The Accidental Fraction Brainbuster



34 thoughts on “The Accidental Fraction Brainbuster

  1. Piquant and zesty are two words that appear all too rarely in relation to math problems in K-12, particularly computational ones. And these are indeed deserving, particularly when approached mentally. I used to give SAT students in the ’80s problems of the non-weapons-grade type because they showed up on real exams (and calculators were not allowed on SAT tests until 1995).

    That said, there does come a point where I’m not sure the mental arithmetic approach doesn’t take more time than other methods. In general, for positive rational numbers of the form a/b and c/d where a and b meet the usual criteria (all integers, b & d not 0), a relatively painless way to compare them is to find d*a and c*b. If the first quantity is larger, then a/b > c/d and vice versa. It’s instructive to ask students why this works and makes sense.

    On a timed test with no calculators, I’d be doing the latter rather than thinking it through mentally once the fractions became less friendly. Given no time pressure, however, the examples you give are excellent practice for good, logical, computational reasoning.

    If nothing else, most students will benefit from learning “piquant.”

    1. I would do exactly your ad and bc comparison, writing ad=(4000-3)(5000+1) and bc=(5000-4)(4000+1). Expanding these, we find we are essentially comparing -3 and -4!

  2. I looked at the progression:
    1000/1001, 1999/2001, 2998/3001, 3997/4001, 4996/5001. In each instance the top is increasing by 999 and the bottom by 1000.

    Since 1000/1001 > 999/1000, 1000/1001 > 1999/2001.

    Following this out logically, the sequence is decreasing and 3997/4001 > 4996/5001.

    I use this logic on a smaller scale often when comparing unit prices:

    700 grams for $5.69 or 900 grams for $7.69?

    The extra jar adds 200 grams for $2. This unit rate is larger than the small jar, so the small jar is the better deal.

  3. Another approach: we need to compare how much less both fractions are than 1. These differences are 4/4001 and 5/5001. If we make both denominators equal by multiplying numerator and denominator by 5001 in the first fraction and by 4001 in the second, we can compare the resulting numerators: 4×5001=20004 and 5×4001=20005. We don’t need to calculate the large number in the denominator. We already know they are the same. But let’s call it n. The result is clear: the fractions are 20004/n and 20005/n, so the second is larger. That means that 4996/5001 differs more from 1 than 3997/4001. Therefore, 3997/4001 > 4996/5001.

    1. Sander: Or we can do the following:

      Invert both the fractions; we get 4001/4 and 5001/5. The former equals 1000 + 1/4, the latter equals 1000 + 1/5; so the former is larger. Hence 4/4001 4996/5001.

      1. I wish I could like this like on Facebook, but I think that this comment will be enough. It’s fascinating how often inverting functions makes the comparison’s easier.

  4. Nice! My first reaction was the following:

    As you said, it boils down to deciding the larger of 4/4001 and 5/5001.

    So just flip the two fractions over. The former flipped is 4001/4 = 1000 + 1/4, and the latter flipped is 5001/5 = 1000 + 1/5.

    So 4001/4 is bigger, so 4/4001 is smaller, so 3997/4001 is bigger.

    1. This is how I solved the problem, too. It’s really cool to see the variety of approaches. As many of us know, math is not devoid of creativity. If only more students felt the same way!

  5. Lovely! After comparing easier smaller pieces or more parts fractions, I’m particularly fond of comparing fewer of larger pieces to more of smaller pieces. 5/12 ? 6/13; 213/450 ? 223/460 But this is another great category of which I’ll make use.

  6. *Pássaro. Poesia**. Por-everson campos&marilha.*


    *Pássaro, responda.*

    *Diga-me se quem muito migra fala todas as línguas*

    *Responda pássaro*

    *Que eu também procuro *

    *Ler os sinais no céu*

    *Arcaindo a tarde Iris *

    *Caminhos de uma plenitude simples*

    *Que só do alto se avista*

    *Peço para me guie*

    *Que eu também procuro pouso*

    *Em um chão sem linhas *

    *e uma longa pausa*

    *na melhor estação.*

    *Responda pássaro*

    *Se que muito muda*

    *Canta todas as canções*

    *Responda pássaro*

    *Como reconhecer a cor*

    *De vossa plumagem*

    *Imagem de um mesmo caminho*

    *Pois ainda procuro*

    *Alinhar-me aos meus*

    *Em um chão sem linhas*

    *Na melhor estação.*

    1. Yeah! I accidentally read the original problem that way.
      Because then when you cross-multiply, you get
      3997 x 5001 = 4997 x 4001.
      Then you use the cool fact that since both sides have terms that sum to the same thing (3997+5001 = 4997+4001 = 8998), the side with terms that are closer together gives a bigger product.
      I’m not sure exactly how to formalize that fact without calculus.

      Now that I think about it, nemoyatpeace’s ‘Marginal Rate’ method upthread works pretty smoothly for this one.
      4997/5001 is kind of like 3997/4001 + 1000/1000.
      1000/1000, or 1, is a better rate than the original, so 4997/5001 is bigger.

  7. Can’t you multiply the denominators and numerators by the opposite fraction’s denominator and figure it out that way?

    I just did it and it works. Isn’t your way overcomplicating things by quite a bit?

    1. That technique works (it always works; you are multiplying both numbers by the product of the denominators, keeping the relative size fixed, but turning both of them into integers which can be compared directly). However, it is much harder to do quickly in your head because keeping track of all the digits in the multiplication requires keeping track of many more digits than this method. If you actually follow the complicated path described above, you see that short-term memory is actually taxed a lot less.

      It *is* more complicated, but with practice, you can solve problems like this one faster than you could using a calculator.

        1. Yeah, me too. It has a lot going for it. You just have to keep straight in your head, on each step, how the comparison relates back to the original question.

  8. So, I said (5000+1)(4000-3) (4000+1)(5000-4)
    Then I looked at
    (4/4001) vs. (5/5001) and said (x/100x + 1)<1/100 for all x, but as x gets bigger it gets closer to 1/100.

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