A negative number to a fractional power -- Is it legit in High school?

[MichPod]

My son (11th grade, Canada school) brought an equation $x^\frac{3}{5}=\frac{x}{4}$ from his class on which the teacher says it has a $x=-32$ root, in addition to $x=0$ and $x=32$, of course.

That was a bit a surprise for me as I was taught in my school time that only a non-negative number may be taken into a fractional power. Even if I accept taking a negative number to a fractional power is ok, I still have a headache trying to understand which equation transformation may be equivalent (not bringing new roots and not loosing existing ones) when operating with such powers.

May I get here a good strict enough definition on a high school level of what $a^\frac{m}{n}$ is when $a$ is negative and maybe some references so that I could see how to solve these sort of equations, which transformations of such equations or expressions are legit etc.

Example: I have no idea whether the following is correct when $a,b\in R$ and $x,y\in Z$

$a^x\cdot b^x = (a\cdot b)^x$
or
$a^x\cdot a^y = a^{x+y}$

Also, is it a normal practice or maybe at least one of the normal options to consider this sort of material in the high school?

Thank you in advance.

 

[andrewkirk]

May I get here a good strict enough definition on a high school level of what $a^\frac{m}{n}$ is when $a$ is negative and maybe some references so that I could see how to solve these sort of equations, which transformations of such equations or expressions are legit etc.

It depends on whether your son's course includes Complex Numbers. In Australia, the highest level maths courses in years 11-12 cover those, but not the other levels. Canada may be similar. If they're covering complex numbers, then you can give a meaning to that expression for any real (or complex) number $a$, be it positive, negative or something else. You need to use the techniques of complex numbers to assign that meaning though.
If we exclude complex numbers then, assuming $m$ and $n$ have no common factors, $a^\frac{m}{n}$ only has a meaning if there exists some real number $b$ such that $b^n=a$. If so, then the answer is $b^m$. Otherwise the expression is meaningless.
In the case here we have $m=3,n=5,a=-32$. Is there a $b$ such that $b^5=-32$? Yes: $b=-2$. So if we set $x=-32$ then the expression $x^{3/5}$ means something.
You might justifiably complain that that's starting with the answer. How does one go about solving the problem?
Well, first note that $x=0$ solves it, so write that down, then assume $x\neq 0$, which allows us to take the next step:
multiply both sides by $\frac 4{x^{3/5}}$ to get equation:
$$4 = x^{2/5}$$
Then raise each side to the power 5 to get:
$$4^5 = x^2$$
The solution of this is either or both of $x=\pm (4^5)^\frac12$, which is $x=\pm 32$.
We try both solutions in the original equation and find they both work.
So we see that the solutions are -32, 0 and 32, and we didn't need to use Complex Number techniques to find them.

 

[hutchphd]

A solution to an equation is valid so long as it can be shown to be a solution to the equation. Is it the contention of the OP that the equation $$x^3 =-27$$ has no "valid" solution? I am somewhat confused by the question.

 

[PeroK]

A fractional power of a negative number is acceptable as long as the denominator is odd. In particular, the following functions are well-defined for all $x \in \mathbb R$: $f(x) = \sqrt[3]{x}$ and $f(x) = \sqrt[5]{x}$.

 

[MichPod]

A solution to an equation is valid so long as it can be shown to be a solution to the equation. Is it the contention of the OP that the equation $$x^3 =-27$$ has no "valid" solution? I am somewhat confused by the question.

Nope, I did not mean that. Taking to an integer power is well defined for the negative numbers.

My question was very specifically about what it is meant to take a negative number to a fractional power. Turns out, I never studied that in my school. ))) Of course, if my memory serves me well.

 

[MichPod]

If we exclude complex numbers then, assuming m and n have no common factors, $a^\frac{m}{n}$ only has a meaning if there exists some real number b such that $b^n=a$. If so, then the answer is $b^m$. Otherwise the expression is meaningless.

Thank you for your detailed answer. My son started studying complex numbers, but in some rudimentary way still. He probably does not know complex logarithms.

Could you clarify your definition above? I understand what it means, but I don't quite see why we cannot define the answer 'x' simply as satisfying $x^n=a^m$ , providing m and n have no common factors.

 

[jack action]

Fascinating.

So I tried this in Wolfram and I have a question about how it handles this equation.

For the original equation, we get https://www.wolframalpha.com/input?i=x^(3/5)=x/4. This gives the solution {0,32}. But it also offers "Assuming the principal root | Use the real‐valued root instead" which gives us the solution presented here, i.e. {-32,0,32}.

