# Understanding an exponent of Zero

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I am looking for reason why anything raised to the power of zero is 1.

On a side note any reason why log's of zero are undefined.

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I am looking for reason why anything raised to the power of zero is 1.

On a side note any reason why log's of zero are undefined.

by definition a^n / a = a^(n-1) or alternatively a^(n+1) = a^n * a

thus a^1 = a*a^0 or rearranged

a^1 / a = a^0

a^1/a = 1 = a^0

And by the same logic you cannot multiple a number (other than zero) by itself any possible number of times (even fractional) and come up with zero

Conway - not a mod note but a thought for you to consider. In your last thread on division by zero you consistently showed that you did not understand basic algebra - even going as far as questioning the meaning and validity of simple rearrangements of equations (which is as low level as algebra gets) on more than a few occasions. If this thread goes the same way I shall be reporting it and asking a mod to lock it .

Questioning the fundamental underpinnings of a subject and trying to create new axiomatic bases is all very well and good; but it cannot be done from a position of wanton ignorance. If you wish to learn why some of the rules you are told by your teachers are taken to be good then this is an excellent place to ask. However any attempt to change these rules must be accompanied by an understanding of why they are there in the first place and the consequences of variation.

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I am looking for reason why anything raised to the power of zero is 1.

Please make sure to note that in the conventional operations, it isn't 'anything' to the power of zero = 1. $0^0$ is taken to be indeterminate, for example. In many cases, it is useful to treat the value as equal to 1 or 0, but formally it is indeterminate.

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Please make sure to note that in the conventional operations, it isn't 'anything' to the power of zero = 1. $0^0$ is taken to be indeterminate, for example. In many cases, it is useful to treat the value as equal to 1 or 0, but formally it is indeterminate.

imatfaals explanation covers this.

0^1/0 = 0^0

Which is indeterminate.

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To All

Why is it considered indeterminate as opposed to undefined? Why the difference? I will accept the answer with out further debate.

Imatfaal

You should know that bringing up dead/locked post is against the rules. As a result I am forced to issue you a citizen's mod note....

"It is against the rules to bring up dead/locked post. Your answer would have sufficed on it's own, you just happened to want to insult me, on what you THINK is my lack of education."

.

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I am looking for reason why anything raised to the power of zero is 1.

On a side note any reason why log's of zero are undefined.

$x=x^1=x^{1+0}=x^1x^0=x(x^0)$. If x is not 0, divide both end terms by x and get $1=x^0$. Note that $0^0$ is undefined.

Limit: as x->0 (for x>0), log(x) -> -infinity.

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mathematic

Right......0 raised to 0 is indeterminate.....my only other question was why this is, instead of undefined, why have two different terms?

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

That would undefined with a zero.

Imatfaal's explanation is way better.

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Why is it considered indeterminate as opposed to undefined? Why the difference?

It is indeterminate because if you look at x^y from the two perspectives, x going to 0 and y going 0, you sort of get two answers. As x goes to 0, x^y goes to 0, but as y goes to 0, x^y goes to 1. So, it is indeterminate because you can't determine whether 0 or 1 is the 'right' answer, but there are two choices.

This is as opposed to 0/0 because, as your other very long thread demonstrated many time, 0/0 can take any and all values and results in the breaking of many other basic rules of the normal mathematical operations. Hence 0/0 is undefined because it can't be defined.

In the end, it is kind of semantics because really the end result is the same. Neither 0^0 or 0/0 are equal to anything, just for slightly different reasons.

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Bignose

Thanks. So indeterminate because it can be done there are just multiple sums. Undefined because it can't be "done". mmmmh....I agree semantics. As opposed to scientific. Thanks again.

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• 1 month later...

This is interesting, I just happened to be reading about this the other day in one of my mathematical physics texts. The case of 0 ^ 0 stood out. As a mathematician, I want to say that it's indeterminant or undefined, yet as a physicist I want to say: nope it's equal to one. If we examine the function of x * ln(x), we see that as x approaches zero from the right, y approaches zero. Thus we can see that x * ln(x) is getting closer and closer to unity as x approaches zero. If we multiply unity by nothing then our value should remain unchanged and we should have one.

Mathematically, start with a ^ m = e ^ ( m * ln(a))

take a = m = x to give x ^ x = e ^ (x * ln(x)) and examine the behaviour of x * ln(x) approaches 0.

Taken directly from the book: "by comparing the representation of the ln(x) as the integral of t ^ -1 with the corresponding integral of t ^ ( -1 + b ) for any positive b, it can be shown that x * ln(x) tends to zero as x tends to zero and so x ^ x tends to zero in the same limit." - from Foundation Mathematics for the Physical Sciences K.F. Riley and M.P. Hobson

I have to admit though, I'm still curious what folks here think about this definition, and if we can really say that we have unity for 0 ^ 0.

Edited by Casey Wood
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I have to admit though, I'm still curious what folks here think about this definition, and if we can really say that we have unity for 0 ^ 0.

You can look at limits of other expressions that are equally as good here and get different answers. There is no canonical way to give a numerical value to 0^0. But sometimes people define it as 0 or 1, depending on what they are doing.

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depending on what they are doing.

It's hard to think of real life situations that demonstrate variable exponents. Whether it's area of a square, volume of a cube or Pythagorean theorem, the exponent is not variable.

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It's hard to think of real life situations that demonstrate variable exponents. Whether it's area of a square, volume of a cube or Pythagorean theorem, the exponent is not variable.

Nature is full of exponential functions - radioactive halflife, spread of disease, standard normal cumulative distribution function, lots of finance etc

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. mmmmh....I agree semantics. As opposed to scientific. Thanks again.

Careful definition of terms is semantics. It is also the underpinning of discussions in science, so your post makes no sense.

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Nature is full of exponential functions - radioactive halflife, spread of disease, standard normal cumulative distribution function, lots of finance etc

Interesting.

On each day, each infected infects x-1 more people, resulting in x infected for each infected the day prior. On day zero only 1 is infected. If x=1 the disease is not infectious. If x=0 the disease cures itself the next day, but it should still hold that there's 1 infected on day zero, thus 0^0=1.

Now x is the amount of radioisotope at start. The isotope remaining is x*(1/2^d) where d is the number of half lives passed. bad example.

Edited by MonDie
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Interesting.

On each day, each infected infects x-1 more people, resulting in x infected for each infected the day prior. On day zero only 1 is infected. If x=1 the disease is not infectious. If x=0 the disease cures itself the next day, but it should still hold that there's 1 infected on day zero, thus 0^0=1.

Now x is the amount of radioisotope at start. The isotope remaining is x*(1/2^d) where d is the number of half lives passed. bad example.

Are you trolling?

You give a counter example of an unworldy disease that doesn't spread cos no one is ill and which doesn't infect anyone - which is normally thought of as "not an infectious disease". Try a simple example - everyone infects two people per day

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Are you trolling?

I'm trying to extrapolate the vale of 0^0 from examples involving variable exponents. The number of days is the exponent.

I'm always trolling.

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