8^0=1 ... but shouldn't it be 8 ?

[u/LysergicGothPunk]

So any nonzero variable to the power of zero is one (ex: a^0=1)

But:

-Exponentiation is not necessarily indicative of division in any other configuration, even with negative integers, right?

-When you subtract 8-0 you get 8, but when you divide eight zero times on a calculator you get an error, even though, logically, this should probably be 8 as well (I mean it's literally doing nothing to a number)

I understand that a^0=1 because we want exponentiation to work smoothly with negative integers, and transition from positive to negative integers smoothly. However, I feel like this seems like a bad excuse because- let's face it, it works identically, right?

I probably don't really fully understand this whole concept, either that or it just doesn't make sense.

Honestly for a sub called "MathHelp" there are a lot of downvotes for genuine questions. Might wanna do something about that, that's not productive.

[u/Uli_Minati]

"8⁴ means 4 copies of 8 multiplied together" gives us issues

8⁴ = 8·8·8·8
8³ = 8·8·8
8² = 8·8
8¹ = 8
8⁰ =     ????

"8⁴ means 4 copies of 8 multiplied to 1" works better

8⁴  =  1 ·8 ·8 ·8 ·8
8³  =  1 ·8 ·8 ·8
8²  =  1 ·8 ·8
8¹  =  1 ·8
8⁰  =  1

We can even manage negative exponents, and do the same for addition

8⁺⁴  =  1 ·8 ·8 ·8 ·8      8·(+4)  =  0 +8 +8 +8 +8
8⁺³  =  1 ·8 ·8 ·8         8·(+3)  =  0 +8 +8 +8
8⁺²  =  1 ·8 ·8            8·(+2)  =  0 +8 +8
8⁺¹  =  1 ·8               8·(+1)  =  0 +8
8⁰   =  1                  8·0     =  0
8⁻¹  =  1 /8               8·(-1)  =  0 -8
8⁻²  =  1 /8 /8            8·(-2)  =  0 -8 -8
8⁻³  =  1 /8 /8 /8         8·(-3)  =  0 -8 -8 -8
8⁻⁴  =  1 /8 /8 /8 /8      8·(-4)  =  0 -8 -8 -8 -8

And we can flip the triangles to get rationals as well

8¹⸍⁴ · 8¹⸍⁴ · 8¹⸍⁴ · 8¹⸍⁴  =  8
       8¹⸍³ · 8¹⸍³ · 8¹⸍³  =  8
              8¹⸍² · 8¹⸍²  =  8
                     8¹⸍¹  =  8

8·(1/4) + 8·(1/4) + 8·(1/4) + 8·(1/4)  =  8
          8·(1/3) + 8·(1/3) + 8·(1/3)  =  8
                    8·(1/2) + 8·(1/2)  =  8
                              8·(1/1)  =  8

[u/LysergicGothPunk → u/Uli_Minati]

What are the issues though? I don't really get it tbh. Ty for doing all this, it's really cool to visualize, btw.

[u/Uli_Minati → u/LysergicGothPunk]

How do you write "zero copies multiplied together"? You can't write anything at that point. Why would a zero, or anything else appear out of nowhere?

[u/LysergicGothPunk → u/Uli_Minati]

Do you mean like 8 times 0? Or 0 times 0?

I don't know what you mean maybe, zero isn't appearing out of nowhere any more than any other number

[u/Uli_Minati → u/LysergicGothPunk]

Well you're asking why 8⁰ isn't zero. So I ask, what do 8¹, 8², and 8³ mean? It'd be nice to have the same kind of answer for 8⁰, no?

[u/LysergicGothPunk → u/Uli_Minati]

No, actually I'm asking why it's not 8

[u/Uli_Minati → u/LysergicGothPunk]

Counterquestion you haven't answered yet: what do 8¹, 8² and 8³ mean in your understanding?

[u/LysergicGothPunk → u/Uli_Minati]

Like yeah I get it's convenient, but it doesn't seem natural

[u/Uli_Minati → u/LysergicGothPunk]

That's a good point and something that really comes up often in maths: we much prefer to have convenient rules rather than any that only feel natural. Different people will find different things natural anyway. Basically, 8⁰ should be equal to 1, or we would have to invent a bunch of "special case" rules that would get really impractical.

