Saturday, March 1, 2014

44: Can two different fractions be the same?

From Gabriel Rosenberg via email:

True or False?
Two-thirds and four-sixths are the same number.

To see why this is tagged Pre-Calculus and Complex Numbers, please see the comments.


  1. It must be false because I was told in school that you can take (-8) and raise it to the two-thirds power, but you can't take it and raise it to the four-sixths power.

  2. How do you figure that? (Sorry, can't tell if you're being sarcastic or if this is a real thing that you were told.)

  3. Can you raise a negative to the 1/6 power and then to the 4th? You can raise it to the 1/3 and then square it. Of course the order matters.

  4. Curmudgeon has the idea. I was told (-8) to the two-thirds is okay because you can raise it to the one-third and then square (now personally I don't think you should be able to raise -8 to the one-third, because I'd rather have two-thirds and four-sixths as the same number, but there it is.) (-8) to the four-sixths seems to require taking (-8) to the one-sixth which I was told is a no-no. Now one might argue do the fourth power first, but if that were okay then (-4) to the two-fourth could be done by squaring to get 16 and then one-fourth power to get 2. So again we have a dilemma. Either one-half and two-fourths are different numbers or else -4 to the one-half is 2. Again I'd personally rather we say fractional powers (and integer roots) are only valid when the base is positive.

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  6. Curmudgeon wrote: "Can you raise a negative to the 1/6 power and then to the 4th? You can raise it to the 1/3 and then square it. Of course the order matters."

    Using complex numbers, sure. -8 = 8cis(π), so you can use the corollary to DeMoivre's Theorem and determine the six sixth roots of 8cis(π). Two of those, when raised to the fourth power, will equal 4.

    Same works if you find the third roots of 8cis(π), one of those squared will equal 4. And in both scenarios, even with the complex roots raised to the appropriate power, the modulus of each result will be 4.

    * * *

    All that said, I figured that's what Galois was getting at, though I wanted to make sure. I don't like when my students say, "Mr./Mrs. So-and-So said you can't do this." Like rational functions, for example. My students come up to me and say, "You can't cross a horizontal asymptote." We cover rational functions in Algebra II and in Precalculus, so they see this topic twice, yet they seem shocked in my Calculus class when I show them that yes, you can cross a horizontal asymptote. Heck, I even show them an example using sine or cosine where you cross a HA an infinite number of times.

  7. As Jason explains, "it must be...because I was told in school" is very dangerous reasoning.

    It's understandable that some problems would creep in with rational roots of negatives though; sticking within the real numbers, failing to use the convention that a^(m/n)=a^(km/kn), or both, makes things ugly. Things are clearer if complex numbers are available and if we allow multivalued functions.

    To actually answer the question: yes, of course 2/3 and 4/6 are the same number. I suppose one could argue that 4 slices out of 6 is quite different from 2 out of 3 (4 being quite different from 2), but that would be confusing two (superficially) different representations of 2/3 with the number 2/3 itself.

  8. I agree the "I was told in school" is dangerous reasoning. For example, Some guy and Jason are now speaking of multivalued functions when "I was told" a function means for each element of the domain there exists a unique element of the range. The concept of multivalued functions opens up more arguments. Can I argue that four to the one-half plus four to the one-half is 0 because one of the second roots of 4cis(0) plus another second root of 4cis(0) is 0. I love complex numbers, but I think neither multivalued functions nor branch cuts necessarily make things "clearer".

    The point, for me, of an argument is not what answer is right, but rather the need for justification can force one to examine more closely one's understanding of a topic. Some guy says "of course" they're the same number, but I find "of course" to be just as dangerous reasoning as "because I was told" Justifying that the numbers are the same requires thinking a little deeper about what is a number and what is a fraction. And what that argument might look like depends on the where a student is in his or her mathematical development.

    Two-thirds mph might mean the average rate when I move 2 mi in 3 hrs just as four-sixths mph is the average rate when I move 4mi in 6 hrs. But in the former case if I then travel a further 1 mi in 1 hr I'll end up with an average rate of three-fourths mph for my journey, and in the latter I'll end up with an average rate of five-sevenths mph. Those rates seem different to me, so there's something interesting going on with rates which I also see when trying to average two rates.

  9. Good point, there is a lot hidden in my "of course". I know the audience here can follow me, and it's not like I'd explain it that way--if you can call it an explanation--to a 4th grader.

    A basketball team shooting 2 for 3 in free throws or three pointers isn't particularly noteworthy, but shooting 20 for 30 in free throws is terrible and shooting 20 for 30 in threes is outstanding. So 20/30 can seem different than 2/3. Being bitten by two mosquitos and two German Shepherds would no doubt seem very different too, so there's no need to even get as sophisticated as fractions; we could just as well ask if 2 is the same as 2. I'd still say "of course" it is, but I would also welcome some discussion. They're not necessarily mutually exclusive options.

  10. Math Arguments ... For The Win!

  11. Jason: You wrote "...yet they seem shocked in my Calculus class when I show them that yes, you can cross a horizontal asymptote." I no longer teach, but I taught precalculus for many years (both in high school and in college), and in the curve sketching part of precalculus one of the standard things we always did was to check to see if (and where) the graph crosses its horizontal asymptote (when a HA exists), right along with finding the x-intercepts, the y-intercept(s), etc. Same for crossing the slant asymptote, or even the curvilinear (polynomial) asymptote that every rational function has, when those were covered.

    1. Dave: We do the same thing regarding intercepts, but not many of our problems assigned in Alg II/Precalc deal with graphs that cross a HA/SA. I get a bit more creative in my Calc classes because they can use derivatives to help describe function behavior, and that's when they often see those for the first time.

      How do you go about having them find those points--make them set the rational function equal to the asymptote function and solve the corresponding equation? Seems like something we could add in, though honestly, our kids struggle enough with determining the intercepts and asymptotes, let alone add that extra wrinkle in. (We are a heterogenerous group. If I had an honors group, I'd definitely cover this. But that's an argument for another day.) :-/

    2. Jason: Set the rational function (stick to rational functions for this) equal to the horizontal asymptote constant. Thus, (6x^2 - 3x + 2)/(3x^2 + 5) has horizontal asymptote y = 6/3 = 2, so solve (6x^2 - 3x + 2)/(3x^2 + 5) = 2. Note that after clearing fractions, the square terms drop out, leaving you with a linear equation to solve. In the case of cubic/cubic, you'll get a quadratic equation, etc. This is easy (for teachers) to see in general, since in solving (ax^n + ...)/(bx^n + ...) = a/b you can see that the x^n terms will drop out. (Note that you have to have equal numerator and denominator degrees to begin with to get a horizontal asymptote.) You can conclude a couple of neat facts from this: (1) (linear)/(linear) never has a horizontal asymptote except in cases where you have (ax)/(bx) to begin with (line with a hole in it), a fact that you (teacher, not students) may already know since any (linear)/(linear) (that doesn't reduce to a constant) can be obtained from y = 1/x by various shifts and vertical stretches (1/x doesn't cross it's HA, and all these particular graph transformations preserve the property of not crossing a horziontal asymptote). (2) (quadratic)/(quadratic)'s usually cross their HA, but in rare cases they will not cross their HA (the "rare cases" are when not only the quadratic terms cancel when you clear fractions as discussed above, but it just so happens that the linear terms also cancel).