Why does 10 raised to the 0 power equal 1? Our pre-cal teacher tells us that we do not need to try to understand it, we just need to know it. If 10^1 = 10, that is because u have 1 (10), where as, 10^0 = 1, but u have 0 (10’s). So, why does 10^0 = 1?
Essentially, exponentiation can be regarded as
x[sup]n[/sup]=1xx*…x with n x’s. So 10[sup]3[/sup], for example, is 1101010=1000. If n=0, then there’s nothing to multiply by the initial 1. As a result, any number raised to the power 0 is 1 (x[sup]0[/sup]=1).
It’s just a matter of definition, really.
Because
10[sup]9[/sup] = 1000000000
10[sup]8[/sup] = 100000000
10[sup]7[/sup] = 10000000
10[sup]6[/sup] = 1000000
10[sup]5[/sup] = 100000
10[sup]4[/sup] = 10000
10[sup]3[/sup] = 1000
10[sup]2[/sup] = 100
10[sup]1[/sup] = 10
so…
10[sup]0[/sup] must be 1.
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friedo is well aware that this is does not constitute mathematical proof of any kind, but it demonstrates that “1” is the logical conclusion to the pattern of exponentiation given.
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In that case 10 [sup] 0 [/sup] would be 0, since anything multiplied by nothing equals nothing.
No, its not 1*0, its 1 with nothing after that. There is nothing to multiply it by, it doesn’t have to have something to be multiplied.
(Poorly worded, I know)
While in addition, 0 is a sort of nothing, in multiplication, 1 is a sort of nothing. It’s what you have if you don’t really have anything. Or, if you put it together with anything, you’ll get the same thing. Just like X + 0 = X, so too X × 1 = X. It’s called the identity. And just like multiplying by 0 gets you the additive identity, 0, exponentiating by 0 gets you the multiplicative identity, 1.
It has to do with the fact that in multiplication the exponents are added, e.g. x[sup]3[/sup]*x[sup]3[/sup]=x[sup]6[/sup]. For this reason, x[sup]3[/sup]*x[sup]0[/sup] must equal x[sup]3[/sup], making x[sup]0[/sup] equal 1
Look at the fact that any number raised to a negative power gives its reciprocal.
Thus we define a positive exponential curve and a negative exponential curve. They both intersect (regardless of base) at zero at one, as we would expect. After all, one is the reciprocal of itself and zero is neither positive nor negative.
Now we need to prove the continuity of the exponential function which can be done using Taylor Expansion and noting that any exponential function can be written as an infinite series of polynomials, and since polynomials and their compositions are continuous, so are exponential functions.
If you want I can explain Taylor Expansions, but that’s pretty long (I actually started doing that and erased it because it was prohibitive in length).
So, exponential function is continuous. The limit on both sides of zero as we approach zero is one
Therefore any number (excepting zero) raised to the zero power is zero.
Why excepting zero? Because of a mathematical concept called “being undefined”, but that’s a hijack for another day.
I should note that most proofs listed here are not rigorous (mine included)… If you really want to know the answer, you gotta get a book on real analysis and build up some theorems about functions, continuity, infinity, countability.
There are no shortcuts for rigor!
Math geek checking in. . . .JS Princeton, I am in wholehearted agreement. Rigour, coming right up!
How do we define a[sup]b[/sup], anyway, for arbitrary real numbers a and b? Intuitively, it’s easy if your b is a positive integer, but we want a more general answer. To get this, we resort to the exponential function exp(x), also known as e[sup]x[/sup], and its inverse function, the natural logarithm (I’ll call it “log” although you might know it as “ln”). The definition that mathematicians use is:
a[sup]b[/sup] = exp(b log a) .
So what’s 10^0? Well, by our formula, we have
10^0 = exp(0 log 10) = exp(0) = 1.
If you want a formal definition of exp(x), it is the limit of the infinite sum
1 + x + x[sup]2[/sup]/2 + x[sup]3[/sup]/6 + . . . + x[sup]n[/sup]/(123* . . . *n) + . . . .
By plugging in x = 0, it should be pretty clear that exp(0) = 1. (To be perfectly rigourous, we also have to check that log 10 is defined and that exp converges. I leave this as an exercise to the reader.)
Good answer, MrDeath… way to throw in those logarithims that I purposely left out (see my note on Taylor expansion)!
Fact of the matter is, 10^0 is a pretty strange concept anyway…