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The most common errors in undergraduate mathematics

math.vanderbilt.edu

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Re: The most common errors in undergraduate mathematics

#21

Post by lutusp »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

It's more of an open question than your reply would seem to indicate:

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

It's less controversial when the terms are integers, but even there an argument can be made for indeterminacy. As the linked article says, "There is no one definition that always works well for 0^0"

I can use 0^0 = 1 to prove that 1 = 2, but I'm sure you can anticipate the argument's form.

Re: The most common errors in undergraduate mathematics

#22

Post by leni536 »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

The main argument that it shouldn't be defined because there will be some situations where any definition will fall short.

Your definition only works for situations where you are dealing with integers.

Re: The most common errors in undergraduate mathematics

#23

Post by titter »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

The debate goes beyond that point, as illustrated by the lengthy Wikipedia section:

https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer...

From an undergraduate math point of view, it's wise to understand that 0^0 is not well defined.

Re: The most common errors in undergraduate mathematics

#24

Post by hansy »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

I posted an article about 0^0 to HN sometime back: http://www.askamathematician.com/2010/12/q-what-does-00-zero...

HN discussion: https://news.ycombinator.com/item?id=7519827

Re: The most common errors in undergraduate mathematics

#25

Post by RiderOfGiraffes »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

It goes deeper than that.

Many things in mathematics are defined in a manner that is consistent and convenient. Defining 0! to be 1 is a similar case. In doing so nothing goes wrong, and the binomial theorem becomes simple and convenient to state. Without defining 0! as 1, it's a dreadful mish-mash of special cases.

Similarly with 0^0. Considering x^y where x and y are complex numbers, there is no consistent single value as x and y each approach 0. So in the reals and the complex numbers we leave 0^0 as undefined. However, in the case of natural numbers there is a case for declaring 0^0 to be 1. That's to make it convenient to talk about the set A^B as the collection of functions from B to A. When we do that we get |A^B| = |A|^|B|.

There is more than expected in mathematics that's defined for convenience.

Re: The most common errors in undergraduate mathematics

#26

Post by [deleted] »

RiderOfGiraffes wrote: post #25
Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

It goes deeper than that. Many things in mathematics are defined in a manner that is consistent and convenient. Defining 0! to be 1 is a similar case. In doing so nothing goes wrong, and the binomial theorem becomes simple and convenient to state. Without defining 0! as 1, it's a dreadful mish-mash of special cases. Similarly with 0^0. Considering x^y where x and y are complex numbers, there is no consistent single v…

[deleted]

Re: The most common errors in undergraduate mathematics

#27

Post by [deleted] »

[deleted] wrote: post #26
RiderOfGiraffes wrote: post #25

Earlier quoted context omitted.

It goes deeper than that. Many things in mathematics are defined in a manner that is consistent and convenient. Defining 0! to be 1 is a similar case. In doing so nothing goes wrong, and the binomial theorem becomes simple and convenient to state. Without defining 0! as 1, it's a dreadful mish-mash of special cases. Similarly with 0^0. Considering x^y where x and y are complex numbers, there is no consistent single v…

[deleted]

[deleted]

Re: The most common errors in undergraduate mathematics

#28

Post by Grue3 »

titter wrote: post #23
Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

The debate goes beyond that point, as illustrated by the lengthy Wikipedia section: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... From an undergraduate math point of view, it's wise to understand that 0^0 is not well defined.

This argument assumes that the function f(x,y)=x^y is continuous near 0 and therefore lim (f(x)^g(x)) = (lim f(x))^(lim g(x)) at zero. But since it's not continous, that formula does not necessarily hold true.

Re: The most common errors in undergraduate mathematics

#29

Post by Xophmeister »

Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

The limit of x^x does equal one when approached from the right (i.e., when x is positive, reducing to zero). From the left, it also approaches one, but is only continuous over the complex numbers. Anyway, while the limit approaches one from both sides, the function cannot be evaluated at x=0 and so is undefined.

Re: The most common errors in undergraduate mathematics

#30

Post by philh »

Xophmeister wrote: post #29
Grue3 wrote: post #17

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

The limit of x^x does equal one when approached from the right (i.e., when x is positive, reducing to zero). From the left, it also approaches one, but is only continuous over the complex numbers. Anyway, while the limit approaches one from both sides, the function cannot be evaluated at x=0 and so is undefined.

Why are you considering 0^0 to be the limit of the function x^x, rather than (say) 0^x or (e^(-1/x))^x?
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