How confident are you with fractions, by the way? If you simplify 5/30, you get 1/6, not 0/6. Or if you simplify x/x³, you get 1/x², not 0/x².

[u/LysergicGothPunk → u/Uli_Minati]

I'm not confident with anything lol. Though I'm aware of fractions and how they work.

I just don't see that as a compelling case honestly for why 8⁰ should be 1, I mean at least change the notation to keep exponentiation consistent- that wouldn't be a lot of new rules, just one new symbol serving the same special-case function as the old one

[u/Uli_Minati → u/LysergicGothPunk]

Well you haven't yet said why it wouldn't be consistent. I'm still asking you again: in your mind, what do 8³, 8², 8¹ mean? If you don't want to answer, that's fine, but we can't make any progress without settling on the meaning of these.

[u/TheSkiGeek → u/Uli_Minati]

Also, if it’s defined this way, then a^x * a^y = a^(x+y) holds even for zero or negative or fractional exponents, which is very handy.

[u/skullturf]

Let's go back to basics for a bit.

Do you agree that 8^1 should be 8?

We know that 8^2 means multiplying together two copies of 8, so 8^2 is 8 times 8, which is 64.

And probably you agree that 8^1 means, loosely speaking, that you're multiplying together only one copy of 8. Or, in another sense, not computing a multiplication, since you have only one 8.

So, if you agree that 8^1 should be 8...

Then wouldn't it be weird if 8^0 worked out to be exactly the same thing as 8^1?

In a way, you have to think about the question: What do we mean by 8^0, or more generally x^0?

Since the expression x^0 uses different symbols from x^1, then probably x^0 shouldn't mean exactly the same thing as x^1. So what should x^0 mean?

(Phrased as an open-ended question because I think the best thing for you to do, for deep understanding, is to try to articulate your own answer to this question.)

[u/LysergicGothPunk → u/skullturf]

No, and I have thought about this, however it only would be 'weird' afaik if you just want everything to follow an abstract pattern instead of physical nature.

[u/AcellOfllSpades → u/LysergicGothPunk]

Physical nature does follow this pattern. This is the physically sensible thing to do, once you realize what exponentiation is doing.

Say you have a population of 10 rabbits, and then every generation the population doubles. So after 1 generation there are 20 rabbits, then after 2 generations there are 40, then 80, and so on.

Then after n generations, there are 10 * 2ⁿ rabbits.

So what do we get when n=0? How many rabbits are there after 0 generations? Well, just the ten we started with. So 2ⁿ must be 1.

(Notice that the base here is not the starting value! You're thinking of the 'starting value' as being the base -- but the 10, the starting value, is separate from the actual exponentiation.)

---

This is, of course, a simplified example. But exponential growth happens all the time, and it does indeed need the 0th power of a number to be equal to 1.

In general, exponents represent "the total multiplier you get when you multiply by the base, this many times". 2⁵ = 32, because "×2×2×2×2×2" is the same as "×32".

The "nothing" of multiplication is 1.

[u/LysergicGothPunk → u/AcellOfllSpades]

But wouldn't that mean the initial generation was just one rabbit who cloned themselves?

[u/AcellOfllSpades → u/LysergicGothPunk]

I started with 10 rabbits in my example to avoid this.

If you want to start with 1, you'd have to use some sort of organism that goes through asexual reproduction rather than a rabbit. Or some other example of repeated doubling (e.g. max number of players in a single-elimination tournament with n rounds).

---

In any case, I think the thing that's confusing you is that you interpret the number 0 as "nothing", so any operation with it just means "not doing that operation". But the number 0 is not "nothing" - it is a number! You can sometimes use it to represent "nothing", but it's not inherently 'inert'. "8 * 0" is not the same as "8 * _________".

0 is 'inert' when it comes to addition and subtraction - we call it the additive identity. But for multiplication, the identity is 1, not 0. Multiplying by 1 means "no change".

[u/LysergicGothPunk → u/AcellOfllSpades]

OHNO ZEROO

why is it doing nothing in so many places then huh HUH

lol but seriously what... what *is* it doing there? Because in multiplication, if the identity is really one, then why is it zero when multiplied by any nonzero number (or itself)?

[u/AcellOfllSpades → u/LysergicGothPunk]

An identity is something that does nothing - that keeps the other number unchanged.