My question arises when I instead ask Wolfram about https://www.wolframalpha.com/input?i=x^0.6=x/4. It is still giving the {0,32} solution, but no longer offers the principal root assumption. Even when "forcing it" by asking https://www.wolframalpha.com/input?i=x^0.6=x/4&assumption="^"+->+"Real", it gives the same {0,32} solution.

How can we justify treating 3/5 differently than 0.6?

 

[MichPod]

Fascinating.

So I tried this in Wolfram and I have a question about how it handles this equation.

For the original equation, we get https://www.wolframalpha.com/input?i=x^(3/5)=x/4. This gives the solution {0,32}. But it also offers "Assuming the principal root | Use the real‐valued root instead" which gives us the solution presented here, i.e. {-32,0,32}.

For what I can see, per their interface, the {0,32} is the answer for the "Assuming the principal root" option, while {-32.0,32} was calculated for the "Assuming the real‐valued root" option.

How can we justify treating 3/5 differently than 0.6?

Well, interesting. How about testing this:
https://www.wolframalpha.com/input?i=x^(6/10)=x/4

They maybe could consider 0.6 as some approximation, but for 6/10 it should be considered as a perfectly legal equivalent of 3/5, and yet they provide the answer {0, 32} and do not offer any other option.

 

[pasmith]

Fascinating.

So I tried this in Wolfram and I have a question about how it handles this equation.

For the original equation, we get https://www.wolframalpha.com/input?i=x^(3/5)=x/4. This gives the solution {0,32}. But it also offers "Assuming the principal root | Use the real‐valued root instead" which gives us the solution presented here, i.e. {-32,0,32}.

My question arises when I instead ask Wolfram about https://www.wolframalpha.com/input?i=x^0.6=x/4. It is still giving the {0,32} solution, but no longer offers the principal root assumption. Even when "forcing it" by asking https://www.wolframalpha.com/input?i=x^0.6=x/4&assumption="^"+->+"Real", it gives the same {0,32} solution.

How can we justify treating 3/5 differently than 0.6?

Suppose you asked for $(-32)^{6/10} = (\sqrt[10]{-32})^6$. $-32$ has no real 10th root, so WA can't offer that option. And since $6/10 = 0.6$, how can we justify treating 6/10 differently than 0.6?

EIDT: I say this, even though WA itself simplifies $(-32)^{6/10}$ to $(-32)^{3/5}$ before giving the principal root; there is no option to use any other branch, although further down the page it does correctly list -8 as one of the five possible values of $(-32)^{3/5}$. However if you ask for $(-32)^{3/5}$ directly it does offer that option.

 

[MichPod]

Suppose you asked for $(-32)^{6/10} = (\sqrt[10]{-32})^6$. $-32$ has no real 10th root, so WA can't offer that option. And since $6/10 = 0.6$, how can we justify treating 6/10 differently than 0.6?

So are we to say that $(-32)^\frac{6}{10}\neq (-32)^\frac{3}{5}$ ???
I think that goes too far. $6/10$ and $3/5$ are the same rational numbers, after all.

 

[pasmith]

So are we to say that $(-32)^\frac{6}{10}\neq (-32)^\frac{3}{5}$ ???
I think that goes too far. $6/10$ and $3/5$ are the same rational numbers, after all.

The correct conclusion is that a symbolic calculation engine is not a mathematician, and actual human intelligence needs to be applied in formulating inputs and interpreting results.

 

[PeroK]

So are we to say that $(-32)^\frac{6}{10}\neq (-32)^\frac{3}{5}$ ???
I think that goes too far. $6/10$ and $3/5$ are the same rational numbers, after all.

Fractional powers really want to be multi-valued functions. We can only make them proper, single-valued functions by specifying a precise rule for deciding between candidates. That rule may include reducing a fraction to its lowest form. There is an inherent ambiguity in $\sqrt 1$ that is resolved by making this non-negative, by definition. If we want to rewrite this is $\sqrt 1 = 1^{\frac 1 2} = 1^{\frac 2 4}$, then we need to define things so that these are all equal.

It gets more complicated with negative numbers, as $(-1)^{\frac 1 3} = -1$. And if we want $(-1)^{\frac 2 6} = -1$, then we need to be careful how we interpret this. If we do the square first, then we get $(-1)^{\frac 2 6} = 1^{\frac 1 6} = 1$. And taking the 6th root first means we must invoke complex numbers.

The only elementary way out of this dilemma is to keep fractional powers (PS for negative numbers, of course) in the lowest quotient form and define them to be only valid where the denominator is odd.

I must admit I don't know what mathematics teaching theory officially has to say about this.

 

[Bystander]

The only elementary way out of this dilemma is to keep fractional powers in the lowest quotient form and define them to be only valid where the denominator is odd.

?

