I would like to argue about (-8)^(1/3) = -2

[Stephen Tashi]

You can find many texts and web pages that teach elementary algebra and say they are only dealing with the real numbers. They state general laws of exponents such as:

$$(x^a)^b = x^{ab}$$ "whenever the exponential operations are defined".

In discussing square roots they are careful to point out that the square root of a negative number is not defined. However, most cannot resist claiming that the cube root of a negative number can be taken. Even "achievement test" material may expect students to compute $$\sqrt[3]{-8} = -2$$.

Using the above mentioned law of exponents, this will lead to disaster when $$X = -8, a = 2/3, b= 3/2$$.

I think most math students have the old fashioned view that mathematics is not required to be logically consistent. They think it refers to a Platonic reality rather than collections of assumptions and definitions. ( I gather than some of the participants in the .9999.. .thread are of this school.) So, perhaps logical contradictions don't upset the social order.

However, from the modern point of view, it would be best if the elementary textbooks could get their story straight on the case of $$\sqrt[3]{-8}$$ and similar arithmetic. Something has to give. Either you have to put some more complicated restrictions on when $$x^a x^b = x^{ab}$$ or you have to say that $$\sqrt[3]{-8}$$ is undefined in the algebra of the real numbers. When a student brings up the above contradiction, what is liable to happen is that the instructor will start getting into some digression about the complex numbers, numbers have 3 cube roots etc. Fine, but if you are going to teach the complex numbers then don't say you are teaching a course restricted to the real numbers.

 

[Hurkyl]

I haven't thought through the general case, but there is a problem with your example -- x^b isn't defined for any real exponential operation I'm aware of.

 

[Stephen Tashi]

I haven't thought through the general case, but there is a problem with your example -- x^b isn't defined for any real exponential operation I'm aware of.

I see your point! I mean to write the law $$(x^a)^b = x^{ab{$$ I edited it.

 

[Mark44]

There's a Wikipedia article -http://en.wikipedia.org/wiki/Exponent. See the section titled "Real powers of positive numbers".

However the identity
$$(a^r)^s = a^{r\cdot s}$$

cannot be extended consistently to where a is a negative real number

 

[Stephen Tashi]

There's a Wikipedia article -http://en.wikipedia.org/wiki/Exponent. See the section titled "Real powers of positive numbers".

I didn't find the article at the link, but I agree that the contradiction can be avoided by putting restrictions on the $$(x^r)^s = x^{rs}$$. My point is that if you look at materials for secondary education, many books and web tutorials don't state any restriction on that law except that the exponentiation operations are defined on both sides. Also, the question arises: would stating restrictions on that particular law fix everything? Can other contradictions arise that force other assumptions to be fixed?

 

[I like Serena]

I didn't find the article at the link, but I agree that the contradiction can be avoided by putting restrictions on the $$(x^r)^s = x^{rs}$$. My point is that if you look at materials for secondary education, many books and web tutorials don't state any restriction on that law except that the exponentiation operations are defined on both sides. Also, the question arises: would stating restrictions on that particular law fix everything? Can other contradictions arise that force other assumptions to be fixed?

Cool!
I wasn't aware of this "inconsistency" yet.

Good wiki article too.
And the explanation on the http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities" is a good one.

As I understand it, even though exponentiation of real numbers look and behave the same as exponentiation of complex numbers, there are a few pitfalls, where you can cross over from real numbers to complex numbers without realizing it.

So what you said: you either have to add counter intuitive restrictions on the exponentiation formula, or you have to restrict the domain of the exponentiation in the real number domain.

 

[Stephen Tashi]

(x^a)^b for rational exponents is perfectly consistent even for negative x

Perfectly consistent with what? Do you mean it can be well defined? I'd agree with that.

We simply leave expressions such as (-2)^(1/4) left undefined, but whenever a rule we have defined applies (as the one I made above) it can be applied consistently.

I assume you agree that the rule $$(x^a)^b = x^{ab}$$ can't be applied consistently when x is negative.

What you say doesn't disagree with my original post. I said that the algebra of the real numbers could be consistently taught by two different sets of assumptions. One way is to allow the odd roots of negative numbers to be uniquely defined and to restrict the laws of exponents - more than the typical secondary textbook bothers to do.

On the Ask Dr. Math site, a teacher pointed out there even a problem with that law of exponents in the case x =-3, a = 2, b = 1/2. Dr. Math favors restrictions:
http://mathforum.org/library/drmath/view/74401.html

 

[disregardthat]

Perfectly consistent with what? Do you mean it can be well defined? I'd agree with that.



I assume you agree that the rule $$(x^a)^b = x^{ab}$$ can't be applied consistently when x is negative.

What you say doesn't disagree with my original post. I said that the algebra of the real numbers could be consistently taught by two different sets of assumptions. One way is to allow the odd roots of negative numbers to be uniquely defined and to restrict the laws of exponents - more than the typical secondary textbook bothers to do.

On the Ask Dr. Math site, a teacher pointed out there even a problem with that law of exponents in the case x =-3, a = 2, b = 1/2. Dr. Math favors restrictions:
http://mathforum.org/library/drmath/view/74401.html

I'm sorry, I realized that after I had posted, so I deleted my post before you replied. Yes, the law of exponents cannot be consistently applied because of your example, even though we can define odd-numbered roots of negative numbers.

 

[coolul007]

The restrictions apply to fractional exponents. We define i, we cannot derive it's use through operations normally used with the integers. Furthermore, we also define the nth root of a number with restrictions as (x^2)^(1/2) = x, is restricted to non-negative values of x. This is where the ambiguities come from. It is really better to define these numbers as solutions to equations, I.E. x^3 + 8 = 0 has 3 solutions, x = -2, 1 + ((3^(1/2))i, 1 - ((3^(1/2))i.

 

[Stephen Tashi]

The restrictions apply to fractional exponents.

I wonder if what should be done with irrational exponents in elementary algebra. For example, If you define them then some student is going to ask whether $$(-1)^\pi$$ is positive or negative.

 

[coolul007]

I wonder if what should be done with irrational exponents in elementary algebra. For example, If you define them then some student is going to ask whether $$(-1)^\pi$$ is positive or negative.

Now we have entered the land of Euler, where there are a different set of rules, those of sums of series.