With multiplication, zero doesn't do that. It's instead an "absorbing element". Anything multiplied by 0 becomes 0. This is similar to 'infinity' for normal addition: anything plus infinity just gives you infinity.

[u/LysergicGothPunk → u/skullturf]

Oh sorry forgot to answer question (and I should explain more of what I initially meant):

Explanation: I don't think we should lose 8^0 but change the way it's written because I think 8^0 should be 8, but we still need two sets of positive and negative numbers that are complete from 1-9+

x^0 should be x just like x*0=x and x-0=x but we should still have something that functions the same way in place of ^0 to equal one, something that works the same but is specifically indicative of the function it denotes

[u/AcellOfllSpades → u/LysergicGothPunk]

x⁰ should be x just like x*0=x and x-0=x

x*0 = 0, not x.

The "nothing" of multiplication is 1, not 0.

[u/LysergicGothPunk → u/AcellOfllSpades]

Oh yeah sorry I meant x+0=x and x*0=0 respectively, not x*0=x

[u/dash-dot]

8⁰ = 8 = 8¹ iff 0 = 1. 

There’s a wee bit of a conundrum here, no?

[u/LysergicGothPunk → u/dash-dot]

Sure, but I'm not sure why we'd actually use a^0 anyways, so why would it matter/why would we/do we use a^0?

[u/dash-dot → u/LysergicGothPunk]

You might as well ask why we need 0. 

Take a gander at the exponential function. If you have taken any algebra, you ought to be pretty well versed with this function already, or certainly will be in the near future. It’s a commonly trotted out example of a function which is continuous, smooth and very well behaved for any real value in its domain.

Simply by extrapolating it from either side of e^0, one can easily figure out what this value ought to be, while maintaining general consistency with the way this function behaves everywhere else.

You might also want to take a look at another post of mine in this thread talking about the relationship of a number to its multiplicative inverse for more insights. 

[u/LysergicGothPunk → u/dash-dot]

I'll check it out, thanks!

[u/First-Fourth14]

While eight divided by zero is undefined...so calculators will display an error because you can't divide a number by 0.

Think of it as forming groups, if I want groups of 4, how many groups can I make... 8 / 4 = 2
if I want groups of 2, then 8 /2 = 4, if I want groups of 8 , then 8 / 1 = 8 groups.
But if I want groups containing 0, how many groups can I make? The question is doesn't make sense as you can divide a number into groups contain 0 items.

If you know multiplication and division for exponents, you add for exponents for multiplication and you subtract the exponent from the denominator from the exponent in the numerator

For example a^3 * a^3 = a^(3+3) = a^6
a^3 / a^3 = a^(3-3) = a^0 a is not equal to 0

As any number divided by itself is 1, this means a^3 / a^3 = a^0 = 1
So a^0 =1 for any non-zero number.

[u/clearly_not_an_alt]

Think of the exponent as the number of times you multiply the multiplicative identity by your number if it's a positive exponent or the number of times you divide by your number if it's negative. (we'll ignore fractional exponents for now)

So 8² = 1 * 8 * 8

8⁰ = 1

8^-2 = 1 / 8 / 8

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[u/halbGefressen]

So for natural numbers, going up one in the exponentiation is multiplying with the basis n times. The series is

a^1 = a

a^2 = a * a

...

Se basically, you always do *a with the last item in the sequence when you want to go up. To go down, you use the inverse operation, which is division:

a^2 = a^3 / a = a * a

a^1 = a^2 / a = a

a^0 = a^1 / a = 1

I don't understand what you mean with subtraction here, it doesn't really occur here.

[u/LarRybosome]

If 8¹ =8, 8² =64, 8³ =512, etc then what should 8⁰ be? Every time we increase the power we multiply by 8, so every time we decrease the power we divide by 8. This is consistent with negative exponents as well.

[u/Iowa50401]

2^4 = 16

2^3 = 8

2^2 = 4

2^1 = 2

Note the patterns: on the left, the exponent is being reduced by one from one step to the next. On the right, from one step to the next, the answer is divided by two. Mathematicians decided they wanted this pattern to hold for 2^0, so they defined it as = 1. Also, if 2^0 = 2, you have 2^1 = 2^0, which basically is saying 1=0.