 

[PeroK]

?

For negative numbers, of course.

 

[Bystander]

For negative numbers, of course.

Forgot the OP.

 

[MichPod]

And, interestingly,

$x^\frac{m}{n} \neq (x^m)^\frac{1}{n}$

Example:

$(-1)^\frac{2}{6} \neq ((-1)^2)^\frac{1}{6}$

$-1\neq 1$

The inequality holds also if we consider all the complex values of the fractional power.
$(-1)^\frac{2}{6}$ and $((-1)^2)^\frac{1}{6}$ generate different set of values.

 

[MichPod]

The correct conclusion is that a symbolic calculation engine is not a mathematician, and actual human intelligence needs to be applied in formulating inputs and interpreting results.

I understand that the Wolfram math is the state of the art, top level product in its field, so I am much puzzled seeing this sort of inconsistency.

 

[gleem]

Some light might be shed on the fractional powers of a negative number by expressing the negative number using the Euler formula. For example, the cube root of -1 is not obtainable with decimal notation . Using Eulers equation
$$(-1)^{1/3} = e^{i\pi /3}$$ $$(-1)^{1/3} = cos(60 ^{\circ } ) +isin(60^{\circ })$$

 

[PeroK]

Some light might be shed on the fractional powers of a negative number by expressing the negative number using the Euler formula. For example, the cube root of -1 is not obtainable with decimal notation . Using Eulers equation
$$(-1)^{1/3} = e^{i\pi /3}$$ $$(-1)^{1/3} = cos(60 ^{\circ } ) +isin(60^{\circ })$$

When we invoke complex numbers then generally roots and rational powers are treated as multi-valued functions. In the real number system, $(-1)^{\frac 1 3} = -1$. And, in fact, $x^{1/3}$ is a well-defined single valued function, which is the inverse of the function $x^3$.

In the complex number system, each non-zero complex number has three cube roots. And, although there is the concept of a principal root, this is not necessarily the real root, where one exists.

This is one example where it's important not to invoke complex numbers. You have to respect real analysis and not assume that complex numbers are always there whenever you need them. In this case, you lose the connection between $x^3$ and $x^{1/3}$ as inverse functions defined on $\mathbb R$.

 

[gleem]

I see your point. One of the reasons in this case to invoke complex numbers is the ambiguity of the order of raising the power and taking the root. for example $(-1)^{2/3} = e^{i\pi 2/3}$ which is clearly a complex number but $[ (-1)^{2 }]^{1/3} = 1$ is real. So how would you determine a rule for what order should be taken regarding powers and roots?

disclaimer: I am,or was an experimental physicist.

 

[Frabjous]

I believe that x^a/b is uniquely defined for the reals when a,b are integers and b is odd. Order of operation will not matter for this case.

 

[gleem]

I believe that xa/b is uniquely defined for the reals when a,b are integers and b is odd. Order of operation will not matter for this case.

See post 20. What am I missing?

 

[Frabjous]

See post 20. What am I missing?

For complex numbers, there are 3 cube roots of 1 (2 complex and 1 real).
1, e^iπ2/3,e^-iπ2/3

 

[PeroK]

I see your point. One of the reasons in this case to invoke complex numbers is the ambiguity of the order of raising the power and taking the root. for example $(-1)^{2/3} = e^{i\pi 2/3}$ which is clearly a complex number but $[ (-1)^{2 }]^{1/3} = 1$ is real. So how would you determine a rule for what order should be taken regarding powers and roots?

disclaimer: I am,or was an experimental physicist.

The convention to reduce to the lowest form eliminates the ambiguity. The only case we need consider is $(-1)^{m/n}$, where $n$ is odd. You can check that, if $m$ is even:
$$((-1)^{m})^{1/n} = ((-1)^{1/n})^m = 1$$And, if $m$ is odd:
$$((-1)^{m})^{1/n} = ((-1)^{1/n})^m = -1$$

 

[PeroK]

See post 20. What am I missing?

I see your point. One of the reasons in this case to invoke complex numbers is the ambiguity of the order of raising the power and taking the root. for example $(-1)^{2/3} = e^{i\pi 2/3}$ which is clearly a complex number but $[ (-1)^{2 }]^{1/3} = 1$ is real. So how would you determine a rule for what order should be taken regarding powers and roots?

If we are dealing with real numbers, then $(-1)^{1/3} = -1$. This is uniquely defined.

If we are dealing with complex numbers, then $(-1)^{1/3} = \frac 1 2 + \frac{\sqrt 3}{2}i$ (this is the principal cube root).

The ambiguity is whether we are working with real numbers of complex numbers. This is why, in mathematics, we need to specify the number system we are working with.