Explain a thing! (Working on Math, Coda HELP!)
 

Exponents are a notation, and not something specifically intrinsic to numbers.

So, when you have 4^3, you’re shorthand writing 1 * 4 * 4 * 4.

Negative exponents are divisions, so 4^-3 is 1 / 4 / 4 / 4, which is the same as 1 / (4 * 4 * 4), or 1/(4^3).


It also might help to consider that 4^4 * 4*^-2 = 4^2, since 4-2 = 2, also that (4 * 4 * 4 * 4) / (4 * 4) = (4 * 4)

While a real explanation doesn't come, google told me that regular exponents are multiplication right, so 9^2 is 81 because 9 times 9 is 81! Negative exponents are the opposite! Because negative. So they're division! And what better to visualize division than fractions! So 9^-2 would be 1 divided by 9 times 9. Or something like that.

Sure, I can explain this pretty easily! :)

All right, so you should already know most of this, but I'm going to build this up in order to show the sequence of steps.

First off, the definition of exponentiation:

aⁿ = a × a × ... × a

That is, exponentiation is repeated multiplication. Next, a useful property:

a^b × a^c = a^b+c

Proof:

(A) a^b = a × a × ... × a (b times) (definition of exponentiation)
(B) a^c = a × a × ... × a (c times) (definition of exponentiation)
(C) a^b × a^c = (a × a × ... × a (b times)) × (a × a × ... × a (c times)) (substitution)
(D) a^b × a^c = (a × a × ... × a (b+c times)) (associativity of multiplication)
(E) a^b × a^c = a^(b+c) (definition of exponentiation)

But what if there's a negative number in the mix? We can't count a -2 times, after all, so we can't write it out as repeated multiplication. But let's apply (D) from the above proof...

(F) a^b-c = (a × a × ... × a (b-c times)) (substitution)
(G) a^b × a^-c = (a × a × ... × a (b-c times)) (as proved above)
(H) (a × a × ... × a (b times)) × a^-c = (a × a × ... × a (b-c times)) (definition of exponentiation)

Okay, so let's just think about this... What could we possibly substitute into a^-c to make this equation true? We need to take away some a's. Let's solve for it:

(I) (a × a × ... × a (b-c times)) = (a × a × ... × a (b times)) × a^-c (rearranging for formatting)
(J) (a × a × ... × a (b-c times)) = a^-c (divide both sides)
____(a × a × ... × a (b times))
(K) 1/(a × a × ... × a (c times)) = a^-c (cancellation)
(L) 1/(a^c) = a^-c (definition of exponentiation)

And there you go -- the property you're looking for!

Said in plain English: Taking a number to a negative exponent is a way to take away multiplications by the number instead of a way to do more multiplications by it. And we take away multiplications by using division!

You rename a thread by editing the first post and changing the topic line.

Pretty easily but look at all that math!

It only looks like a lot of math because I was showing a step-by-step -- showing why it works instead of just saying it does.

Wait, heck, that makes sense. IDK if I just never questioned it, or had another way of thinking about it.

I, for one, had never questioned it. I just followed the properties handed to me and they all seemed to work, so I didn't bother thinking any harder about it. That proof there is actually something I came up with on the fly.

Although technically, that's not a proof of correctness -- step (F) does not follow from the rest. The definition of exponentiation that was asserted at the beginning implies that only natural numbers are valid, so if this were being treated as a rigorous proof this would be an error that disproves it, or at least it would establish that it's only true if b > c.

Instead, this is a proof that if you extend exponentiation so negative integers are allowed in addition to positive ones, then this is what a negative exponent must be in order to be consistent with the property we want it to have. You have to follow a similar procedure if you want to define an extension that allows fractional exponents. (Extending it from fractions to real numbers, on the other hand, can't be done with this procedure, and you have to introduce logarithms.)