponent to get 2^8. We can extend this, for example what would 2^3 * 2^0 be? If we want our rule to be consistent, it would be 2^3, implying that 2^0 = 1. This works fine for all numbers except for 0. Like you pointed out, 0^5 = 0, 0^1 = 0, so it would make sense that 0^0 = 0? but if x^0 = 1, we have an issue.

badrabbit 6y ago
Why is x^0=1? How can something be generated out of any equation where the only number is 0 (nothing)
- conradludgate 6y ago
  > We can extend this, for example what would 2^3 * 2^0 be? If we want our rule to be consistent, it would be 2^3, implying that 2^0 = 1 replace 2 with x. Another way to see it is 2^2 = 4 = 8 / 2 = 2^3 / 2 2^1 = 2 = 4 / 2 = 2^2 / 2 2^0 = ? how about 2^1 / 2 = 1 Again, replace 2 with x and the same follows. With this example though, when x = 0, it's obvious that it no longer works since 0^0 would be 0^1 / 0 = 0/0