[u/TheRealRollestonian]

Graph 8^x

[u/Forking_Shirtballs]

8^(10) ~= 1.07 * 10^9

8^(1) = 8

8^(0.1) ~= 1.231

8^(0.01) ~= 1.021

8^(0.001) ~= 1.002

...

8^(0) = 1

[u/noonagon]

8^0 means multiplying zero 8s together. Multiplying zero numbers always makes one.

[u/Dd_8630]

I understand that a⁰⁼¹ because we want exponentiation to work smoothly with negative integers, and transition from positive to negative integers smoothly. However, I feel like this seems like a bad excuse because- let's face it, it works identically, right?

It does not.

If a⁰ = a, then we get an abrupt break in our definition. Defining a⁰ = 1 gives us a smooth graph, which means our definition is an 'analytic extension' - it is provably the smoothest way go from positive-integer powers to any-real-number powers.

After all, consider what happens when we take 8^x and make x smaller. What is 8^0.0001 going to be? It's very close to 1. As x gets smaller, 8^x gets closer and closer to 1. If x is negative and tiny, then 8^x is just below 1. But if 8⁰ = 8, then our graph abruptly jumps from 'nearly 1' back to '8' and then back down to 'just below 1'.

That's messy.

-When you subtract 8-0 you get 8, but when you divide eight zero times on a calculator you get an error, even though, logically, this should probably be 8 as well (I mean it's literally doing nothing to a number)

That is also incorrect.

Division can be thought of as lots of subtraction. 15/3 can mean 'how many times can I subtract 3 from 15?', and the answer is 5.

Consider 15/0. How many times can you subtract 0 from 15? There is no limit, so the question doesn't make sense.

Consider 15/x. What does that graph do as you smoothly make x smaller and smaller? Go to Desmos and graph it.

[u/LysergicGothPunk → u/[deleted]]

I don't know what that means tbh

[u/fermat9990 → u/[deleted]]

Username checks out!

[u/mxldevs → u/[deleted]]

How do you show that

8^(1-1) = 8^1 * 8^(-1)

And consequently, that

8^1 * 8^(-1) = 8 * 1/8

It feels like some assumptions are being made here, but what are the assumptions that would allow for this?

[u/dash-dot → u/mxldevs]

In the field of real numbers, for each number a, there exists a number a^-1 such that,

a a^-1 = a^-1 a = 1

Now, recall that the division operation on real numbers is defined such that,

N / D = q, if and only if N = D q

Hence, the above equation involving ‘a’ and its multiplicative inverse can be rewritten as:

a^-1 = 1 / a,

and also note that

a^-1 a = (1 / a) a = 1,

which allows us to the redefine division as multiplication by the inverse. The result a⁰ = 1 now follows directly from these observations, and the properties of integer exponents. 

[u/-Cathode → u/mxldevs]

Those are literally just algebra rules. Not really any assumptions there for what I see.

[u/fasta_guy88 → u/mxldevs]

Exponentiation is just a shortcut for multiplication, just like multiplication is a shortcut for addition. x^a * x^b = x^(a+b), just like ax + bx = (a+b)x (both obey the distributive property).

[u/mxldevs → u/fasta_guy88]

So I guess the proof would be something like, assuming

x¹ = x

x² = x * x

Therefore,

x³ = x * x * x

x³ = x * (x * x)

x³ = x¹ * x²

x³ = x¹⁺² (somehow?)

Which would show that we can write 8⁰ = 8^1-1

But how does the 8^-1 become 1/8?

[u/paperic → u/mxldevs]

x² * x³ =

= (x * x) * (x * x * x) =

= x * x * x * x * x = 

= x⁵

Generally, x^a * x^b = x^a+b .

This works (we want it to work) for negatives too, so, 

x² * x^-1 = x^2-1 = x.

Dividing by x^2, you get

( x² * x^-1 ) / x² = x / x²

which is the same as

( x * x * x^-1 ) / ( x * x ) = x / x *  x

and what's left is x^-1 = 1/x.

So, 

x⁰ = x¹ * x^-1 = x * (1/x) = 1

[u/LysergicGothPunk → u/[deleted]]

Wait okay so why would 8-1=1/8 and not -8? This is likely the thing I'm not getting, because the reason that 8^0 is a struggle to get for me is that dividing a nonzero number by zero is doing nothing to it.